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

The RD Sharma solutions which contain a huge number of solved examples and illustrations provide stepwise explanations of various difficult concepts and contains a wide variety of questions for practice. This chapter introduces the concept of formulae expressing the values of trigonometric functions at multiples and submultiples of ‘x’. RD Sharma Solutions are completely based on the exam-oriented approach to help students score well in their board examination. The pdf of RD Sharma Class 11 Solutions Maths Chapter 9 Values of Trigonometric Functions at Multiples and Submultiples of an Angle is provided here. Practising these questions will ensure that they can easily excel in their final exams. Students can refer and download this chapter pdf from the below-given links.

Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle contains three exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. 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.

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Exercise 9.1 Solutions

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

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.

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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,

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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,

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= 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,

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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,

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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,

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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.

RD Sharma Solutions for Class 11 Maths Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 16

RD Sharma Solutions for Class 11 Maths Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle image - 17

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.

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