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 up on their problem-solving abilities, which is an important aspect of 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 PDF format to solve problems of RD Sharma Class 11 textbook easily from the below-mentioned links.

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

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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 equations (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 equations (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 10^o + sin³ 20^o) = 3 (cos 10^o + sin 20^o)

Solution:

Let us consider LHS:

4 (cos³ 10^o + sin³ 20^o)

We know that, sin 60^o = √3/2 = cos 30^o

Sin 30^o = cos 60^o = 1/2

So,

Sin (3×20^o) = cos (3×10^o)

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 equations (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 equations (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 equations (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 equations (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 equations (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.