# RD Sharma Solutions for Class 11 Chapter 9 - Values of Trigonometric Functions at Multiples and Submultiples of an Angle Exercise 9.2

In this Exercise 9.2 of Chapter 9, we shall discuss problems based on the values of trigonometric functions at ‘3x’ in terms of values at ‘x’. A set of examples are given before the exercise problems to make it possible for students to solve questions effortlessly. It also helps students in checking upon their problem-solving abilities, which is an important aspect in solving problems. To score well in their board exams students are advised to practice the solutions regularly without fail. Students can also access exercise-wise solutions, which are available in the pdf format to solve problems of RD Sharma Class 11 textbook easily from the below-mentioned links.

## Download the Pdf of RD Sharma Solutions for Class 11 Maths Exercise 9.2 Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle

### Also, access other exercises of RD Sharma Solutions for Class 11 Maths Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle

Exercise 9.1 Solutions

Exercise 9.3 solutions

### Access answers to RD Sharma Solutions for Class 11 Maths Exercise 9.2 Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle

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.