{"id":543259,"date":"2021-09-29T11:13:54","date_gmt":"2021-09-29T11:13:54","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543259"},"modified":"2022-12-26T10:15:36","modified_gmt":"2022-12-26T10:15:36","slug":"rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae"},"content":{"rendered":"\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 11: Maths Chapter 8 solutions. Complete Class 11 Maths Chapter 8 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\">RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae<\/h2>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">RD Sharma 11th Maths Chapter 8, Class 11 Maths Chapter 8 solutions<\/p>\n\n\n\n<p>EXERCISE 8.1 PAGE NO: 8.6<\/p>\n\n\n\n<p><strong>1. Express each of the following as the sum or difference of sines and cosines:<br>(i) 2 sin 3x cos x<br>(ii) 2 cos 3x sin 2x<br>(iii) 2 sin 4x sin 3x<br>(iv) 2 cos 7x cos 3x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)<\/strong>&nbsp;2 sin 3x cos x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 sin A cos B = sin (A + B) + sin (A \u2013 B)<\/p>\n\n\n\n<p>2 sin 3x cos x = sin (3x + x) + sin (3x \u2013 x)<\/p>\n\n\n\n<p>= sin (4x) + sin (2x)<\/p>\n\n\n\n<p>= sin 4x + sin 2x<\/p>\n\n\n\n<p><strong>(ii)<\/strong>&nbsp;2 cos 3x sin 2x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 cos A sin B = sin (A + B) \u2013 sin (A \u2013 B)<\/p>\n\n\n\n<p>2 cos 3x sin 2x = sin (3x + 2x) \u2013 sin (3x \u2013 2x)<\/p>\n\n\n\n<p>= sin (5x) \u2013 sin (x)<\/p>\n\n\n\n<p>= sin 5x \u2013 sin x<\/p>\n\n\n\n<p><strong>(iii)<\/strong>&nbsp;2 sin 4x sin 3x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 sin A sin B = cos (A \u2013 B) \u2013 cos (A + B)<\/p>\n\n\n\n<p>2 sin 4x sin 3x = cos (4x \u2013 3x) \u2013 cos (4x + 3x)<\/p>\n\n\n\n<p>= cos (x) \u2013 cos (7x)<\/p>\n\n\n\n<p>= cos x \u2013 cos 7x<\/p>\n\n\n\n<p><strong>(iv)<\/strong>&nbsp;2 cos 7x cos 3x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 cos A cos B = cos (A + B) + cos (A \u2013 B)<\/p>\n\n\n\n<p>2 sin 3x cos x = cos (7x + 3x) + cos (7x \u2013 3x)<\/p>\n\n\n\n<p>= cos (10x) + cos (4x)<\/p>\n\n\n\n<p>= cos 10x + cos 4x<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;<strong>Prove that:<br>(i) 2 sin 5\u03c0\/12 sin \u03c0\/12 = 1\/2<\/strong><\/p>\n\n\n\n<p><strong>(ii) 2 cos 5\u03c0\/12 cos \u03c0\/12 = 1\/2<\/strong><\/p>\n\n\n\n<p><strong>(iii) 2 sin 5\u03c0\/12 cos \u03c0\/12 = (\u221a3 + 2)\/2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>2 sin 5\u03c0\/12 sin \u03c0\/12 = 1\/2<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 sin A sin B = cos (A \u2013 B) \u2013 cos (A + B)<\/p>\n\n\n\n<p>2 sin 5\u03c0\/12 sin \u03c0\/12 = cos (5\u03c0\/12 \u2013 \u03c0\/12) \u2013 cos (5\u03c0\/12 + \u03c0\/12)<\/p>\n\n\n\n<p>= cos (4\u03c0\/12) \u2013 cos (6\u03c0\/12)<\/p>\n\n\n\n<p>= cos (\u03c0\/3) \u2013 cos (\u03c0\/2)<\/p>\n\n\n\n<p>= cos (180<sup>o<\/sup>\/3) \u2013 cos (180<sup>o<\/sup>\/2)<\/p>\n\n\n\n<p>= cos 60\u00b0 \u2013 cos 90\u00b0<\/p>\n\n\n\n<p>= 1\/2 \u2013 0<\/p>\n\n\n\n<p>= 1\/2<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>2 cos 5\u03c0\/12 cos \u03c0\/12 = 1\/2<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 cos A cos B = cos (A + B) + cos (A \u2013 B)<\/p>\n\n\n\n<p>2 cos 5\u03c0\/12 cos \u03c0\/12 = cos (5\u03c0\/12 + \u03c0\/12) + cos (5\u03c0\/12 \u2013 \u03c0\/12)<\/p>\n\n\n\n<p>= cos (6\u03c0\/12) + cos (4\u03c0\/12)<\/p>\n\n\n\n<p>= cos (\u03c0\/2) + cos (\u03c0\/3)<\/p>\n\n\n\n<p>= cos (180<sup>o<\/sup>\/2) + cos (180<sup>o<\/sup>\/3)<\/p>\n\n\n\n<p>= cos 90\u00b0 + cos 60\u00b0<\/p>\n\n\n\n<p>= 0 + 1\/2<\/p>\n\n\n\n<p>= 1\/2<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>2 sin 5\u03c0\/12 cos \u03c0\/12 = (\u221a3 + 2)\/2<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 sin A cos B = sin (A + B) + sin (A \u2013 B)<\/p>\n\n\n\n<p>2 sin 5\u03c0\/12 cos \u03c0\/12 = sin (5\u03c0\/12 + \u03c0\/12) + sin (5\u03c0\/12 \u2013 \u03c0\/12)<\/p>\n\n\n\n<p>= sin (6\u03c0\/12) + sin (4\u03c0\/12)<\/p>\n\n\n\n<p>= sin (\u03c0\/2) + sin (\u03c0\/3)<\/p>\n\n\n\n<p>= sin (180<sup>o<\/sup>\/2) + sin (180<sup>o<\/sup>\/3)<\/p>\n\n\n\n<p>= sin 90\u00b0 + sin 60\u00b0<\/p>\n\n\n\n<p>= 1 + \u221a3<\/p>\n\n\n\n<p>= (2 + \u221a3)\/2<\/p>\n\n\n\n<p>= (\u221a3 + 2)\/2<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>3. show that:<\/strong><\/p>\n\n\n\n<p><strong>(i) sin 50<sup>o<\/sup>&nbsp;cos 85<sup>o<\/sup>&nbsp;= (1 \u2013 \u221a2sin 35<sup>o<\/sup>)\/2\u221a2<\/strong><\/p>\n\n\n\n<p><strong>(ii) sin 25<sup>o<\/sup>&nbsp;cos 115<sup>o<\/sup>&nbsp;= 1\/2 {sin 140<sup>o<\/sup>&nbsp;\u2013 1}<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin 50<sup>o<\/sup>&nbsp;cos 85<sup>o<\/sup>&nbsp;= (1 \u2013 \u221a2sin 35<sup>o<\/sup>)\/2\u221a2<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 sin A cos B = sin (A + B) + sin (A \u2013 B)<\/p>\n\n\n\n<p>sin A cos B = [sin (A + B) + sin (A \u2013 B)] \/ 2<\/p>\n\n\n\n<p>sin 50<sup>o<\/sup>&nbsp;cos 85<sup>o<\/sup>&nbsp;= [sin(50<sup>o<\/sup>&nbsp;+ 85<sup>o<\/sup>) + sin (50<sup>o<\/sup>&nbsp;\u2013 85<sup>o<\/sup>)] \/ 2<\/p>\n\n\n\n<p>= [sin (135<sup>o<\/sup>) + sin (-35<sup>o<\/sup>)] \/ 2<\/p>\n\n\n\n<p>= [sin (135<sup>o<\/sup>) \u2013 sin (35<sup>o<\/sup>)] \/ 2 (since, sin (-x) = -sin x)<\/p>\n\n\n\n<p>= [sin (180<sup>o<\/sup>&nbsp;\u2013 45<sup>o<\/sup>) \u2013 sin 35<sup>o<\/sup>] \/ 2<\/p>\n\n\n\n<p>= [sin 45<sup>o<\/sup>&nbsp;\u2013 sin 35<sup>o<\/sup>] \/ 2<\/p>\n\n\n\n<p>= [(1\/\u221a2) \u2013 sin 35<sup>o<\/sup>] \/ 2<\/p>\n\n\n\n<p>= [(1 \u2013 sin 35<sup>o<\/sup>)\/<strong>\u221a<\/strong>2] \/ 2<\/p>\n\n\n\n<p>= (1 \u2013 sin 35<sup>o<\/sup>) \/ 2<strong>\u221a<\/strong>2<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>sin 25<sup>o<\/sup>&nbsp;cos 115<sup>o<\/sup>&nbsp;= 1\/2 {sin 140<sup>o<\/sup>&nbsp;\u2013 1}<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 sin A cos B = sin (A + B) + sin (A \u2013 B)<\/p>\n\n\n\n<p>sin A cos B = [sin (A + B) + sin (A \u2013 B)] \/ 2<\/p>\n\n\n\n<p>sin 20<sup>o<\/sup>&nbsp;cos 115<sup>o<\/sup>&nbsp;= [sin(25<sup>o<\/sup>&nbsp;+ 115<sup>o<\/sup>) + sin (25<sup>o<\/sup>&nbsp;\u2013 115<sup>o<\/sup>)] \/ 2<\/p>\n\n\n\n<p>= [sin (140<sup>o<\/sup>) + sin (-90<sup>o<\/sup>)] \/ 2<\/p>\n\n\n\n<p>= [sin (140<sup>o<\/sup>) \u2013 sin (90<sup>o<\/sup>)] \/ 2 (since, sin (-x) = -sin x)<\/p>\n\n\n\n<p>= 1\/2 {sin 140<sup>o<\/sup>&nbsp;\u2013 1}<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>4. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>4 cos x cos (\u03c0\/3 + x) cos (\u03c0\/3 \u2013 x) = cos 3x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>4 cos x cos (\u03c0\/3 + x) cos (\u03c0\/3 \u2013 x) = 2 cos x (2 cos (\u03c0\/3 + x) cos (\u03c0\/3 \u2013 x))<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 cos A cos B = cos (A + B) + cos (A \u2013 B)<\/p>\n\n\n\n<p>2 cos x (2 cos (\u03c0\/3+x) cos (\u03c0\/3 \u2013 x)) = 2 cos x (cos (\u03c0\/3+x + \u03c0\/3-x) + cos (\u03c0\/3+x \u2013 \u03c0\/3+ x))<\/p>\n\n\n\n<p>= 2 cos x (cos (2\u03c0\/3) + cos (2x))<\/p>\n\n\n\n<p>= 2 cos x {cos 120\u00b0 + cos 2x}<\/p>\n\n\n\n<p>= 2 cos x {cos (180\u00b0 \u2013 60\u00b0) + cos 2x}<\/p>\n\n\n\n<p>= 2 cos x (cos 2x \u2013 cos 60\u00b0) (since, {cos (180\u00b0 \u2013 A) = \u2013 cos A})<\/p>\n\n\n\n<p>= 2 cos 2x cos x \u2013 2 cos x cos 60\u00b0<\/p>\n\n\n\n<p>= (cos (x + 2x) + cos (2x \u2013 x)) \u2013 (2cos x)\/2<\/p>\n\n\n\n<p>= cos 3x + cos x \u2013 cos x<\/p>\n\n\n\n<p>= cos 3x<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p>EXERCISE 8.2 PAGE NO: 8.17<\/p>\n\n\n\n<p><strong>1. Express each of the following as the product of sines and cosines:<br>(i) sin 12x + sin 4x<br>(ii) sin 5x \u2013 sin x<br>(iii) cos 12x + cos 8x<br>(iv) cos 12x \u2013 cos 4x<\/strong><\/p>\n\n\n\n<p><strong>(v) sin 2x + cos 4x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin 12x + sin 4x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 12x + sin 4x = 2 sin (12x + 4x)\/2 cos (12x \u2013 4x)\/2<\/p>\n\n\n\n<p>= 2 sin 16x\/2 cos 8x\/2<\/p>\n\n\n\n<p>= 2 sin 8x cos 4x<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>sin 5x \u2013 sin x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>sin 5x \u2013 sin x = 2 cos (5x + x)\/2 sin (5x \u2013 x)\/2<\/p>\n\n\n\n<p>= 2 cos 6x\/2 sin 4x\/2<\/p>\n\n\n\n<p>= 2 cos 3x sin 2x<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>cos 12x + cos 8x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 12x + cos 8x = 2 cos (12x + 8x)\/2 cos (12x \u2013 8x)\/2<\/p>\n\n\n\n<p>= 2 cos 20x\/2 cos 4x\/2<\/p>\n\n\n\n<p>= 2 cos 10x cos 2x<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>cos 12x \u2013 cos 4x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A \u2013 cos B = -2 sin (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>cos 12x \u2013 cos 4x = -2 sin (12x + 4x)\/2 sin (12x \u2013 4x)\/2<\/p>\n\n\n\n<p>= -2 sin 16x\/2 sin 8x\/2<\/p>\n\n\n\n<p>= -2 sin 8x sin 4x<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>sin 2x + cos 4x<\/p>\n\n\n\n<p>sin 2x + cos 4x = sin 2x + sin (90<sup>o<\/sup>&nbsp;\u2013 4x)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 2x + sin (90<sup>o<\/sup>&nbsp;\u2013 4x) = 2 sin (2x + 90<sup>o<\/sup>&nbsp;\u2013 4x)\/2 cos (2x \u2013 90<sup>o<\/sup>&nbsp;+ 4x)\/2<\/p>\n\n\n\n<p>= 2 sin (90<sup>o<\/sup>&nbsp;\u2013 2x)\/2 cos (6x \u2013 90<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin (45\u00b0 \u2013 x) cos (3x \u2013 45\u00b0)<\/p>\n\n\n\n<p>= 2 sin (45\u00b0 \u2013 x) cos {-(45\u00b0 \u2013 3x)} (since, {cos (-x) = cos x})<\/p>\n\n\n\n<p>= 2 sin (45\u00b0 \u2013 x) cos (45\u00b0 \u2013 3x)<\/p>\n\n\n\n<p>= 2 sin (\u03c0\/4 \u2013 x) cos (\u03c0\/4 \u2013 3x)<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;<strong>Prove that :<br>(i) sin 38\u00b0 + sin 22\u00b0 = sin 82\u00b0<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos 100\u00b0 + cos 20\u00b0 = cos 40\u00b0<\/strong><\/p>\n\n\n\n<p><strong>(iii) sin 50\u00b0 + sin 10\u00b0 = cos 20\u00b0<\/strong><\/p>\n\n\n\n<p><strong>(iv) sin 23\u00b0 + sin 37\u00b0 = cos 7\u00b0<\/strong><\/p>\n\n\n\n<p><strong>(v) sin 105\u00b0 + cos 105\u00b0 = cos 45\u00b0<\/strong><\/p>\n\n\n\n<p><strong>(vi) sin 40\u00b0 + sin 20\u00b0 = cos 10\u00b0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin 38\u00b0 + sin 22\u00b0 = sin 82\u00b0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 38\u00b0 + sin 22\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 38\u00b0 + sin 22\u00b0 = 2 sin (38<sup>o<\/sup>&nbsp;+ 22<sup>o<\/sup>)\/2 cos (38<sup>o<\/sup>&nbsp;\u2013 22<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 60<sup>o<\/sup>\/2 cos 16<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 30<sup>o<\/sup>&nbsp;cos 8<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 8<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 8<sup>o<\/sup><\/p>\n\n\n\n<p>= cos (90\u00b0 \u2013 82\u00b0)<\/p>\n\n\n\n<p>= sin 82\u00b0 (since, {cos (90\u00b0 \u2013 A) = sin A})<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>cos 100\u00b0 + cos 20\u00b0 = cos 40\u00b0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos 100\u00b0 + cos 20\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 100\u00b0 + cos 20\u00b0 = 2 cos (100<sup>o<\/sup>&nbsp;+ 20<sup>o<\/sup>)\/2 cos (100<sup>o<\/sup>&nbsp;\u2013 20<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 cos 80<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 cos 60<sup>o<\/sup>&nbsp;cos 4<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 40<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 40<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>sin 50\u00b0 + sin 10\u00b0 = cos 20\u00b0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 50\u00b0 + sin 10\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 50\u00b0 + sin 10\u00b0 = 2 sin (50<sup>o<\/sup>&nbsp;+ 10<sup>o<\/sup>)\/2 cos (50<sup>o<\/sup>&nbsp;\u2013 10<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 60<sup>o<\/sup>\/2 cos 40<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 30<sup>o<\/sup>&nbsp;cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>sin 23\u00b0 + sin 37\u00b0 = cos 7\u00b0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 23\u00b0 + sin 37\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 23\u00b0 + sin 37\u00b0 = 2 sin (23<sup>o<\/sup>&nbsp;+ 37<sup>o<\/sup>)\/2 cos (23<sup>o<\/sup>&nbsp;\u2013 37<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 60<sup>o<\/sup>\/2 cos -14<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 30<sup>o<\/sup>&nbsp;cos -7<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos -7<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 7<sup>o<\/sup>&nbsp;(since, {cos (-A) = cos A})<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>sin 105\u00b0 + cos 105\u00b0 = cos 45\u00b0<\/p>\n\n\n\n<p>Let us consider LHS: sin 105\u00b0 + cos 105\u00b0<\/p>\n\n\n\n<p>sin 105\u00b0 + cos 105\u00b0 = sin 105<sup>o<\/sup>&nbsp;+ sin (90<sup>o<\/sup>&nbsp;\u2013 105<sup>o<\/sup>) [since, {sin (90\u00b0 \u2013 A) = cos A}]<\/p>\n\n\n\n<p>= sin 105<sup>o<\/sup>&nbsp;+ sin (-15<sup>o<\/sup>)<\/p>\n\n\n\n<p>= sin 105<sup>o<\/sup>&nbsp;\u2013 sin 15<sup>o<\/sup>&nbsp;[{sin(-A) = \u2013 sin A}]<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>Sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>sin 105<sup>o<\/sup>&nbsp;\u2013 sin 15<sup>o<\/sup>&nbsp;= 2 cos (105<sup>o<\/sup>&nbsp;+ 15<sup>o<\/sup>)\/2 sin (105<sup>o<\/sup>&nbsp;\u2013 15<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 sin 90<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 cos 60<sup>o<\/sup>&nbsp;sin 45<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 1\/<strong>\u221a<\/strong>2<\/p>\n\n\n\n<p>= 1\/<strong>\u221a<\/strong>2<\/p>\n\n\n\n<p>= cos 45<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(vi)&nbsp;<\/strong>sin 40\u00b0 + sin 20\u00b0 = cos 10\u00b0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 40\u00b0 + sin 20\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 40\u00b0 + sin 20\u00b0 = 2 sin (40<sup>o<\/sup>&nbsp;+ 20<sup>o<\/sup>)\/2 cos (40<sup>o<\/sup>&nbsp;\u2013 20<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 60<sup>o<\/sup>\/2 cos 20<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 30<sup>o<\/sup>&nbsp;cos 10<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 10<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 10<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>3. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i) cos 55\u00b0 + cos 65\u00b0 + cos 175\u00b0 = 0<\/strong><\/p>\n\n\n\n<p><strong>(ii) sin 50\u00b0 \u2013 sin 70\u00b0 + sin 10\u00b0 = 0<\/strong><\/p>\n\n\n\n<p><strong>(iii) cos 80\u00b0 + cos 40\u00b0 \u2013 cos 20\u00b0 = 0<\/strong><\/p>\n\n\n\n<p><strong>(iv) cos 20\u00b0 + cos 100\u00b0 + cos 140\u00b0 = 0<\/strong><\/p>\n\n\n\n<p><strong>(v) sin 5\u03c0\/18 \u2013 cos 4\u03c0\/9 = \u221a3 sin \u03c0\/9<\/strong><\/p>\n\n\n\n<p><strong>(vi) cos \u03c0\/12 \u2013 sin \u03c0\/12 = 1\/\u221a2<\/strong><\/p>\n\n\n\n<p><strong>(vii) sin 80\u00b0 \u2013 cos 70\u00b0 = cos 50\u00b0<\/strong><\/p>\n\n\n\n<p><strong>(viii) sin 51\u00b0 + cos 81\u00b0 = cos 21\u00b0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>cos 55\u00b0 + cos 65\u00b0 + cos 175\u00b0 = 0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos 55\u00b0 + cos 65\u00b0 + cos 175\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 55\u00b0 + cos 65\u00b0 + cos 175\u00b0 = 2 cos (55<sup>o<\/sup>&nbsp;+ 65<sup>o<\/sup>)\/2 cos (55<sup>o<\/sup>&nbsp;\u2013 65<sup>o<\/sup>) + cos (180<sup>o<\/sup>&nbsp;\u2013 5<sup>o<\/sup>)<\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 cos (-10<sup>o<\/sup>)\/2 \u2013 cos 5<sup>o<\/sup>&nbsp;(since, {cos (180\u00b0 \u2013 A) = \u2013 cos A})<\/p>\n\n\n\n<p>= 2 cos 60\u00b0 cos (-5\u00b0) \u2013 cos 5\u00b0 (since, {cos (-A) = cos A})<\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 5<sup>o<\/sup>&nbsp;\u2013 cos 5<sup>o<\/sup><\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>sin 50\u00b0 \u2013 sin 70\u00b0 + sin 10\u00b0 = 0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 50\u00b0 \u2013 sin 70\u00b0 + sin 10\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>sin 50\u00b0 \u2013 sin 70\u00b0 + sin 10\u00b0 = 2 cos (50<sup>o<\/sup>&nbsp;+ 70<sup>o<\/sup>)\/2 sin (50<sup>o<\/sup>&nbsp;\u2013 70<sup>o<\/sup>) + sin 10<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 sin (-20<sup>o<\/sup>)\/2 + sin 10<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 cos 60<sup>o<\/sup>&nbsp;(- sin 10<sup>o<\/sup>) + sin 10<sup>o<\/sup>&nbsp;[since,{sin (-A) = -sin (A)}]<\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 \u2013 sin 10<sup>o<\/sup>&nbsp;+ sin 10<sup>o<\/sup><\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>cos 80\u00b0 + cos 40\u00b0 \u2013 cos 20\u00b0 = 0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos 80\u00b0 + cos 40\u00b0 \u2013 cos 20\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 80\u00b0 + cos 40\u00b0 \u2013 cos 20\u00b0 = 2 cos (80<sup>o<\/sup>&nbsp;+ 40<sup>o<\/sup>)\/2 cos (80<sup>o<\/sup>&nbsp;\u2013 40<sup>o<\/sup>) \u2013 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 cos 40<sup>o<\/sup>\/2 \u2013 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 cos 60\u00b0 cos 20<sup>o<\/sup>&nbsp;\u2013 cos 20\u00b0<\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 20<sup>o<\/sup>&nbsp;\u2013 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>cos 20\u00b0 + cos 100\u00b0 + cos 140\u00b0 = 0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos 20\u00b0 + cos 100\u00b0 + cos 140\u00b0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 20\u00b0 + cos 100\u00b0 + cos 140\u00b0 = 2 cos (20<sup>o<\/sup>&nbsp;+ 100<sup>o<\/sup>)\/2 cos (20<sup>o<\/sup>&nbsp;\u2013 100<sup>o<\/sup>) + cos (180<sup>o<\/sup>&nbsp;\u2013 40<sup>o<\/sup>)<\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 cos (-80<sup>o<\/sup>)\/2 \u2013 cos 40<sup>o<\/sup>&nbsp;(since, {cos (180\u00b0 \u2013 A) = \u2013 cos A})<\/p>\n\n\n\n<p>= 2 cos 60\u00b0 cos (-40\u00b0) \u2013 cos 40\u00b0 (since, {cos (-A) = cos A})<\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 40<sup>o<\/sup>&nbsp;\u2013 cos 40<sup>o<\/sup><\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>sin 5\u03c0\/18 \u2013 cos 4\u03c0\/9 = \u221a3 sin \u03c0\/9<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 5\u03c0\/18 \u2013 cos 4\u03c0\/9 = sin 5\u03c0\/18 \u2013 sin (\u03c0\/2 \u2013 4\u03c0\/9) (since, cos A = sin (90<sup>o<\/sup>&nbsp;\u2013 A))<\/p>\n\n\n\n<p>= sin 5\u03c0\/18 \u2013 sin (9\u03c0 \u2013 8\u03c0)\/18<\/p>\n\n\n\n<p>= sin 5\u03c0\/18 \u2013 sin \u03c0\/18<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"417\" height=\"55\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8.gif\" alt=\"\" class=\"wp-image-543267\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 1\"\/><\/figure>\n\n\n\n<p>= 2 cos (6\u03c0\/36) sin (4\u03c0\/36)<\/p>\n\n\n\n<p>= 2 cos \u03c0\/6 sin \u03c0\/9<\/p>\n\n\n\n<p>= 2 cos 30<sup>o<\/sup>&nbsp;sin \u03c0\/9<\/p>\n\n\n\n<p>= 2 \u00d7 \u221a3\/2 \u00d7 sin \u03c0\/9<\/p>\n\n\n\n<p>= \u221a3 sin \u03c0\/9<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(vi)<\/strong>&nbsp;cos \u03c0\/12 \u2013 sin \u03c0\/12 = 1\/\u221a2<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos \u03c0\/12 \u2013 sin \u03c0\/12 = sin (\u03c0\/2 \u2013 \u03c0\/12) \u2013 sin \u03c0\/12 (since, cos A = sin(90<sup>o<\/sup>&nbsp;\u2013 A))<\/p>\n\n\n\n<p>= sin (6\u03c0 \u2013 5\u03c0)\/12 \u2013 sin \u03c0\/12<\/p>\n\n\n\n<p>= sin 5\u03c0\/12 \u2013 sin \u03c0\/12<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"417\" height=\"55\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-1.gif\" alt=\"\" class=\"wp-image-543269\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 2\"\/><\/figure>\n\n\n\n<p>= 2 cos (6\u03c0\/24) sin (4\u03c0\/24)<\/p>\n\n\n\n<p>= 2 cos \u03c0\/4 sin \u03c0\/6<\/p>\n\n\n\n<p>= 2 cos 45<sup>o<\/sup>&nbsp;sin 30<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/\u221a2 \u00d7 1\/2<\/p>\n\n\n\n<p>= 1\/\u221a2<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(vii)&nbsp;<\/strong>sin 80\u00b0 \u2013 cos 70\u00b0 = cos 50\u00b0<\/p>\n\n\n\n<p>sin 80\u00b0 = cos 50\u00b0 + cos 70<sup>o<\/sup><\/p>\n\n\n\n<p>So, now let us consider RHS<\/p>\n\n\n\n<p>cos 50\u00b0 + cos 70<sup>o<\/sup><\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 50\u00b0 + cos 70<sup>o<\/sup>&nbsp;= 2 cos (50<sup>o<\/sup>&nbsp;+ 70<sup>o<\/sup>)\/2 cos (50<sup>o<\/sup>&nbsp;\u2013 70<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 cos 120<sup>o<\/sup>\/2 cos (-20<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 cos 60<sup>o<\/sup>&nbsp;cos (-10<sup>o<\/sup>)<\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 10<sup>o<\/sup>&nbsp;(since, cos (-A) = cos A)<\/p>\n\n\n\n<p>= cos 10<sup>o<\/sup><\/p>\n\n\n\n<p>= cos (90\u00b0 \u2013 80\u00b0)<\/p>\n\n\n\n<p>= sin 80\u00b0 (since, cos (90\u00b0 \u2013 A) = sin A)<\/p>\n\n\n\n<p>= LHS<\/p>\n\n\n\n<p>Hence Proved.<\/p>\n\n\n\n<p><strong>(viii)&nbsp;<\/strong>sin 51\u00b0 + cos 81\u00b0 = cos 21\u00b0<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 51\u00b0 + cos 81\u00b0 = sin 51<sup>o<\/sup>&nbsp;+ sin (90<sup>o<\/sup>&nbsp;\u2013 81<sup>o<\/sup>)<\/p>\n\n\n\n<p>= sin 51<sup>o<\/sup>&nbsp;+ sin 9<sup>o<\/sup>&nbsp;(since, sin (90\u00b0 \u2013 A) = cos A)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 51<sup>o<\/sup>&nbsp;+ sin 9<sup>o<\/sup>&nbsp;= 2 sin (51<sup>o<\/sup>&nbsp;+ 9<sup>o<\/sup>)\/2 cos (51<sup>o<\/sup>&nbsp;\u2013 9<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 60<sup>o<\/sup>\/2 cos 42<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 30<sup>o<\/sup>&nbsp;cos 21<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 21<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 21<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>4. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i) cos (3\u03c0\/4 + x) \u2013 cos (3\u03c0\/4 \u2013 x) = -\u221a2 sin x<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos (\u03c0\/4 + x) + cos (\u03c0\/4 \u2013 x) = \u221a2 cos x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>cos (3\u03c0\/4 + x) \u2013 cos (3\u03c0\/4 \u2013 x) = -\u221a2 sin x<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos (3\u03c0\/4 + x) \u2013 cos (3\u03c0\/4 \u2013 x)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A \u2013 cos B = -2 sin (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>cos (3\u03c0\/4 + x) \u2013 cos (3\u03c0\/4 \u2013 x) = -2 sin (3\u03c0\/4 + x + 3\u03c0\/4 \u2013 x)\/2 sin (3\u03c0\/4 + x \u2013 3\u03c0\/4 + x)\/2<\/p>\n\n\n\n<p>= -2 sin (6\u03c0\/4)\/2 sin 2x\/2<\/p>\n\n\n\n<p>= -2 sin 6\u03c0\/8 sin x<\/p>\n\n\n\n<p>= -2 sin 3\u03c0\/4 sin x<\/p>\n\n\n\n<p>= -2 sin (\u03c0 \u2013 \u03c0\/4) sin x<\/p>\n\n\n\n<p>= -2 sin \u03c0\/4 sin x (since, (\u03c0-A) = sin A)<\/p>\n\n\n\n<p>= -2 \u00d7 1\/\u221a2 \u00d7 sin x<\/p>\n\n\n\n<p>= -\u221a2 sin x<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>cos (\u03c0\/4 + x) + cos (\u03c0\/4 \u2013 x) = \u221a2 cos x<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos (\u03c0\/4 + x) + cos (\u03c0\/4 \u2013 x)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos (\u03c0\/4 + x) + cos (\u03c0\/4 \u2013 x) = 2 cos (\u03c0\/4 + x + \u03c0\/4 \u2013 x)\/2 cos (\u03c0\/4 + x \u2013 \u03c0\/4 + x)\/2<\/p>\n\n\n\n<p>= 2 cos (2\u03c0\/4)\/2 cos 2x\/2<\/p>\n\n\n\n<p>= 2 cos 2\u03c0\/8 cos x<\/p>\n\n\n\n<p>= 2 sin \u03c0\/4 cos x<\/p>\n\n\n\n<p>= 2 \u00d7 1\/\u221a2 \u00d7 cos x<\/p>\n\n\n\n<p>= \u221a2 cos x<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>5. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i) sin 65<sup>o<\/sup>&nbsp;+ cos 65<sup>o<\/sup>&nbsp;= \u221a2 cos 20<sup>o<\/sup><\/strong><\/p>\n\n\n\n<p><strong>(ii) sin 47<sup>o<\/sup>&nbsp;+ cos 77<sup>o<\/sup>&nbsp;= cos 17<sup>o<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin 65<sup>o<\/sup>&nbsp;+ cos 65<sup>o<\/sup>&nbsp;= \u221a2 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 65<sup>o<\/sup>&nbsp;+ cos 65<sup>o<\/sup>&nbsp;= sin 65<sup>o<\/sup>&nbsp;+ sin (90<sup>o<\/sup>&nbsp;\u2013 65<sup>o<\/sup>)<\/p>\n\n\n\n<p>= sin 65<sup>o<\/sup>&nbsp;+ sin 25<sup>o<\/sup><\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 65<sup>o<\/sup>&nbsp;+ sin 25<sup>o<\/sup>&nbsp;= 2 sin (65<sup>o<\/sup>&nbsp;+ 25<sup>o<\/sup>)\/2 cos (65<sup>o<\/sup>&nbsp;\u2013 25<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 90<sup>o<\/sup>\/2 cos 40<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 45<sup>o<\/sup>&nbsp;cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/\u221a2 \u00d7 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= \u221a2 cos 20<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>sin 47<sup>o<\/sup>&nbsp;+ cos 77<sup>o<\/sup>&nbsp;= cos 17<sup>o<\/sup><\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 47<sup>o<\/sup>&nbsp;+ cos 77<sup>o<\/sup>&nbsp;= sin 47<sup>o<\/sup>&nbsp;+ sin (90<sup>o<\/sup>&nbsp;\u2013 77<sup>o<\/sup>)<\/p>\n\n\n\n<p>= sin 47<sup>o<\/sup>&nbsp;+ sin 13<sup>o<\/sup><\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>sin 47<sup>o<\/sup>&nbsp;+ sin 13<sup>o<\/sup>&nbsp;= 2 sin (47<sup>o<\/sup>&nbsp;+ 13<sup>o<\/sup>)\/2 cos (47<sup>o<\/sup>&nbsp;\u2013 13<sup>o<\/sup>)\/2<\/p>\n\n\n\n<p>= 2 sin 60<sup>o<\/sup>\/2 cos 34<sup>o<\/sup>\/2<\/p>\n\n\n\n<p>= 2 sin 30<sup>o<\/sup>&nbsp;cos 17<sup>o<\/sup><\/p>\n\n\n\n<p>= 2 \u00d7 1\/2 \u00d7 cos 17<sup>o<\/sup><\/p>\n\n\n\n<p>= cos 17<sup>o<\/sup><\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>6.<\/strong>&nbsp;<strong>Prove that:<br>(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A<\/strong><\/p>\n\n\n\n<p><strong>(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A\/2 cos 3A\/2 sin 3A<\/strong><\/p>\n\n\n\n<p><strong>(iv) sin 3A + sin 2A \u2013 sin A = 4 sin A cos A\/2 cos 3A\/2<\/strong><\/p>\n\n\n\n<p><strong>(v) cos 20<sup>o<\/sup>&nbsp;cos 100<sup>o<\/sup>&nbsp;+ cos 100<sup>o<\/sup>&nbsp;cos 140<sup>o<\/sup>&nbsp;\u2013 cos 140<sup>o<\/sup>&nbsp;cos 200<sup>o<\/sup>&nbsp;= \u2013 3\/4<\/strong><\/p>\n\n\n\n<p><strong>(vi) sin x\/2 sin 7x\/2 + sin 3x\/2 sin 11x\/2 = sin 2x sin 5x<\/strong><\/p>\n\n\n\n<p><strong>(vii) cos x cos x\/2 \u2013 cos 3x cos 9x\/2 = sin 4x sin 7x\/2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos 3A + cos 5A + cos 7A + cos 15A<\/p>\n\n\n\n<p>So now,<\/p>\n\n\n\n<p>(cos 5A + cos 3A) + (cos 15A + cos 7A)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>(cos 5A + cos 3A) + (cos 15A + cos 7A)<\/p>\n\n\n\n<p>= [2 cos (5A+3A)\/2 cos (5A-3A)\/2] + [2 cos (15A+7A)\/2 cos (15A-7A)\/2]<\/p>\n\n\n\n<p>= [2 cos 8A\/2 cos 2A\/2] + [2 cos 22A\/2 cos 8A\/2]<\/p>\n\n\n\n<p>= [2 cos 4A cos A] + [2 cos 11A cos 4A]<\/p>\n\n\n\n<p>= 2 cos 4A (cos 11A + cos A)<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>2 cos 4A (cos 11A + cos A) = 2 cos 4A [2 cos (11A+A)\/2 cos (11A-A)\/2]<\/p>\n\n\n\n<p>= 2 cos 4A [2 cos 12A\/2 cos 10A\/2]<\/p>\n\n\n\n<p>= 2 cos 4A [2 cos 6A cos 5A]<\/p>\n\n\n\n<p>= 4 cos 4A cos 5A cos 6A<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(ii)<\/strong>&nbsp;cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos A + cos 3A + cos 5A + cos 7A<\/p>\n\n\n\n<p>So now,<\/p>\n\n\n\n<p>(cos 3A + cos A) + (cos 7A + cos 5A)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>(cos 3A + cos A) + (cos 7A + cos 5A)<\/p>\n\n\n\n<p>= [2 cos (3A+A)\/2 cos (3A-A)\/2] + [2 cos (7A+5A)\/2 cos (7A-5A)\/2]<\/p>\n\n\n\n<p>= [2 cos 4A\/2 cos 2A\/2] + [2 cos 12A\/2 cos 2A\/2]<\/p>\n\n\n\n<p>= [2 cos 2A cos A] + [2 cos 6A cos A]<\/p>\n\n\n\n<p>= 2 cos A (cos 6A + cos 2A)<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>2 cos A (cos 6A + cos 2A) = 2 cos A [2 cos (6A+2A)\/2 cos (6A-2A)\/2]<\/p>\n\n\n\n<p>= 2 cos A [2 cos 8A\/2 cos 4A\/2]<\/p>\n\n\n\n<p>= 2 cos A [2 cos 4A cos 2A]<\/p>\n\n\n\n<p>= 4 cos A cos 2A cos 4A<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(iii)<\/strong>&nbsp;sin A + sin 2A + sin 4A + sin 5A = 4 cos A\/2 cos 3A\/2 sin 3A<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin A + sin 2A + sin 4A + sin 5A<\/p>\n\n\n\n<p>So now,<\/p>\n\n\n\n<p>(sin 2A + sin A) + (sin 5A + sin 4A)<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>(sin 2A + sin A) + (sin 5A + sin 4A) =<\/p>\n\n\n\n<p>= [2 sin (2A+A)\/2 cos (2A-A)\/2] + [2 sin (5A+4A)\/2 cos (5A-4A)\/2]<\/p>\n\n\n\n<p>= [2 sin 3A\/2 cos A\/2] + [2 sin 9A\/2 cos A\/2]<\/p>\n\n\n\n<p>= 2 cos A\/2 (sin 9A\/2 + sin 3A\/2)<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>2 cos A\/2 (sin 9A\/2 + sin 3A\/2) = 2 cos A\/2 [2 sin (9A\/2 + 3A\/2)\/2 cos (9A\/2 \u2013 3A\/2)\/2]<\/p>\n\n\n\n<p>= 2 cos A\/2 [2 sin ((9A+3A)\/2)\/2 cos ((9A-3A)\/2)\/2]<\/p>\n\n\n\n<p>= 2 cos A\/2 [2 sin 12A\/4 cos 6A\/4]<\/p>\n\n\n\n<p>= 2 cos A\/2 [2 sin 3A cos 3A\/2]<\/p>\n\n\n\n<p>= 4 cos A\/2 cos 3A\/2 sin 3A<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(iv)<\/strong>&nbsp;sin 3A + sin 2A \u2013 sin A = 4 sin A cos A\/2 cos 3A\/2<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin 3A + sin 2A \u2013 sin A<\/p>\n\n\n\n<p>So now,<\/p>\n\n\n\n<p>(sin 3A \u2013 sin A) + sin 2A<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>(sin 3A \u2013 sin A) + sin 2A = 2 cos (3A + A)\/2 sin (3A \u2013 A)\/2 + sin 2A<\/p>\n\n\n\n<p>= 2 cos 4A\/2 sin 2A\/2 + sin 2A<\/p>\n\n\n\n<p>We know that, sin 2A = 2 sin A cos A<\/p>\n\n\n\n<p>= 2 cos 2A Sin A + 2 sin A cos A<\/p>\n\n\n\n<p>= 2 sin A (cos 2A + cos A)<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>2 sin A (cos 2A + cos A) = 2 sin A [2 cos (2A+A)\/2 cos (2A-A)\/2]<\/p>\n\n\n\n<p>= 2 sin A [2 cos 3A\/2 cos A\/2]<\/p>\n\n\n\n<p>= 4 sin A cos A\/2 cos 3A\/2<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>cos 20<sup>o<\/sup>&nbsp;cos 100<sup>o<\/sup>&nbsp;+ cos 100<sup>o<\/sup>&nbsp;cos 140<sup>o<\/sup>&nbsp;\u2013 cos 140<sup>o<\/sup>&nbsp;cos 200<sup>o<\/sup>&nbsp;= \u2013 3\/4<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos 20<sup>o<\/sup>&nbsp;cos 100<sup>o<\/sup>&nbsp;+ cos 100<sup>o<\/sup>&nbsp;cos 140<sup>o<\/sup>&nbsp;\u2013 cos 140<sup>o<\/sup>&nbsp;cos 200<sup>o<\/sup>&nbsp;=<\/p>\n\n\n\n<p>We shall multiply and divide by 2 we get,<\/p>\n\n\n\n<p>= 1\/2 [2 cos 100<sup>o<\/sup>&nbsp;cos 20<sup>o<\/sup>&nbsp;+ 2 cos 140<sup>o<\/sup>&nbsp;cos 100<sup>o<\/sup>&nbsp;\u2013 2 cos 200<sup>o<\/sup>&nbsp;cos 140<sup>o<\/sup>]<\/p>\n\n\n\n<p>We know that 2 cos A cos B = cos (A+B) + cos (A-B)<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>= 1\/2 [cos (100<sup>o&nbsp;<\/sup>+ 20<sup>o<\/sup>) + cos (100<sup>o&nbsp;<\/sup>\u2013 20<sup>o<\/sup>) + cos (140<sup>o&nbsp;<\/sup>+ 100<sup>o<\/sup>) + cos (140<sup>o&nbsp;<\/sup>\u2013 100<sup>o<\/sup>) \u2013 cos (200<sup>o&nbsp;<\/sup>+ 140<sup>o<\/sup>) \u2013 cos (200<sup>o&nbsp;<\/sup>\u2013 140<sup>o<\/sup>)]]<\/p>\n\n\n\n<p>= 1\/2 [cos 120<sup>o<\/sup>&nbsp;+ cos 80<sup>o<\/sup>&nbsp;+ cos 240<sup>o<\/sup>&nbsp;+ cos 40<sup>o<\/sup>&nbsp;\u2013 cos 340<sup>o<\/sup>&nbsp;\u2013 cos 60<sup>o<\/sup>]<\/p>\n\n\n\n<p>= 1\/2 [cos (90<sup>o<\/sup>&nbsp;+ 30<sup>o<\/sup>) + cos 80<sup>o<\/sup>&nbsp;+ cos (180<sup>o<\/sup>&nbsp;+ 60<sup>o<\/sup>) + cos 40<sup>o<\/sup>&nbsp;\u2013 cos (360<sup>o<\/sup>&nbsp;\u2013 20<sup>o<\/sup>) \u2013 cos 60<sup>o<\/sup>]<\/p>\n\n\n\n<p>We know, cos (180<sup>o<\/sup>&nbsp;+ A) = \u2013 cos A, cos (90<sup>o<\/sup>&nbsp;+ A) = \u2013 sin A, cos (360<sup>o<\/sup>&nbsp;\u2013 A) = cos A<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>= 1\/2 [- sin 30<sup>o<\/sup>&nbsp;+ cos 80<sup>o<\/sup>&nbsp;\u2013 cos 60<sup>o<\/sup>&nbsp;+ cos 40<sup>o<\/sup>&nbsp;\u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 cos 60<sup>o<\/sup>]<\/p>\n\n\n\n<p>= 1\/2 [- sin 30<sup>o<\/sup>&nbsp;+ cos 80<sup>o<\/sup>&nbsp;+ cos 40<sup>o<\/sup>&nbsp;\u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 2 cos 60<sup>o<\/sup>]<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>= 1\/2 [- sin 30<sup>o<\/sup>&nbsp;+ 2 cos (80<sup>o<\/sup>+40<sup>o<\/sup>)\/2 cos (80<sup>o<\/sup>-40<sup>o<\/sup>)\/2 \u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 2 \u00d7 1\/2]<\/p>\n\n\n\n<p>= 1\/2 [- sin 30<sup>o<\/sup>&nbsp;+ 2 cos 120<sup>o<\/sup>\/2 cos 40<sup>o<\/sup>\/2 \u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 1]<\/p>\n\n\n\n<p>= 1\/2 [- sin 30<sup>o<\/sup>&nbsp;+ 2 cos 60<sup>o<\/sup>&nbsp;cos 20<sup>o<\/sup>&nbsp;\u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 1]<\/p>\n\n\n\n<p>= 1\/2 [- 1\/2 + 2\u00d71\/2\u00d7cos 20<sup>o<\/sup>&nbsp;\u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 1]<\/p>\n\n\n\n<p>= 1\/2 [-1\/2 + cos 20<sup>o<\/sup>&nbsp;\u2013 cos 20<sup>o<\/sup>&nbsp;\u2013 1]<\/p>\n\n\n\n<p>= 1\/2 [-1\/2 -1]<\/p>\n\n\n\n<p>= 1\/2 [-3\/2]<\/p>\n\n\n\n<p>= -3\/4<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(vi)<\/strong>&nbsp;sin x\/2 sin 7x\/2 + sin 3x\/2 sin 11x\/2 = sin 2x sin 5x<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin x\/2 sin 7x\/2 + sin 3x\/2 sin 11x\/2 =<\/p>\n\n\n\n<p>We shall multiply and divide by 2 we get,<\/p>\n\n\n\n<p>= 1\/2 [2 sin 7x\/2 sin x\/2 + 2 sin 11x\/2 sin 3x\/2]<\/p>\n\n\n\n<p>We know that 2 sin A sin B = cos (A-B) \u2013 cos (A+B)<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>= 1\/2 [cos (7x\/2 \u2013 x\/2) \u2013 cos (7x\/2 + x\/2) + cos (11x\/2 \u2013 3x\/2) \u2013 cos (11x\/2 + 3x\/2)]<\/p>\n\n\n\n<p>= 1\/2 [cos (7x-x)\/2 \u2013 cos (7x+x)\/2 + cos (11x-3x)\/2 \u2013 cos (11x+3x)\/2]<\/p>\n\n\n\n<p>= 1\/2 [cos 6x\/2 \u2013 cos 8x\/2 + cos 8x\/2 \u2013 cos 14x\/2]<\/p>\n\n\n\n<p>= 1\/2 [cos 3x \u2013 cos 7x]<\/p>\n\n\n\n<p>= \u2013 1\/2 [cos 7x \u2013 cos 3x]<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>cos A \u2013 cos B = -2 sin (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>= -1\/2 [-2 sin (7x+3x)\/2 sin (7x-3x)\/2]<\/p>\n\n\n\n<p>= -1\/2 [-2 sin 10x\/2 sin 4x\/2]<\/p>\n\n\n\n<p>= -1\/2 [-2 sin 5x sin 2x]<\/p>\n\n\n\n<p>= -2\/-2 sin 5x sin 2x<\/p>\n\n\n\n<p>= sin 2x sin 5x<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(vii)&nbsp;<\/strong>cos x cos x\/2 \u2013 cos 3x cos 9x\/2 = sin 4x sin 7x\/2<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos x cos x\/2 \u2013 cos 3x cos 9x\/2 =<\/p>\n\n\n\n<p>We shall multiply and divide by 2 we get,<\/p>\n\n\n\n<p>= 1\/2 [2 cos x cos x\/2 \u2013 2 cos 9x\/2 cos 3x]<\/p>\n\n\n\n<p>We know that 2 cos A cos B = cos (A+B) + cos (A-B)<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>= 1\/2 [cos (x + x\/2) + cos (x \u2013 x\/2) \u2013 cos (9x\/2 + 3x) \u2013 cos (9x\/2 \u2013 3x)]<\/p>\n\n\n\n<p>= 1\/2 [cos (2x+x)\/2 + cos (2x-x)\/2 \u2013 cos (9x+6x)\/2 \u2013 cos (9x-6x)\/2]<\/p>\n\n\n\n<p>= 1\/2 [cos 3x\/2 + cos x\/2 \u2013 cos 15x\/2 \u2013 cos 3x\/2]<\/p>\n\n\n\n<p>= 1\/2 [cos x\/2 \u2013 cos 15x\/2]<\/p>\n\n\n\n<p>= \u2013 1\/2 [cos 15x\/2 \u2013 cos x\/2]<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>cos A \u2013 cos B = -2 sin (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>= \u2013 1\/2 [-2 sin (15x\/2 + x\/2)\/2 sin (15x\/2 \u2013 x\/2)\/2]<\/p>\n\n\n\n<p>= -1\/2 [-2 sin (16x\/2)\/2 sin (14x\/2)\/2]<\/p>\n\n\n\n<p>= -1\/2 [-2 sin 16x\/4 sin 7x\/2]<\/p>\n\n\n\n<p>= \u2013 1\/2 [-2 sin 4x sin 7x\/2]<\/p>\n\n\n\n<p>= -2\/-2 [sin 4x sin 7x\/2]<\/p>\n\n\n\n<p>= sin 4x sin 7x\/2<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>7. Prove that:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"249\" height=\"179\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8.png\" alt=\"\" class=\"wp-image-543270\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 3\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"352\" height=\"113\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-1.png\" alt=\"\" class=\"wp-image-543271\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 4\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-1.png 352w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-1-300x96.png 300w\" sizes=\"auto, (max-width: 352px) 100vw, 352px\" \/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"434\" height=\"513\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-2.png\" alt=\"\" class=\"wp-image-543272\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 5\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-2.png 434w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-2-254x300.png 254w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-2-400x473.png 400w\" sizes=\"auto, (max-width: 434px) 100vw, 434px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"235\" height=\"170\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-3.png\" alt=\"\" class=\"wp-image-543273\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 6\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"500\" height=\"360\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-4.png\" alt=\"\" class=\"wp-image-543274\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 7\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-4.png 500w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-4-300x216.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-4-400x288.png 400w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"342\" height=\"452\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-5.png\" alt=\"\" class=\"wp-image-543275\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 8\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-5.png 342w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-5-227x300.png 227w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"351\" height=\"462\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-6.png\" alt=\"\" class=\"wp-image-543276\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 9\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-6.png 351w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-6-228x300.png 228w\" sizes=\"auto, (max-width: 351px) 100vw, 351px\" \/><\/figure>\n\n\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"510\" height=\"165\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-8.png\" alt=\"\" class=\"wp-image-543277\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 11\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-8.png 510w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-8-300x97.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-8-400x129.png 400w\" sizes=\"auto, (max-width: 510px) 100vw, 510px\" \/><\/figure>\n\n\n\n<p><strong>8. Prove that:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"414\" height=\"502\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-9.png\" alt=\"\" class=\"wp-image-543278\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 12\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-9.png 414w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-9-247x300.png 247w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-9-400x485.png 400w\" sizes=\"auto, (max-width: 414px) 100vw, 414px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"361\" height=\"114\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-10.png\" alt=\"\" class=\"wp-image-543279\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 13\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-10.png 361w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-10-300x95.png 300w\" sizes=\"auto, (max-width: 361px) 100vw, 361px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"412\" height=\"54\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-11.png\" alt=\"\" class=\"wp-image-543280\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 14\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-11.png 412w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-11-300x39.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-11-400x52.png 400w\" sizes=\"auto, (max-width: 412px) 100vw, 412px\" \/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"523\" height=\"500\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-12.png\" alt=\"\" class=\"wp-image-543281\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 15\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-12.png 523w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-12-300x287.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-12-400x382.png 400w\" sizes=\"auto, (max-width: 523px) 100vw, 523px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"410\" height=\"123\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-13.png\" alt=\"\" class=\"wp-image-543282\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 16\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-13.png 410w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-13-300x90.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-13-400x120.png 400w\" sizes=\"auto, (max-width: 410px) 100vw, 410px\" \/><\/figure>\n\n\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"587\" height=\"504\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-15.png\" alt=\"\" class=\"wp-image-543283\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 18\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-15.png 587w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-15-300x258.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-15-400x343.png 400w\" sizes=\"auto, (max-width: 587px) 100vw, 587px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"476\" height=\"269\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-16.png\" alt=\"\" class=\"wp-image-543284\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 19\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-16.png 476w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-16-300x170.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-16-400x226.png 400w\" sizes=\"auto, (max-width: 476px) 100vw, 476px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"553\" height=\"347\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-17.png\" alt=\"\" class=\"wp-image-543285\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 20\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-17.png 553w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-17-300x188.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-17-400x251.png 400w\" sizes=\"auto, (max-width: 553px) 100vw, 553px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"587\" height=\"509\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-18.png\" alt=\"\" class=\"wp-image-543286\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 21\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-18.png 587w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-18-300x260.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-18-400x347.png 400w\" sizes=\"auto, (max-width: 587px) 100vw, 587px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"392\" height=\"227\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-19.png\" alt=\"\" class=\"wp-image-543287\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 22\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-19.png 392w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-19-300x174.png 300w\" sizes=\"auto, (max-width: 392px) 100vw, 392px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"602\" height=\"508\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-20.png\" alt=\"\" class=\"wp-image-543288\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 23\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-20.png 602w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-20-300x253.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-20-400x338.png 400w\" sizes=\"auto, (max-width: 602px) 100vw, 602px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"240\" height=\"100\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-21.png\" alt=\"\" class=\"wp-image-543289\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 24\"\/><\/figure>\n\n\n\n<p>= cot 6A<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"560\" height=\"497\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-22.png\" alt=\"\" class=\"wp-image-543290\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 25\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-22.png 560w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-22-300x266.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-22-400x355.png 400w\" sizes=\"auto, (max-width: 560px) 100vw, 560px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"342\" height=\"279\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-23.png\" alt=\"\" class=\"wp-image-543291\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 26\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-23.png 342w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-23-300x245.png 300w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"235\" height=\"291\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-24.png\" alt=\"\" class=\"wp-image-543292\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 27\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"581\" height=\"469\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-25.png\" alt=\"\" class=\"wp-image-543293\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 28\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-25.png 581w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-25-300x242.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-25-400x323.png 400w\" sizes=\"auto, (max-width: 581px) 100vw, 581px\" \/><\/figure>\n\n\n\n<p>By using the formulas,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>cos A \u2013 cos B = -2 sin (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"290\" height=\"312\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-26.png\" alt=\"\" class=\"wp-image-543294\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 29\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-26.png 290w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-26-279x300.png 279w\" sizes=\"auto, (max-width: 290px) 100vw, 290px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"549\" height=\"509\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-27.png\" alt=\"\" class=\"wp-image-543295\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 30\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-27.png 549w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-27-300x278.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-27-400x371.png 400w\" sizes=\"auto, (max-width: 549px) 100vw, 549px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"342\" height=\"423\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-28.png\" alt=\"\" class=\"wp-image-543296\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 31\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-28.png 342w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-28-243x300.png 243w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"557\" height=\"386\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-29.png\" alt=\"\" class=\"wp-image-543297\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 32\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-29.png 557w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-29-300x208.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-29-400x277.png 400w\" sizes=\"auto, (max-width: 557px) 100vw, 557px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"518\" height=\"541\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-30.png\" alt=\"\" class=\"wp-image-543298\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 33\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-30.png 518w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-30-287x300.png 287w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-30-400x418.png 400w\" sizes=\"auto, (max-width: 518px) 100vw, 518px\" \/><\/figure>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"484\" height=\"251\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-31.png\" alt=\"\" class=\"wp-image-543299\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 34\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-31.png 484w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-31-300x156.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-31-400x207.png 400w\" sizes=\"auto, (max-width: 484px) 100vw, 484px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"343\" height=\"374\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-32.png\" alt=\"\" class=\"wp-image-543300\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 35\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-32.png 343w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-32-275x300.png 275w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"598\" height=\"438\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-33.png\" alt=\"\" class=\"wp-image-543301\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 36\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-33.png 598w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-33-300x220.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-33-400x293.png 400w\" sizes=\"auto, (max-width: 598px) 100vw, 598px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"225\" height=\"242\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-34.png\" alt=\"\" class=\"wp-image-543302\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 37\"\/><\/figure>\n\n\n\n<p><strong>9. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i) sin \u03b1 + sin \u03b2 + sin \u03b3 \u2013 sin (\u03b1 + \u03b2 + \u03b3) = 4 sin (\u03b1 + \u03b2)\/2 sin (\u03b2 + \u03b3)\/2 sin (\u03b1 + \u03b3)\/2<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos (A + B + C) + cos (A \u2013 B + C) + cos (A + B \u2013 C) + cos (-A + B + C) = 4 cos A cos B cos C<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin \u03b1 + sin \u03b2 + sin \u03b3 \u2013 sin (\u03b1 + \u03b2 + \u03b3) = 4 sin (\u03b1 + \u03b2)\/2 sin (\u03b2 + \u03b3)\/2 sin (\u03b1 + \u03b3)\/2<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>sin \u03b1 + sin \u03b2 + sin \u03b3 \u2013 sin (\u03b1 + \u03b2 + \u03b3)<\/p>\n\n\n\n<p>By using the formulas,<\/p>\n\n\n\n<p>Sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>Sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"600\" height=\"349\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-35.png\" alt=\"\" class=\"wp-image-543303\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 38\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-35.png 600w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-35-300x175.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-35-400x233.png 400w\" sizes=\"auto, (max-width: 600px) 100vw, 600px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"580\" height=\"559\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-36.png\" alt=\"\" class=\"wp-image-543304\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 39\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-36.png 580w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-36-300x289.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-36-400x386.png 400w\" sizes=\"auto, (max-width: 580px) 100vw, 580px\" \/><\/figure>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>cos (A + B + C) + cos (A \u2013 B + C) + cos (A + B \u2013 C) + cos (-A + B + C) = 4 cos A cos B cos C<\/p>\n\n\n\n<p>Let us consider LHS:<\/p>\n\n\n\n<p>cos (A + B + C) + cos (A \u2013 B + C) + cos (A + B \u2013 C) + cos (-A + B + C)<\/p>\n\n\n\n<p>so,<\/p>\n\n\n\n<p>cos (A + B + C) + cos (A \u2013 B + C) + cos (A + B \u2013 C) + cos (-A + B + C) =<\/p>\n\n\n\n<p>={cos (A + B + C) + cos (A \u2013 B + C)} + {cos (A + B \u2013 C) + cos (-A + B + C)}<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>Cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"602\" height=\"541\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-37.png\" alt=\"\" class=\"wp-image-543305\" title=\"RD Sharma Solutions for Class 11 Maths Chapter 8 \u2013 Transformation Formulae image - 40\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-37.png 602w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-37-300x270.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-8-37-400x359.png 400w\" sizes=\"auto, (max-width: 602px) 100vw, 602px\" \/><\/figure>\n\n\n\n<p>= 4 cos A cos B cos C<\/p>\n\n\n\n<p>= RHS<\/p>\n\n\n\n<p>Hence proved.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-11-maths-chapter-8-download-pdf\">RD Sharma Solutions for Class 11 Maths Chapter 8:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RD-Sharma-Solutions-for-Class-11-Maths-Chapter-8\u2013Transformation-Formulae.pdf\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise RD Sharma Solutions for Class 11&nbsp;Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-1-sets\/\">Chapter 1\u2013Sets<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-2-relations\/\">Chapter 2\u2013Relations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-3-functions\/\">Chapter 3\u2013Functions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-4-measurement-of-angles\/\">Chapter 4\u2013Measurement of Angles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-5-trigonometric-functions\/\">Chapter 5\u2013Trigonometric Functions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-6-graphs-of-trigonometric-functions\/\">Chapter 6\u2013Graphs of Trigonometric Functions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-7-values-of-trigonometric-functions-at-sum-or-difference-of-angles\/\">Chapter 7\u2013Values of Trigonometric Functions at Sum or Difference of Angles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\">Chapter 8\u2013Transformation Formulae<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-9-values-of-trigonometric-functions-at-multiples-and-submultiples-of-an-angle\/\">Chapter 9\u2013Values of Trigonometric Functions at Multiples and Submultiples of an Angle<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-10-sine-and-cosine-formulae-and-their-applications\/\">Chapter 10\u2013Sine and Cosine Formulae and their Applications<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\">Chapter 11\u2013Trigonometric Equations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-12-mathematical-induction\/\">Chapter 12\u2013Mathematical Induction<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers\/\">Chapter 13\u2013Complex Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations\/\">Chapter 14\u2013Quadratic Equations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-15-linear-inequations\/\">Chapter 15\u2013Linear Inequations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-16-permutations\/\">Chapter 16\u2013Permutations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-17-combinations\/\">Chapter 17\u2013Combinations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem\/\">Chapter 18\u2013Binomial Theorem<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-19-arithmetic-progressions\/\">Chapter 19\u2013Arithmetic Progressions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-20-geometric-progressions\/\">Chapter 20\u2013Geometric Progressions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-21-some-special-series\/\">Chapter 21\u2013Some Special Series<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-22-brief-review-of-cartesian-system-of-rectangular-coordinates\/\">Chapter 22\u2013Brief review of Cartesian System of Rectangular Coordinates<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-23-the-straight-lines\/\">Chapter 23\u2013The Straight Lines<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle\/\">Chapter 24\u2013The Circle<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-25-parabola\/\">Chapter 25\u2013Parabola<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-26-ellipse\/\">Chapter 26\u2013Ellipse<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-27-hyperbola\/\">Chapter 27\u2013Hyperbola<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-28-introduction-to-three-dimensional-coordinate-geometry\/\">Chapter 28\u2013Introduction to Three Dimensional Coordinate Geometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-29-limits\/\">Chapter 29\u2013Limits<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-30-derivatives\/\">Chapter 30\u2013Derivatives<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-31-mathematical-reasoning\/\">Chapter 31\u2013Mathematical Reasoning<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-32-statistics\/\">Chapter 32\u2013Statistics<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-33-probability\/\">Chapter 33\u2013Probability<\/a><\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About RD Sharma<\/h2>\n\n\n\n<p>RD Sharma i<em>sn&#8217;t the kind of author you&#8217;d bump into at lit fests. But his bestselling books have helped many&nbsp;<\/em>CBSE<em>&nbsp;students lose their dread of&nbsp;<\/em>maths<em>. Sunday Times profiles the tutor turned internet star<\/em><br>He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like &#8216;series solution of linear differential equations&#8217;. Meet Dr&nbsp;Ravi Dutt Sharma&nbsp;\u2014&nbsp;mathematics&nbsp;teacher and author of 25 reference books \u2014 whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it&#8217;s only recently that a spoof video turned the tutor into a YouTube star.<\/p>\n\n\n\n<p>R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. &#8220;I like to spend all my time thinking and writing about maths problems. I find it relaxing,&#8221; he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government&#8217;s Guru Nanak Dev Institute of Technology.<\/p>\n\n\n\n<div class=\"wp-block-buttons is-layout-flex wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions\/\" style=\"background-color:#cd5c5c\">RD Sharma Solutions<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-class-xi\/\" style=\"background-color:#cd5c5c\">NCERT Class 11 Solutions<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-class-11\/\" style=\"background-color:#cd5c5c\">RD Sharma Class 11 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Class 11: Maths Chapter 8 solutions. Complete Class 11 Maths Chapter 8 Notes. RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae RD Sharma 11th Maths Chapter 8, Class 11 Maths Chapter 8 solutions EXERCISE 8.1 PAGE NO: 8.6 1. Express each of the following as the sum or difference of sines and cosines:(i) [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":543262,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,919],"tags":[1962],"boards":[],"class_list":["post-543259","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-11","tag-rd-sharma-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>RD Sharma Solutions for Class 11, maths Chapter 8 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae | Browse Class 11 Maths Chapters RD Sharma books - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae\" \/>\n<meta property=\"og:description\" content=\"Class 11: Maths Chapter 8 solutions. Complete Class 11 Maths Chapter 8 Notes. RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae RD\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2021-09-29T11:13:54+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-12-26T10:15:36+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/class11m8.png\" \/>\n\t<meta property=\"og:image:width\" content=\"1200\" \/>\n\t<meta property=\"og:image:height\" content=\"675\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"31 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae\",\"datePublished\":\"2021-09-29T11:13:54+00:00\",\"dateModified\":\"2022-12-26T10:15:36+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\"},\"wordCount\":3935,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/class11m8.png\",\"keywords\":[\"RD Sharma Solutions\"],\"articleSection\":[\"Book Solutions\",\"class 11\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\",\"name\":\"RD Sharma Solutions for Class 11, maths Chapter 8 - 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Complete Class 11 Maths Chapter 8 Notes. 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