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

RD Sharma Solutions Class 11 Maths Chapter 9 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle are provided here for students to study and score good marks in the board exams. RD Sharma Class 11 Solutions Maths designed by our subject expert faculty in an illustrative manner help students obtain in-depth knowledge of concepts. This chapter introduces the concept of formulae expressing the values of trigonometric functions at multiples and submultiples of ‘x’.

Class 11 Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle contains three exercises. RD Sharma Solutions offers precise answers to each question of the textbook in a simple and understandable language. Solutions are well structured by expert teachers based on the current 2022-23 syllabus of the CBSE board to help students score well in their board examinations. Now, let us have a look at the concepts discussed in this chapter.

• Values of trigonometric functions at ‘2x’ in terms of values at ‘x’.
  • Values of trigonometric functions at ‘x’ in terms of values at ‘x/2’.
  • Values of trigonometric functions at ‘x/2’ in terms of values at ‘cos x’.
• Values of trigonometric functions at ‘3x’ in terms of values at ‘x’.
• Values of trigonometric functions at ‘x’ in terms of values at ‘x/3’.
• Values of trigonometric functions at some important points.

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

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Exercise 9.1 Solutions

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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:

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:

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 sin²x)

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 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 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 24^o – sin² 6^o = (√5 – 1)/8

Solution:

Let us consider LHS:

sin² 24^o – sin² 6^o

we know, sin (A + B) sin (A – B) = sin²A – sin²B

Then the above equation becomes,

sin² 24^o – sin² 6^o = sin (24^o + 6^o) – sin (24^o – 6^o)

= sin 30^o – sin 18^o

= sin 30^o – (√5 – 1)/4 [since, sin 18^o = (√5 – 1)/4]

= 1/2 × (√5 – 1)/4

= (√5 – 1)/8

= RHS

Hence proved.

3. sin² 42^o – cos² 78^o = (√5 + 1)/8

Solution:

Let us consider LHS:

sin² 42^o – cos² 78^o = sin² (90^o – 48^o) – cos² (90^o – 12^o)

= cos² 48^o – sin² 12^o [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 (48^o + 12^o) cos (48^o – 12^o)

= cos 60^o cos 36^o [since, cos 36^o = (√5 + 1)/4]

= 1/2 × (√5 + 1)/4

= (√5 + 1)/8

= RHS

Hence proved.

4. cos 78^o cos 42^o cos 36^o = 1/8

Solution:

Let us consider LHS:

cos 78^o cos 42^o cos 36^o

Let us multiply and divide by 2 we get,

cos 78^o cos 42^o cos 36^o = 1/2 (2 cos 78^o cos 42^o cos 36^o)

We know, 2 cos A cos B = cos (A + B) + cos (A – B)

Then the above equation becomes,

= 1/2 (cos (78^o + 42^o) + cos (78^o – 42^o)) × cos 36^o

= 1/2 (cos 120^o + cos 36^o) × cos 36^o

= 1/2 (cos (180^o – 60^o) + cos 36^o) × cos 36^o

= 1/2 (-cos (60^o) + cos 36^o) × cos 36^o [since, cos(180° – A) = – A]

= 1/2 (-1/2 + (√5 + 1)/4) ((√5 + 1)/4) [since, cos 36^o = (√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.