Class 11: Maths Chapter 6 solutions. Complete Class 11 Maths Chapter 6 Notes.
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RD Sharma Solutions for Class 11 Maths Chapter 6–Graphs of Trigonometric Functions
RD Sharma 11th Maths Chapter 6, Class 11 Maths Chapter 6 solutions
EXERCISE 6.1 PAGE NO: 6.5
1. Sketch the graphs of the following functions:
(i) f (x) = 2 sin x, 0 ≤ x ≤ π
(ii) g (x) = 3 sin (x – π/4), 0 ≤ x ≤ 5π/4
(iii) h (x) = 2 sin 3x, 0 ≤ x ≤ 2π/3
(iv) ϕ (x) = 2 sin (2x – π/3), 0 ≤ x ≤ 7π/3
(v) Ψ (x) = 4 sin 3 (x – π/4), 0 ≤ x ≤ 2π
(vi) θ (x) = sin (x/2 – π/4), 0 ≤ x ≤ 4π
(vii) u (x) = sin2 x, 0 ≤ x ≤ 2π υ (x) = |sin x|, 0 ≤ x ≤ 2π
(viii) f (x) = 2 sin πx, 0 ≤ x ≤ 2
Solution:
(i) f (x) = 2 sin x, 0 ≤ x ≤ π
We know that g (x) = sin x is a periodic function with period π.
So, f (x) = 2 sin x is a periodic function with period π. So, we will draw the graph of f (x) = 2 sin x in the interval [0, π]. The values of f (x) = 2 sin x at various points in [0, π] are listed in the following table:
x | 0(A) | π/6 (B) | π/3 (C) | π/2 (D) | 2π/3 (E) | 5π/6 (F) | Π (G) |
f (x) = 2 sin x | 0 | 1 | √3 = 1.73 | 2 | √3 = 1.73 | 1 | 0 |
The required curve is:

(ii) g (x) = 3 sin (x – π/4), 0 ≤ x ≤ 5π/4
We know that if f (x) is a periodic function with period T, then f (ax + b) is periodic with period T/|a|.
So, g (x) = 3 sin (x – π/4) is a periodic function with period π. So, we will draw the graph of g (x) = 3 sin (x – π/4) in the interval [0, 5π/4]. The values of g (x) = 3 sin (x – π/4) at various points in [0, 5π/4] are listed in the following table:
x | 0(A) | π/4 (B) | π/2 (C) | 3π/4 (D) | π (E) | 5π/4 (F) |
g (x) = 3 sin (x – π/4) | -3/√2 = -2.1 | 0 | 3/√2 = 2.12 | 3 | 3/√2 = 2.12 | 0 |
The required curve is:

(iii) h (x) = 2 sin 3x, 0 ≤ x ≤ 2π/3
We know that g (x) = sin x is a periodic function with period 2π.
So, h (x) = 2 sin 3x is a periodic function with period 2π/3. So, we will draw the graph of h (x) = 2 sin 3x in the interval [0, 2π/3]. The values of h (x) = 2 sin 3x at various points in [0, 2π/3] are listed in the following table:
x | 0 (A) | π/6 (B) | π/3 (C) | π/2 (D) | 2π/3 (E) |
h (x) = 2 sin 3x | 0 | 2 | 0 | -2 | 0 |
The required curve is:

(iv) ϕ (x) = 2 sin (2x – π/3), 0 ≤ x ≤ 7π/3
We know that if f(x) is a periodic function with period T, then f (ax + b) is periodic with period T/|a|.
So, ϕ (x) = 2 sin (2x – π/3) is a periodic function with period π. So, we will draw the graph of ϕ (x) = 2 sin (2x – π/3), in the interval [0, 7π/5]. The values of ϕ (x) = 2 sin (2x – π/3), at various points in [0, 7π/5] are listed in the following table:
x | 0 (A) | π/6 (B) | 2π/3 (C) | 7π/6 (D) | 7π/5 (E) |
ϕ (x) = 2 sin (2x – π/3) | -√3 = -1.73 | 0 | 0 | 0 | 1.98 |
The required curve is:

(v) Ψ (x) = 4 sin 3 (x – π/4), 0 ≤ x ≤ 2π
We know that if f(x) is a periodic function with period T, then f (ax + b) is periodic with period T/|a|.
So, Ψ (x) = 4 sin 3 (x – π/4) is a periodic function with period 2π. So, we will draw the graph of Ψ (x) = 4 sin 3 (x – π/4) in the interval [0, 2π]. The values of Ψ (x) = 4 sin 3 (x – π/4) at various points in [0, 2π] are listed in the following table:
x | 0 (A) | π/4 (B) | π/2 (C) | π (D) | 5π/4 (E) | 2π (F) |
Ψ (x) = 4 sin 3 (x – π/4) | -2√2 = -2.82 | 0 | 2√2 = 2.82 | 0 | 1.98 | -2√2 = -2.82 |
The required curve is:

(vi) θ (x) = sin (x/2 – π/4), 0 ≤ x ≤ 4π
We know that if f(x) is a periodic function with period T, then f (ax + b) is periodic with period T/|a|.
So, θ (x) = sin (x/2 – π/4) is a periodic function with period 4π. So, we will draw the graph of θ (x) = sin (x/2 – π/4) in the interval [0, 4π]. The values of θ (x) = sin (x/2 – π/4) at various points in [0, 4π] are listed in the following table:
x | 0 (A) | π/2 (B) | π (C) | 2π (D) | 5π/2 (E) | 3π (F) | 4π (G) |
θ (x) = sin (x/2 – π/4) | -1/√2 = -0.7 | 0 | 1/√2 = 0.7 | 1/√2 = 0.7 | 0 | -1/√2 = -0.7 | -1/√2 = -0.7 |
The required curve is:

(vii) u (x) = sin2 x, 0 ≤ x ≤ 2π υ (x) = |sin x|, 0 ≤ x ≤ 2π
We know that g (x) = sin x is a periodic function with period π.
So, u (x) = sin2 x is a periodic function with period 2π. So, we will draw the graph of u (x) = sin2 x in the interval [0, 2π]. The values of u (x) = sin2 x at various points in [0, 2π] are listed in the following table:
x | 0 (A) | π/2 (B) | Π (C) | 3π/2 (D) | 2π (E) |
u (x) = sin2 x | 0 | 1 | 0 | 1 | 0 |
The required curve is:

(viii) f (x) = 2 sin πx, 0 ≤ x ≤ 2
We know that g (x) = sin x is a periodic function with period 2π.
So, f (x) = 2 sin πx is a periodic function with period 2. So, we will draw the graph of f (x) = 2 sin πx in the interval [0, 2]. The values of f (x) = 2 sin πx at various points in [0, 2] are listed in the following table:
x | 0 (A) | 1/2 (B) | 1 (C) | 3/2 (D) | 2 (E) |
f (x) = 2 sin πx | 0 | 2 | 0 | -2 | 0 |
The required curve is:

2. Sketch the graphs of the following pairs of functions on the same axes:
(i) f (x) = sin x, g (x) = sin (x + π/4)
(ii) f (x) = sin x, g (x) = sin 2x
(iii) f (x) = sin 2x, g (x) = 2 sin x
(iv) f (x) = sin x/2, g (x) = sin x
Solution:
(i) f (x) = sin x, g (x) = sin (x + π/4)
We know that the functions f (x) = sin x and g (x) = sin (x + π/4) are periodic functions with periods 2π and 7π/4.
The values of these functions are tabulated below:
Values of f (x) = sin x in [0, 2π]
x | 0 | π/2 | π | 3π/2 | 2π |
f (x) = sin x | 0 | 1 | 0 | -1 | 0 |
Values of g (x) = sin (x + π/4) in [0, 7π/4]
x | 0 | π/4 | 3π/4 | 5π/4 | 7π/4 |
g (x) = sin (x + π/4) | 1/√2 = 0.7 | 1 | 0 | -1 | 0 |
The required curve is:
(ii) f (x) = sin x, g (x) = sin 2x
We know that the functions f(x) = sin x and g (x) = sin 2x are periodic functions with periods 2π and π.
The values of these functions are tabulated below:
Values of f (x) = sin x in [0, 2π]
x | 0 | π/2 | π | 3π/2 | 2π |
f (x) = sin x | 0 | 1 | 0 | -1 | 0 |
Values of g (x) = sin (2x) in [0, π]
x | 0 | π/4 | π/2 | 3π/4 | π | 5π/4 | 3π/2 | 7π/4 | 2π |
g (x) = sin (2x) | 0 | 1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 |
The required curve is:

(iii) f (x) = sin 2x, g (x) = 2 sin x
We know that the functions f(x) = sin 2x and g (x) = 2 sin x are periodic functions with periods π and π.
The values of these functions are tabulated below:
Values of f (x) = sin (2x) in [0, π]
x | 0 | π/4 | π/2 | 3π/4 | π | 5π/4 | 3π/2 | 7π/4 | 2π |
f (x) = sin (2x) | 0 | 1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 |
Values of g (x) = 2 sin x in [0, π]
x | 0 | π/2 | π | 3π/2 | 2π |
g (x) = 2 sin x | 0 | 1 | 0 | -1 | 0 |
The required curve is:

(iv) f (x) = sin x/2, g (x) = sin x
We know that the functions f(x) = sin x/2 and g (x) = sin x are periodic functions with periods π and 2π.
The values of these functions are tabulated below:
Values of f (x) = sin x/2 in [0, π]
x | 0 | π | 2π | 3π | 4π |
f (x) = sin x/2 | 0 | 1 | 0 | -1 | 0 |
Values of g (x) = sin (x) in [0, 2π]
x | 0 | π/2 | π | 3π/2 | 2π | 5π/2 | 3π | 7π/2 | 4π |
g (x) = sin (x) | 0 | 1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 |
The required curve is:

EXERCISE 6.2 PAGE NO: 6.8
1. Sketch the graphs of the following trigonometric functions:
(i) f (x) = cos (x – π/4)
(ii) g (x) = cos (x + π/4)
(iii) h (x) = cos2 2x
(iv) ϕ (x) = 2 cos (x – π/6)
(v) ψ (x) = cos (3x)
(vi) u (x) = cos2 x/2
(vii) f (x) = cos πx
(viii) g (x) = cos 2π x
Solution:
(i) f (x) = cos (x – π/4)
We know that g (x) = cos x is a periodic function with period 2π.
So, f (x) = cos (x – π/4) is a periodic function with period π. So, we will draw the graph of f (x) = cos (x – π/4) in the interval [0, π]. The values of f (x) = cos (x – π/4) at various points in [0, π] are listed in the following table:
x | 0 (A) | π/4 (B) | π/2 (C) | 3π/4 (D) | π (E) | 5π/4 (F) | 3π/2 (G) | 7π/4 (H) |
f (x) = cos (x – π/4) | 1/√2 = 0.7 | 1 | 1/√2 = 0.7 | 0 | -1/√2 = -0.7 | -1 | -1/√2 = -0.7 | 0 |
The required curve is:

(ii) g (x) = cos (x + π/4)
We know that f (x) = cos x is a periodic function with period 2π.
So, g (x) = cos (x + π/4) is a periodic function with period π. So, we will draw the graph of g (x) = cos (x + π/4) in the interval [0, π]. The values of g (x) = cos (x + π/4) at various points in [0, π] are listed in the following table:
x | 0 (A) | π/4 (B) | π/2 (C) | 3π/4 (D) | π (E) | 5π/4 (F) | 3π/2 (G) | 7π/4 (H) |
g (x) = cos (x + π/4) | 1/√2 = 0.7 | 0 | -1/√2 = -0.7 | -1 | -1/√2 = -0.7 | 0 | 1/√2 = 0.7 | 1 |
The required curve is:

(iii) h (x) = cos2 2x
We know that f (x) = cos x is a periodic function with period 2π.
So, h (x) = cos2 2x is a periodic function with period π. So, we will draw the graph of h (x) = cos2 2x in the interval [0, π]. The values of h (x) = cos2 2x at various points in [0, π] are listed in the following table:
x | 0 (A) | π/4 (B) | π/2 (C) | 3π/4 (D) | π (E) | 5π/4 (F) | 3π/2 (G) |
h (x) = cos2 2x | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
The required curve is:

(iv) ϕ (x) = 2 cos (x – π/6)
We know that f (x) = cos x is a periodic function with period 2π.
So, ϕ (x) = 2cos (x – π/6) is a periodic function with period π. So, we will draw the graph of ϕ (x) = 2cos (x – π/6) in the interval [0, π]. The values of ϕ (x) = 2cos (x – π/6) at various points in [0, π] are listed in the following table:
x | 0 (A) | π/3 (B) | 2π/3 (C) | π (D) | 4π/3 (E) | 5π/3 (F) |
ϕ (x) = 2 cos (x – π/6) | √3 = 1.73 | √3 = 1.73 | 0 | -√3 = -1.73 | -√3 = -1.73 | 0 |
The required curve is:

(v) ψ (x) = cos (3x)
We know that f (x) = cos x is a periodic function with period 2π.
So, ψ (x) = cos (3x) is a periodic function with period 2π/3. So, we will draw the graph of ψ (x) = cos (3x) in the interval [0, 2π/3]. The values of ψ (x) = cos (3x) at various points in [0, 2π/3] are listed in the following table:
x | 0 (A) | π/6 (B) | π/3 (C) | π/2 (D) | 2π/3 (E) | 5π/6 (F) |
ψ (x) = cos (3x) | 1 | 0 | -1 | 0 | 1 | 0 |
The required curve is:

(vi) u (x) = cos2 x/2
We know that f (x) = cos x is a periodic function with period 2π.
So, u (x) = cos2 (x/2) is a periodic function with period π. So, we will draw the graph of u (x) = cos2 (x/2) in the interval [0, π]. The values of u (x) = cos2 (x/2) at various points in [0, π] are listed in the following table:
x | 0 (A) | π (B) | 2π (C) | 3π (D) |
u (x) = cos2 x/2 | 1 | 0 | 1 | 0 |
The required curve is:

(vii) f (x) = cos πx
We know that g (x) = cos x is a periodic function with period 2π.
So, f (x) = cos (πx) is a periodic function with period 2. So, we will draw the graph of f (x) = cos (πx) in the interval [0, 2]. The values of f (x) = cos (πx) at various points in [0, 2] are listed in the following table:
x | 0 (A) | 1/2 (B) | 1 (C) | 3/2 (D) | 2 (E) | 5/2 (F) |
f (x) = cos πx | 1 | 0 | -1 | 0 | 1 | 0 |
The required curve is:

(viii) g (x) = cos 2π x
We know that f (x) = cos x is a periodic function with period 2π.
So, g (x) = cos (2πx) is a periodic function with period 1. So, we will draw the graph of g (x) = cos (2πx) in the interval [0, 1]. The values of g (x) = cos (2πx) at various points in [0, 1] are listed in the following table:
x | 0 (A) | 1/4 (B) | 1/2 (C) | 3/4 (D) | 1 (E) | 5/4 (F) | 3/2 (G) | 7/4 (H) | 2 |
g (x) = cos 2π x | 1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 | 1 |
The required curve is:

2. Sketch the graphs of the following curves on the same scale and the same axes:
(i) y = cos x and y = cos (x – π/4)
(ii) y = cos 2x and y = cos (x – π/4)
(iii) y = cos x and y = cos x/2
(iv) y = cos2 x and y = cos x
Solution:
(i) y = cos x and y = cos (x – π/4)
We know that the functions y = cos x and y = cos (x – π/4) are periodic functions with periods π and π.
The values of these functions are tabulated below:
x | 0 | π/4 | π/2 | 3π/4 | π | 5π/4 | 3π/2 | 7π/4 |
y = cos x | 1 | 1/√2 = 0.7 | 0 | -1/√2 = -0.7 | -1 | -1/√2 = -0.7 | 0 | 1 |
y = cos (x – π/4) | 1/√2 = 0.7 | 1 | 1/√2 = 0.7 | 0 | -1/√2 = -0.7 | -1 | -1/√2 = -0.7 | 0 |
The required curve is:

(ii) y = cos 2x and y = cos 2(x – π/4)
We know that the functions y = cos 2x and y = cos 2(x – π/4) are periodic functions with periods π and π.
The values of these functions are tabulated below:
x | 0 | π/4 | π/2 | 3π/4 | π | 5π/4 | 3π/2 | 7π/4 |
y = cos x | 1 | 0 | -1 | 0 | 1 | 0 | -1 | 0 |
y = cos 2 (x – π/4) | 0 | 1 | 0 | -1 | 0 | 1 | 0 | -1 |
The required curve is:

(iii) y = cos x and y = cos x/2
We know that the functions y = cos x and y = cos (x/2) are periodic functions with periods π and π.
The values of these functions are tabulated below:
x | 0 | π/2 | π | 3π/2 | 2π |
y = cos x | 1 | 0 | -1 | 0 | 1 |
y = cos x/2 | 1 | 1/√2 = 0.7 | 0 | -1/√2 = -0.7 | -1 |
The required curve is:

(iv) y = cos2 x and y = cos x
We know that the functions y = cos2 x and y = cos x are periodic functions with period 2π.
The values of these functions are tabulated below:
x | 0 | π/2 | π | 3π/2 | 2π |
y = cos2 x | 1 | 0 | 1 | 0 | 1 |
y = cos x | 1 | 0 | -1 | 0 | 1 |
The required curve is:

EXERCISE 6.3 PAGE NO: 6.13
Sketch the graphs of the following functions:
1. f (x) = 2 cosec πx
Solution:
We know that f (x) = cosec x is a periodic function with period 2π.
So, f (x) = 2 cosec (πx) is a periodic function with period 2. So, we will draw the graph of f (x) = 2 cosec (πx) in the interval [0, 2]. The values of f (x) = 2 cosec (πx) at various points in [0, 2] are listed in the following table:
x | 0 (A) | 1/2 (B) | 1 (C) | -1 (D) | 3/2 (E) | -2 (F) | 2 (G) | 5/2 (H) |
f (x) = 2 cosec (πx) | ∞ | 2 | ∞ | -∞ | -2 | -∞ | ∞ | 2 |
The required curve is:
2. f (x) = 3 sec x
Solution:
We know that f (x) = sec x is a periodic function with period π.
So, f (x) = 3 sec (x) is a periodic function with period π. So, we will draw the graph of f (x) = 3 sec (x) in the interval [0, π]. The values of f (x) = 3 sec (x) at various points in [0, π] are listed in the following table:
x | 0 (A) | π/2 (B) | -π/2 (C) | π (D) | -3π/2 (E) | 3π/2 (F) | 2π (G) | 5π/2 (H) |
f (x) = sec x | 3 | ∞ | -∞ | -3 | -∞ | ∞ | 3 | ∞ |
The required curve is:

3. f (x) = cot 2x
Solution:
We know that f (x) = cot x is a periodic function with period π.
So, f (x) = cot (2x) is a periodic function with period π. So, we will draw the graph of f (x) = cot (2x) in the interval [0, π]. The values of f (x) = cot (2x) at various points in [0, π] are listed in the following table:
x | 0 (A) | π/4 (B) | -π/2 (C) | π/2 (D) | 3π/4 (E) | -π (F) |
f (x) = cot x | →∞ | 0 | -∞ | →∞ | 0 | -∞ |
The required curve is:

4. f (x) = 2 sec πx
Solution:
We know that f (x) = sec x is a periodic function with period π.
So, f (x) = 2 sec (πx) is a periodic function with period 1. So, we will draw the graph of f (x) = 2 sec (πx) in the interval [0, 1]. The values of f (x) = 2 sec (πx) at various points in [0, 1] are listed in the following table:
x | 0 | 1/2 | -1/2 | 1 | -3/2 | 3/2 | 2 |
f (x) = 2 sec (πx) | 2 | ∞ | →-∞ | -2 | -∞ | ∞ | 2 |
The required curve is:

5. f (x) = tan2 x
Solution:
We know that f (x) = tan x is a periodic function with period π.
So, f (x) = tan2 (x) is a periodic function with period π. So, we will draw the graph of f (x) = tan2 (x) in the interval [0, π]. The values of f (x) = tan2 (x) at various points in [0, π] are listed in the following table:
x | 0 (A) | π/2 (B) | π/2 (C) | π (D) | 3π/2 (E) | 3π/2 (F) | 2 π |
f (x) = tan2 (x) | 0 | ∞ | →∞ | 0 | ∞ | →∞ | 0 |
The required curve is:

RD Sharma Solutions for Class 11 Maths Chapter 6: Download PDF
RD Sharma Solutions for Class 11 Maths Chapter 6–Graphs of Trigonometric Functions
Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 6–Graphs of Trigonometric Functions 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.