Class 11: Maths Chapter 9 solutions. Complete Class 11 Maths Chapter 9 Notes.
Contents
RD Sharma Solutions for Class 11 Maths Chapter 9–Values of Trigonometric Functions at Multiples and Submultiples of an Angle
RD Sharma 11th Maths Chapter 9, Class 11 Maths Chapter 9 solutions
EXERCISE 9.1 PAGE NO: 9.28
Prove the following identities:
1. √[(1 – cos 2x) / (1 + cos 2x)] = tan x
Solution:
Let us consider LHS:
√[(1 – cos 2x) / (1 + cos 2x)]
We know that cos 2x = 1 – 2 sin2 x
= 2 cos2 x – 1
So,
√[(1 – cos 2x) / (1 + cos 2x)] = √[(1 – (1 – 2sin2 x)) / (1 + (2cos2x – 1))]
= √[(1 – 1 + 2sin2 x) / (1 + 2cos2 x – 1)]
= √[2 sin2 x / 2 cos2 x]
= sin x/cos x
= tan x
= RHS
Hence proved.
2. sin 2x / (1 – cos 2x) = cot x
Solution:
Let us consider LHS:
sin 2x / (1 – cos 2x)
We know that cos 2x = 1 – 2 sin2 x
Sin 2x = 2 sin x cos x
So,
sin 2x / (1 – cos 2x) = (2 sin x cos x) / (1 – (1 – 2sin2 x))
= (2 sin x cos x) / (1 – 1 + 2sin2 x)]
= [2 sin x cos x / 2 sin2 x]
= cos x/sin x
= cot x
= RHS
Hence proved.
3. sin 2x / (1 + cos 2x) = tan x
Solution:
Let us consider LHS:
sin 2x / (1 + cos 2x)
We know that cos 2x = 1 – 2 sin2 x
= 2 cos2 x – 1
Sin 2x = 2 sin x cos x
So,
sin 2x / (1 + cos 2x) = [2 sin x cos x / (1 + (2cos2x – 1))]
= [2 sin x cos x / (1 + 2cos2 x – 1)]
= [2 sin x cos x / 2 cos2 x]
= sin x/cos x
= tan x
= RHS
Hence proved.

5. [1 – cos 2x + sin 2x] / [1 + cos 2x + sin 2x] = tan x
Solution:
Let us consider LHS:[1 – cos 2x + sin 2x] / [1 + cos 2x + sin 2x]
We know that, cos 2x = 1 – 2 sin2 x
= 2 cos2 x – 1
Sin 2x = 2 sin x cos x
So,

6. [sin x + sin 2x] / [1 + cos x + cos 2x] = tan x
Solution:
Let us consider LHS:[sin x + sin 2x] / [1 + cos x + cos 2x]
We know that, cos 2x = cos2 x – sin2 x
Sin 2x = 2 sin x cos x
So,

= RHS
Hence proved.
7. cos 2x / (1 + sin 2x) = tan (π/4 – x)
Solution:
Let us consider LHS:
cos 2x / (1 + sin 2x)
We know that, cos 2x = cos2 x – sin2 x
Sin 2x = 2 sin x cos x
So,



8. cos x / (1 – sin x) = tan (π/4 + x/2)
Solution:
Let us consider LHS:
cos x / (1 – sin x)
We know that, cos 2x = cos2 x – sin2 x
Cos x = cos2 x/2 – sin2 x/2
Sin 2x = 2 sin x cos x
Sin x = 2 sin x/2 cos x/2
So,





11. (cos α + cos β) 2 + (sin α + sin β) 2 = 4 cos2 (α – β)/2
Solution:
Let us consider LHS:
(cos α + cos β)2 + (sin α + sin β)2
Upon expansion, we get,
(cos α + cos β)2 + (sin α + sin β)2 =
= cos2 α + cos2 β + 2 cos α cos β + sin2 α + sin2 β + 2 sin α sin β
= 2 + 2 cos α cos β + 2 sin α sin β
= 2 (1 + cos α cos β + sin α sin β)
= 2 (1 + cos (α – β)) [since, cos (A – B) = cos A cos B + sin A sin B]
= 2 (1 + 2 cos2 (α – β)/2 – 1) [since, cos2x = 2cos2 x – 1]
= 2 (2 cos2 (α – β)/2)
= 4 cos2 (α – β)/2
= RHS
Hence Proved.
12. sin2 (π/8 + x/2) – sin2 (π/8 – x/2) = 1/√2 sin x
Solution:
Let us consider LHS:
sin2 (π/8 + x/2) – sin2 (π/8 – x/2)
we know, sin2 A – sin2 B = sin (A+B) sin (A-B)
so,
sin2 (π/8 + x/2) – sin2 (π/8 – x/2) = sin (π/8 + x/2 + π/8 – x/2) sin (π/8 + x/2 – (π/8 – x/2))
= sin (π/8 + π/8) sin (π/8 + x/2 – π/8 + x/2)
= sin π/4 sin x
= 1/√2 sin x [since, since π/4 = 1/√2]
= RHS
Hence proved.
13. 1 + cos2 2x = 2 (cos4 x + sin4 x)
Solution:
Let us consider LHS:
1 + cos2 2x
We know, cos2x = cos2 x – sin2 x
cos2 x + sin2 x = 1
so,
1 + cos2 2x = (cos2 x + sin2 x) 2 + (cos2 x – sin2 x) 2
= (cos4 x + sin4 x + 2 cos2 x sin2 x) + (cos4 x + sin4 x – 2 cos2 x sin2 x)
= cos4 x + sin4 x + cos4 x + sin4 x
= 2 cos4 x + 2 sin4 x
= 2 (cos4 x + sin4 x)
= RHS
Hence proved.
14. cos3 2x + 3 cos 2x = 4 (cos6 x – sin6 x)
Solution:
Let us consider RHS:
4 (cos6 x – sin6 x)
Upon expansion we get,
4 (cos6 x – sin6 x) = 4 [(cos2 x)3 – (sin2 x)3]
= 4 (cos2 x – sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)
By using the formula,
a3 – b3 = (a-b) (a2 + b2 + ab)
= 4 cos 2x (cos4 x + sin4 x + cos2 x sin2 x + cos2 x sin2 x – cos2 x sin
We know, cos 2x = cos2 x – sin2 x
So,
= 4 cos 2x (cos4 x + sin4 x + 2 cos2 x sin2 x – cos2 x sin2 x)
= 4 cos 2x [(cos2 x)2 + (sin2 x)2 + 2 cos2 x sin2 x – cos2 x sin2 x]
We know, a2 + b2 + 2ab = (a + b)2
= 4 cos 2x [(1)2 – 1/4 (4 cos2 x sin2 x)]
= 4 cos 2x [(1)2 – 1/4 (2 cos x sin x)2]
We know, sin 2x = 2sin x cos x
= 4 cos 2x [(12) – 1/4 (sin 2x)2]
= 4 cos 2x (1 – 1/4 sin2 2x)
We know, sin2 x = 1 – cos2 x
= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]
= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]
= 4 cos 2x [3/4 + 1/4 cos2 2x]
= 4 (3/4 cos 2x + 1/4 cos3 2x)
= 3 cos 2x + cos3 2x
= cos3 2x + 3 cos 2x
= LHS
Hence proved.
15. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Solution:
Let us consider LHS:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= (sin 3x) (sin x) + sin2 x + (cos 3x) (cos x) – cos2 x
= [(sin 3x) (sin x) + (cos 3x) (cos x)] + (sin2 x – cos2 x)
= [(sin 3x) (sin x) + (cos 3x) (cos x)] – (cos2 x – sin2 x)
= cos (3x – x) – cos 2x
We know, cos 2x = cos2 x – sin2 x
cos A cos B + sin A sin B = cos(A – B)
So,
= cos 2x – cos 2x
= 0
= RHS
Hence Proved.
16. cos2 (π/4 – x) – sin2 (π/4 – x) = sin 2x
Solution:
Let us consider LHS:
cos2 (π/4 – x) – sin2 (π/4 – x)
We know, cos2 A – sin2 A = cos 2A
So,
cos2 (π/4 – x) – sin2 (π/4 – x) = cos 2 (π/4 – x)
= cos (π/2 – 2x)
= sin 2x [since, cos (π/2 – A) = sin A]
= RHS
Hence proved.
17. cos 4x = 1 – 8 cos2 x + 8 cos4 x
Solution:
Let us consider LHS:
cos 4x
We know, cos 2x = 2 cos2 x – 1
So,
cos 4x = 2 cos2 2x – 1
= 2(2 cos2 2x – 1)2 – 1
= 2[(2 cos2 2x) 2 + 12 – 2×2 cos2 x] – 1
= 2(4 cos4 2x + 1 – 4 cos2 x) – 1
= 8 cos4 2x + 2 – 8 cos2 x – 1
= 8 cos4 2x + 1 – 8 cos2 x
= RHS
Hence Proved.
18. sin 4x = 4 sin x cos3 x – 4 cos x sin3 x
Solution:
Let us consider LHS:
sin 4x
We know, sin 2x = 2 sin x cos x
cos 2x = cos2 x – sin2 x
So,
sin 4x = 2 sin 2x cos 2x
= 2 (2 sin x cos x) (cos2 x – sin2 x)
= 4 sin x cos x (cos2 x – sin2 x)
= 4 sin x cos3 x – 4 sin3 x cos x
= RHS
Hence proved.
19. 3(sin x – cos x) 4 + 6 (sin x + cos x) 2 + 4 (sin6 x + cos6 x) = 13
Solution:
Let us consider LHS:
3(sin x – cos x) 4 + 6 (sin x + cos x) 2 + 4 (sin6 x + cos6 x)
We know, (a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab
a3 + b3 = (a + b) (a2 + b2 – ab)
So,
3(sin x – cos x) 4 + 6 (sin x + cos x) 2 + 4 (sin6 x + cos6 x) = 3{(sin x – cos x) 2}2 + 6 {(sin x)2 + (cos x)2 + 2 sin x cos x)} + 4 {(sin2 x)3 + (cos2 x)3}
= 3{(sin x) 2 + (cos x)2 – 2 sin x cos x)}2 + 6 (sin2 x + cos2 x + 2 sin x cos x) + 4{(sin2 x + cos2 x) (sin4 x + cos4 x – sin2 x cos2 x)}
= 3(1 – 2 sin x cos x) 2 + 6 (1 + 2 sin x cos x) + 4{(1) (sin4 x + cos4 x – sin2 x cos2 x)}
We know, sin2 x + cos2 x = 1
So,
= 3{12 + (2 sin x cos x) 2 – 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2 x)2 + (cos2 x)2 + 2 sin2 x cos2 x – 3 sin2 x cos2 x)}
= 3{1 + 4 sin2 x cos2 x – 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2 x + cos2 x) 2 – 3 sin2 x cos2 x)}
= 3 + 12 sin2 x cos2 x – 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)2 – 3 sin2 x cos2 x)}
= 9 + 12 sin2 x cos2 x + 4(1 – 3 sin2 x cos2 x)
= 9 + 12 sin2 x cos2 x + 4 – 12 sin2 x cos2 x
= 13
= RHS
Hence proved.
20. 2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1 = 0
Solution:
Let us consider LHS:
2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1
We know, (a + b)2 = a2 + b2 + 2ab
a3 + b3 = (a + b) (a2 + b2 – ab)
So,
2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1 = 2{(sin2 x) 3 + (cos2 x) 3} – 3{(sin2 x) 2 + (cos2 x) 2} + 1
= 2{(sin2 x + cos2 x) (sin4 x + cos4 x – sin2 x cos2 x} – 3{(sin2 x) 2 + (cos2 x) 2 + 2sin2 x cos2 x – 2sin2 x cos2 x} + 1
= 2{(1) (sin4 x + cos4 x + 2 sin2 x cos2 x – 3 sin2 x cos2 x} – 3{(sin2 x + cos2 x) 2 – 2sin2 x cos2 x} + 1
We know, sin2 x + cos2 x = 1
= 2{(sin2 x + cos2 x) 2 – 3 sin2 x cos2 x} – 3{(1)2 – 2sin2 x cos2 x} + 1
= 2{(1)2 – 3 sin2 x cos2 x} – 3(1 – 2sin2 x cos2 x) + 1
= 2(1 – 3 sin2 x cos2 x) – 3 + 6 sin2 x cos2 x + 1
= 2 – 6 sin2 x cos2 x – 2 + 6 sin2 x cos2 x
= 0
= RHS
Hence proved.
21. cos6 x – sin6 x = cos 2x (1 – 1/4 sin2 2x)
Solution:
Let us consider LHS:
cos6 x – sin6 x
We know, (a + b) 2 = a2 + b2 + 2ab
a3 – b3 = (a – b) (a2 + b2 + ab)
So,
cos6 x – sin6 x = (cos2 x)3 – (sin2 x)3
= (cos2 x – sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)
We know, cos 2x = cos2 x – sin2 x
So,
= cos 2x [(cos2 x) 2 + (sin2 x) 2 + 2 cos2 x sin2 x – cos2 x sin2 x]
= cos 2x [(cos2 x) 2 + (sin2 x) 2 – 1/4 × 4 cos2 x sin2 x]
We know, sin2 x + cos2 x = 1
So,
= cos 2x [(1)2 – 1/4 × (2 cos x sin x) 2]
We know, sin 2x = 2 sin x cos x
So,
= cos 2x [1 – 1/4 × (sin 2x) 2]
= cos 2x [1 – 1/4 × sin2 2x]
= RHS
Hence proved.
22. tan (π/4 + x) + tan (π/4 – x) = 2 sec 2x
Solution:
Let us consider LHS:
tan (π/4 + x) + tan (π/4 – x)
We know,
tan (A+B) = (tan A + tan B)/(1- tan A tan B)
tan (A-B) = (tan A – tan B)/(1+ tan A tan B)
So,



EXERCISE 9.2 PAGE NO: 9.36
Prove that:
1. sin 5x = 5 sin x – 20 sin3 x + 16 sin5 x
Solution:
Let us consider LHS:
sin 5x
Now,
sin 5x = sin (3x + 2x)
But we know,
Sin (x + y) = sin x cos y + cos x sin y…..(i)
So,
sin 5x = sin 3x cos 2x + cos 3x sin 2x
= sin (2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)
And
cos (x + y) = cos x cos y – sin x sin y……(iii)
Now substituting equation (i) and (iii) in equation (ii), we get
sin 5x = (sin 2x cos x + cos 2x sin x ) cos 2x + ( cos 2x cos x – sin 2x sin x) sin 2x
= sin 2x cos 2x cos x + cos2 2x sin x + (sin 2x cos 2x cos x – sin2 2x sin x)
= 2sin 2x cos 2x cos x + cos2 2x sin x – sin2 2x sin x …….(iv)
Now sin 2x = 2sin x cos x………(v)
And cos 2x = cos2x – sin2x………(vi)
Substituting equation (v) and (vi) in equation (iv), we get
sin 5x = 2(2sin x cos x) (cos2x –sin2x) cos x + (cos2x – sin2x)2 sin x – (2sin x cos x)2 sin x
= 4(sin x cos2 x) ([1– sin2x] – sin2x) + ([1–sin2x] – sin2x)2 sin x – (4sin2 x cos2 x)sin x
(as cos2x + sin2x = 1 ⇒ cos2x = 1– sin2x)
sin 5x = 4(sin x [1 – sin2x]) (1 – 2sin2x) + (1 – 2sin2x)2 sin x – 4sin3 x [1 – sin2x]
= 4sin x (1 – sin2x) (1 – 2sin2 x) + (1 – 4sin2x + 4sin4x) sin x – 4sin3 x + 4sin5x
= (4sin x – 4sin3x) (1 – 2sin2x) + sin x – 4sin3x + 4sin5x – 4sin3 x + 4sin5x
= 4sin x – 8sin3x – 4sin3x + 8sin5x + sin x – 8sin3x + 8sin5x
= 5sin x – 20sin3x + 16sin5x
= RHS
Hence proved.
2. 4 (cos3 10o + sin3 20o) = 3 (cos 10o + sin 20o)
Solution:
Let us consider LHS:
4 (cos3 10o + sin3 20o)
We know that, sin 60o = √3/2 = cos 30o
Sin 30o = cos 60o = 1/2
So,
Sin (3×20o) = cos (3×10o)
3sin 20°– 4sin320° = 4cos310° – 3cos 10°
(we know, sin 3θ = 3sin θ – 4sin3 θ and cos 3θ = 4cos3θ – 3cosθ)
So,
4(cos310°+sin320°) = 3(sin 20°+cos 10°)
= RHS
Hence proved.
3. cos3 x sin 3x + sin3 x cos 3x = 3/4 sin 4x
Solution:
We know that,
cos 3θ = 4cos3θ – 3cosθ
So, 4 cos3θ = cos3θ + 3cosθ
cos3 θ = [cos3θ + 3cosθ]/4 …… (i)
Similarly,
sin 3θ = 3sin θ – 4sin3 θ
4 sin3θ = 3sinθ – sin 3θ
sin3θ = [3sinθ – sin 3θ]/4 …….. (ii)
Now,
Let us consider LHS:
cos3 x sin 3x + sin3 x cos 3x
Substituting the values from equation (i) and (ii), we get
cos3 x sin 3x + sin3 x cos 3x = (cos 3x + 3 cos x)/4 sin 3x + (3sin x – sin 3x)/4 cos 3x
= 1/4 (sin 3x cos 3x + 3 sin 3x cox x + 3sin x cos 3x – sin 3x cos 3x)
= 1/4 (3(sin 3x cos x + sin x cos 3x) + 0)
= 1/4 (3 sin (3x + x))
(We know, sin(x + y) = sin x cos y + cos x sin y)
= 3/4 sin 4x
= RHS
Hence proved.
4. sin 5x = 5 cos4 x sin x – 10 cos2 x sin3 x + sin5 x
Solution:
Let us consider LHS:
sin 5x
Now,
sin 5x = sin (3x + 2x)
But we know,
Sin (x + y) = sin x cos y + cos x sin y…..(i)
So,
sin 5x = sin 3x cos 2x + cos 3x sin 2x
= sin (2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)
And
cos (x + y) = cos x cos y – sin x sin y……(iii)
Now substituting equation (i) and (iii) in equation (ii), we get
sin 5x = (sin 2x cos x + cos 2x sin x ) cos 2x + ( cos 2x cos x – sin 2x sin x) sin 2x … (iv)
Now sin 2x = 2sin x cos x………(v)
And cos 2x = cos2x – sin2x………(vi)
Substituting equation (v) and (vi) in equation (iv), we get
sin 5x = [(2 sin x cos x) cos x + (cos2x – sin2x) sin x] (cos2x – sin2x) + [(cos2x – sin2x) cos x – (2 sin x cos x) sin x)] (2 sin x cos x)
= [2 sin x cos2 x + sin x cos2x – sin3x] (cos2x – sin2x) + [cos3x – sin2x cos x – 2 sin2 x cos x] (2 sin x cos x)
= cos2x [3 sin x cos2 x – sin3x] – sin2x [3 sin x cos2 x – sin3x] + 2 sin x cos4x – 2 sin3 x cos2 x – 4 sin3 x cos2 x
= 3 sin x cos4 x – sin3x cos2x – 3 sin3 x cos2 x – sin5x + 2 sin x cos4x – 2 sin3 x cos2 x – 4 sin3 x cos2 x
= 5 sin x cos4 x –10sin3xcos2x +sin5x
= RHS
Hence proved.
5. sin 5x = 5 sin x – 20 sin3 x + 16 sin5 x
Solution:
Let us consider LHS:
sin 5x
Now,
sin 5x = sin (3x + 2x)
But we know,
Sin (x + y) = sin x cos y + cos x sin y…..(i)
So,
sin 5x = sin 3x cos 2x + cos 3x sin 2x
= sin (2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)
And
cos (x + y) = cos x cos y – sin x sin y……(iii)
Now substituting equation (i) and (iii) in equation (ii), we get
sin 5x = (sin 2x cos x + cos 2x sin x ) cos 2x + ( cos 2x cos x – sin 2x sin x) sin 2x
= sin 2x cos 2x cos x + cos2 2x sin x + (sin 2x cos 2x cos x – sin2 2x sin x)
= 2sin 2x cos 2x cos x + cos2 2x sin x – sin2 2x sin x …….(iv)
Now sin 2x = 2sin x cos x………(v)
And cos 2x = cos2x – sin2x………(vi)
Substituting equation (v) and (vi) in equation (iv), we get
sin 5x = 2(2sin x cos x) (cos2x –sin2x) cos x + (cos2x – sin2x)2 sin x – (2sin x cos x)2 sin x
= 4(sin x cos2 x) ([1– sin2x] – sin2x) + ([1–sin2x] – sin2x)2 sin x – (4sin2 x cos2 x)sin x
(as cos2x + sin2x = 1 ⇒ cos2x = 1– sin2x)
sin 5x = 4(sin x [1 – sin2x]) (1 – 2sin2x) + (1 – 2sin2x)2 sin x – 4sin3 x [1 – sin2x]
= 4sin x (1 – sin2x) (1 – 2sin2 x) + (1 – 4sin2x + 4sin4x) sin x – 4sin3 x + 4sin5x
= (4sin x – 4sin3x) (1 – 2sin2x) + sin x – 4sin3x + 4sin5x – 4sin3 x + 4sin5x
= 4sin x – 8sin3x – 4sin3x + 8sin5x + sin x – 8sin3x + 8sin5x
= 5sin x – 20sin3x + 16sin5x
= RHS
Hence proved.


EXERCISE 9.3 PAGE NO: 9.42
Prove that:
1. sin2 2π/5 – sin2 π/3 = (√5 – 1)/8
Solution:
Let us consider LHS:
sin2 2π/5 – sin2 π/3 = sin2 (π/2 – π/10) – sin2 π/3
we know, sin (90°– A) = cos A
So, sin2 (π/2 – π/10) = cos2 π/10
Sin π/3 = √3/2
Then the above equation becomes,
= Cos2 π/10 – (√3/2)2
We know, cos π/10 = √(10+2√5)/4
the above equation becomes,
= [√(10+2√5)/4]2 – 3/4
= [10 + 2√5]/16 – 3/4
= [10 + 2√5 – 12]/16
= [2√5 – 2]/16
= [√5 – 1]/8
= RHS
Hence proved.
2. sin2 24o – sin2 6o = (√5 – 1)/8
Solution:
Let us consider LHS:
sin2 24o – sin2 6o
we know, sin (A + B) sin (A – B) = sin2A – sin2B
Then the above equation becomes,
sin2 24o – sin2 6o = sin (24o + 6o) – sin (24o – 6o)
= sin 30o – sin 18o
= sin 30o – (√5 – 1)/4 [since, sin 18o = (√5 – 1)/4]
= 1/2 × (√5 – 1)/4
= (√5 – 1)/8
= RHS
Hence proved.
3. sin2 42o – cos2 78o = (√5 + 1)/8
Solution:
Let us consider LHS:
sin2 42o – cos2 78o = sin2 (90o – 48o) – cos2 (90o – 12o)
= cos2 48o – sin2 12o [since, sin (90 – A) = cos A and cos (90 – A) = sin A]
We know, cos (A + B) cos (A – B) = cos2A – sin2B
Then the above equation becomes,
= cos2 (48o + 12o) cos (48o – 12o)
= cos 60o cos 36o [since, cos 36o = (√5 + 1)/4]
= 1/2 × (√5 + 1)/4
= (√5 + 1)/8
= RHS
Hence proved.
4. cos 78o cos 42o cos 36o = 1/8
Solution:
Let us consider LHS:
cos 78o cos 42o cos 36o
Let us multiply and divide by 2 we get,
cos 78o cos 42o cos 36o = 1/2 (2 cos 78o cos 42o cos 36o)
We know, 2 cos A cos B = cos (A + B) + cos (A – B)
Then the above equation becomes,
= 1/2 (cos (78o + 42o) + cos (78o – 42o)) × cos 36o
= 1/2 (cos 120o + cos 36o) × cos 36o
= 1/2 (cos (180o – 60o) + cos 36o) × cos 36o
= 1/2 (-cos (60o) + cos 36o) × cos 36o [since, cos(180° – A) = – A]
= 1/2 (-1/2 + (√5 + 1)/4) ((√5 + 1)/4) [since, cos 36o = (√5 + 1)/4]
= 1/2 (√5 + 1 – 2)/4 ((√5 + 1)/4)
= 1/2 (√5 – 1)/4) ((√5 + 1)/4)
= 1/2 ((√5)2 – 12)/16
= 1/2 (5-1)/16
= 1/2 (4/16)
= 1/8
= RHS
Hence proved.
5. cos π/15 cos 2π/15 cos 4π/15 cos 7π/15 = 1/16
Solution:
Let us consider LHS:
cos π/15 cos 2π/15 cos 4π/15 cos 7π/15
Let us multiply and divide by 2 sin π/15, we get,
= [2 sin π/15 cos π/15] cos 2π/15 cos 4π/15 cos 7π/15] / 2 sin π/15
We know, 2sin A cos A = sin 2A
Then the above equation becomes,
= [(sin 2π/15) cos 2π/15 cos 4π/15 cos 7π/15] / 2 sin π/15
Now, multiply and divide by 2 we get,
= [(2 sin 2π/15 cos 2π/15) cos 4π/15 cos 7π/15] / 2 × 2 sin π/15
We know, 2sin A cos A = sin 2A
Then the above equation becomes,
= [(sin 4π/15) cos 4π/15 cos 7π/15] / 4 sin π/15
Now, multiply and divide by 2 we get,
= [(2 sin 4π/15 cos 4π/15) cos 7π/15] / 2 × 4 sin π/15
We know, 2sin A cos A = sin 2A
Then the above equation becomes,
= [(sin 8π/15) cos 7π/15] / 8 sin π/15
Now, multiply and divide by 2 we get,
= [2 sin 8π/15 cos 7π/15] / 2 × 8 sin π/15
We know, 2sin A cos B = sin (A+B) + sin (A–B)
Then the above equation becomes,
= [sin (8π/15 + 7π/15) + sin (8π/15 – 7π/15)] / 16 sin π/15
= [sin (π) + sin (π/15)] / 16 sin π/15
= [0 + sin (π/15)] / 16 sin π/15
= sin (π/15) / 16 sin π/15
= 1/16
= RHS
Hence proved.
RD Sharma Solutions for Class 11 Maths Chapter 9: Download PDF
RD Sharma Solutions for Class 11 Maths Chapter 9–Values of Trigonometric Functions at Multiples and Submultiples of an Angle
Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 1–Sets PDF
Chapterwise RD Sharma Solutions for Class 11 Maths :
- Chapter 1–Sets
- Chapter 2–Relations
- Chapter 3–Functions
- Chapter 4–Measurement of Angles
- Chapter 5–Trigonometric Functions
- Chapter 6–Graphs of Trigonometric Functions
- Chapter 7–Values of Trigonometric Functions at Sum or Difference of Angles
- Chapter 8–Transformation Formulae
- Chapter 9–Values of Trigonometric Functions at Multiples and Submultiples of an Angle
- Chapter 10–Sine and Cosine Formulae and their Applications
- Chapter 11–Trigonometric Equations
- Chapter 12–Mathematical Induction
- Chapter 13–Complex Numbers
- Chapter 14–Quadratic Equations
- Chapter 15–Linear Inequations
- Chapter 16–Permutations
- Chapter 17–Combinations
- Chapter 18–Binomial Theorem
- Chapter 19–Arithmetic Progressions
- Chapter 20–Geometric Progressions
- Chapter 21–Some Special Series
- Chapter 22–Brief review of Cartesian System of Rectangular Coordinates
- Chapter 23–The Straight Lines
- Chapter 24–The Circle
- Chapter 25–Parabola
- Chapter 26–Ellipse
- Chapter 27–Hyperbola
- Chapter 28–Introduction to Three Dimensional Coordinate Geometry
- Chapter 29–Limits
- Chapter 30–Derivatives
- Chapter 31–Mathematical Reasoning
- Chapter 32–Statistics
- Chapter 33–Probability
About RD Sharma
RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — 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’s only recently that a spoof video turned the tutor into a YouTube star.
R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” 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’s Guru Nanak Dev Institute of Technology.