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 â€“ Download Free PDF Updated for 2021 – 2022

RD Sharma Solutions for Class 11 Maths Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an AngleÂ are provided here for students to study and score good marks in the board exams. 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 â€˜xâ€™.

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 â€˜2xâ€™ in terms of values at â€˜xâ€™.
• Values of trigonometric functions at â€˜xâ€™ in terms of values at â€˜x/2â€™.
• Values of trigonometric functions at â€˜x/2â€™ in terms of values at â€˜cos xâ€™.
• Values of trigonometric functions at â€˜3xâ€™ in terms of values at â€˜xâ€™.
• Values of trigonometric functions at â€˜xâ€™ in terms of values at â€˜x/3â€™.
• Values of trigonometric functions at some important points.

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Exercise 9.1 Solutions

Exercise 9.2 Solutions

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EXERCISE 9.1 PAGE NO: 9.28

Prove the following identities:

1. âˆš[(1 â€“ cos 2x) / (1 + cos 2x)] = tan x

Solution:

Let us consider LHS:

âˆš[(1 â€“ cos 2x) / (1 + cos 2x)]

We know that cos 2x = 1 â€“ 2 sin2 x

= 2 cos2 x – 1

So,

âˆš[(1 â€“ cos 2x) / (1 + cos 2x)] = âˆš[(1 â€“ (1 – 2sin2 x)) / (1 + (2cos2x – 1))]

= âˆš[(1 â€“ 1 + 2sin2 x) / (1 + 2cos2 x – 1)]

= âˆš[2 sin2 x / 2 cos2 x]

= sin x/cos x

= tan x

= RHS

Hence proved.

2. sin 2x / (1 â€“ cos 2x) = cot x

Solution:

Let us consider LHS:

sin 2x / (1 â€“ cos 2x)

We know that cos 2x = 1 â€“ 2 sin2 x

Sin 2x = 2 sin x cos x

So,

sin 2x / (1 â€“ cos 2x) = (2 sin x cos x) / (1 – (1 â€“ 2sin2 x))

= (2 sin x cos x) / (1 â€“ 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 â€“ 2 sin2 x

= 2 cos2 x â€“ 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 â€“ cos 2x + sin 2x] / [1 + cos 2x + sin 2x] = tan x

Solution:

Let us consider LHS:

[1 â€“ cos 2x + sin 2x] / [1 + cos 2x + sin 2x]

We know that, cos 2x = 1 â€“ 2 sin2 x

= 2 cos2 x â€“ 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:

[sin x + sin 2x] / [1 + cos x + cos 2x]

We know that, cos 2x = cos2 x â€“ sin2 x

Sin 2x = 2 sin x cos x

So,

= RHS

Hence proved.

7. cos 2x / (1 + sin 2x) = tan (Ï€/4 – x)

Solution:

Let us consider LHS:

cos 2x / (1 + sin 2x)

We know that, cos 2x = cos2 x â€“ sin2 x

Sin 2x = 2 sin x cos x

So,

8. cos x / (1 – sin x) = tan (Ï€/4 + x/2)

Solution:

Let us consider LHS:

cos x / (1 – sin x)

We know that, cos 2x = cos2 x â€“ sin2 x

Cos x = cos2 x/2 â€“ sin2 x/2

Sin 2x = 2 sin x cos x

Sin x = 2 sin x/2 cos x/2

So,

11. (cos Î± + cos Î²) 2 + (sin Î± + sin Î²) 2 = 4 cos2 (Î± – Î²)/2

Solution:

Let us consider LHS:

(cos Î± + cos Î²)2 + (sin Î± + sin Î²)2

Upon expansion, we get,

(cos Î± + cos Î²)2 + (sin Î± + sin Î²)2 =

= cos2 Î± + cos2 Î² + 2 cos Î± cos Î² + sin2 Î± + sin2 Î² + 2 sin Î± sin Î²

= 2 + 2 cos Î± cos Î² + 2 sin Î± sin Î²

= 2 (1 + cos Î± cos Î² + sin Î± sin Î²)

= 2 (1 + cos (Î± – Î²)) [since, cos (A â€“ B) = cos A cos B + sin A sin B]

= 2 (1 + 2 cos2 (Î± – Î²)/2 – 1) [since, cos2x = 2cos2Â x â€“ 1]

= 2 (2 cos2 (Î± – Î²)/2)

= 4 cos2 (Î± – Î²)/2

= RHS

Hence Proved.

12. sin2 (Ï€/8 + x/2) â€“ sin2 (Ï€/8 â€“ x/2) = 1/âˆš2 sin x

Solution:

Let us consider LHS:

sin2 (Ï€/8 + x/2) â€“ sin2 (Ï€/8 â€“ x/2)

we know, sin2 A â€“ sin2 B = sin (A+B) sin (A-B)

so,

sin2 (Ï€/8 + x/2) â€“ sin2 (Ï€/8 â€“ x/2) = sin (Ï€/8 + x/2 + Ï€/8 â€“ x/2) sin (Ï€/8 + x/2 â€“ (Ï€/8 â€“ x/2))

= sin (Ï€/8 + Ï€/8) sin (Ï€/8 + x/2 – Ï€/8 + x/2)

= sin Ï€/4 sin x

= 1/âˆš2 sin x [since, since Ï€/4 = 1/âˆš2]

= RHS

Hence proved.

13. 1 + cos2 2x = 2 (cos4 x + sin4 x)

Solution:

Let us consider LHS:

1 + cos2 2x

We know, cos2x = cos2Â x â€“ sin2Â x

cos2Â x + sin2Â x = 1

so,

1 + cos2 2x = (cos2 x + sin2 x) 2 + (cos2 x â€“ sin2 x) 2

= (cos4 x + sin4 x + 2 cos2 x sin2 x) + (cos4 x + sin4 x â€“ 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 â€“ sin6 x)

Solution:

Let us consider RHS:

4 (cos6 x â€“ sin6 x)

Upon expansion we get,

4 (cos6 x â€“ sin6 x) = 4 [(cos2 x)3 â€“ (sin2 x)3]

= 4 (cos2 x â€“ sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)

By using the formula,

a3 â€“ b3 = (a-b) (a2 + b2 + ab)

= 4 cos 2x (cos4 x + sin4 x + cos2 x sin2 x + cos2 x sin2 x â€“ cos2 x sin

We know, cos 2x = cos2 x â€“ 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, a2Â + b2Â + 2ab = (a + b)2

= 4 cos 2x [(1)2 â€“ 1/4 (4 cos2 x sin2 x)]

= 4 cos 2x [(1)2 â€“ 1/4 (2 cos x sin x)2]

We know, sin 2x = 2sin x cos x

= 4 cos 2x [(12) â€“ 1/4 (sin 2x)2]

= 4 cos 2x (1 â€“ 1/4 sin2 2x)

We know, sin2 x = 1 â€“ cos2 x

= 4 cos 2x [1 â€“ 1/4 (1 â€“ cos2 2x)]

= 4 cos 2x [1 â€“ 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 â€“ cos x) cos x = 0

Solution:

Let us consider LHS:

(sin 3x + sin x) sin x + (cos 3x â€“ cos x) cos x

= (sin 3x) (sin x) + sin2Â x + (cos 3x) (cos x) â€“ cos2Â x

= [(sin 3x) (sin x) + (cos 3x) (cos x)] + (sin2Â x â€“ cos2Â x)

= [(sin 3x) (sin x) + (cos 3x) (cos x)] â€“ (cos2Â x â€“ sin2Â x)

= cos (3x â€“ x) â€“ cos 2x

We know,Â cos 2x =Â cos2Â x â€“ sin2Â x

cos A cos B + sin A sin B = cos(A â€“ B)

So,

= cos 2x â€“ cos 2x

= 0

= RHS

Hence Proved.

16. cos2 (Ï€/4 – x) â€“ sin2 (Ï€/4 – x) = sin 2x

Solution:

Let us consider LHS:

cos2 (Ï€/4 – x) â€“ sin2 (Ï€/4 – x)

We know, cos2 A â€“ sin2 A = cos 2A

So,

cos2 (Ï€/4 – x) â€“ sin2 (Ï€/4 – x) = cos 2 (Ï€/4 – x)

= cos (Ï€/2 – 2x)

= sin 2x [since, cos (Ï€/2 – A) = sin A]

= RHS

Hence proved.

17. cos 4x = 1 â€“ 8 cos2Â x + 8 cos4Â x

Solution:

Let us consider LHS:

cos 4x

We know, cos 2x = 2 cos2Â x â€“ 1

So,

cos 4x = 2 cos2Â 2x â€“ 1

= 2(2 cos2Â 2x â€“ 1)2Â â€“ 1

= 2[(2 cos2Â 2x) 2Â + 12Â â€“ 2Ã—2 cos2Â x] â€“ 1

= 2(4 cos4Â 2x + 1 â€“ 4 cos2Â x) â€“ 1

= 8 cos4Â 2x + 2 â€“ 8 cos2Â x â€“ 1

= 8 cos4Â 2x + 1 â€“ 8 cos2Â x

= RHS

Hence Proved.

18. sin 4x = 4 sin x cos3Â x â€“ 4 cos x sin3Â x

Solution:

Let us consider LHS:

sin 4x

We know, sin 2x = 2 sin x cos x

cos 2x = cos2Â x â€“ sin2Â x

So,

sin 4x = 2 sin 2x cos 2x

= 2 (2 sin x cos x) (cos2Â x â€“ sin2Â x)

= 4 sin x cos x (cos2Â x â€“ sin2Â x)

= 4 sin x cos3Â x â€“ 4 sin3Â x cos x

= RHS

Hence proved.

19. 3(sin x â€“ cos x) 4Â + 6 (sin x + cos x) 2Â + 4 (sin6Â x + cos6Â x) = 13

Solution:

Let us consider LHS:

3(sin x â€“ cos x) 4Â + 6 (sin x + cos x) 2Â + 4 (sin6Â x + cos6Â x)

We know, (a + b)2Â = a2Â + b2Â + 2ab

(a â€“ b)2Â = a2Â + b2Â â€“ 2ab

a3Â + b3Â = (a + b) (a2Â + b2Â â€“ ab)

So,

3(sin x â€“ cos x) 4Â + 6 (sin x + cos x) 2Â + 4 (sin6Â x + cos6Â x) = 3{(sin x â€“ cos x) 2}2Â + 6 {(sin x)2Â + (cos x)2Â + 2 sin x cos x)} + 4 {(sin2Â x)3Â + (cos2Â x)3}

= 3{(sin x) 2Â + (cos x)2Â â€“ 2 sin x cos x)}2Â + 6 (sin2Â x + cos2Â x + 2 sin x cos x) + 4{(sin2Â x + cos2Â x) (sin4Â x + cos4Â x â€“ sin2Â x cos2Â x)}

= 3(1 â€“ 2 sin x cos x)Â 2Â + 6 (1 + 2 sin x cos x) + 4{(1) (sin4Â x + cos4Â x â€“ sin2Â x cos2Â x)}

We know,Â sin2Â x + cos2Â x = 1

So,

= 3{12Â + (2 sin x cos x) 2Â â€“ 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2Â x)2Â + (cos2Â x)2Â + 2 sin2Â x cos2Â x â€“ 3 sin2Â x cos2Â x)}

= 3{1 + 4 sin2Â x cos2Â x â€“ 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2Â x + cos2Â x) 2Â â€“ 3 sin2Â x cos2Â x)}

= 3 + 12 sin2Â x cos2Â x â€“ 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)2Â â€“ 3 sin2Â x cos2Â x)}

= 9 + 12 sin2Â x cos2Â x + 4(1 â€“ 3 sin2Â x cos2Â x)

= 9 + 12 sin2Â x cos2Â x + 4 â€“ 12 sin2Â x cos2Â x

= 13

= RHS

Hence proved.

20. 2(sin6Â x + cos6Â x) â€“ 3(sin4Â x + cos4Â x) + 1 = 0

Solution:

Let us consider LHS:

2(sin6Â x + cos6Â x) â€“ 3(sin4Â x + cos4Â x) + 1

We know, (a + b)2Â = a2Â + b2Â + 2ab

a3Â + b3Â = (a + b) (a2Â + b2Â â€“ ab)

So,

2(sin6Â x + cos6Â x) â€“ 3(sin4Â x + cos4Â x) + 1 =Â 2{(sin2Â x) 3Â + (cos2Â x) 3} â€“ 3{(sin2Â x) 2Â + (cos2Â x) 2} + 1

=Â 2{(sin2Â x + cos2Â x) (sin4Â x + cos4Â x â€“ sin2Â x cos2Â x} â€“ 3{(sin2Â x) 2Â + (cos2Â x) 2Â + 2sin2Â x cos2Â x â€“ 2sin2Â x cos2Â x} + 1

=Â 2{(1) (sin4Â x + cos4Â x + 2 sin2Â x cos2Â x â€“ 3 sin2Â x cos2Â x} â€“ 3{(sin2Â x + cos2Â x) 2Â â€“ 2sin2Â x cos2Â x} + 1

We know, sin2Â x + cos2Â x = 1

=Â 2{(sin2Â x + cos2Â x) 2Â â€“ 3 sin2Â x cos2Â x} â€“ 3{(1)2Â â€“ 2sin2Â x cos2Â x} + 1

=Â 2{(1)2Â â€“ 3 sin2Â x cos2Â x} â€“ 3(1 â€“ 2sin2Â x cos2Â x) + 1

=Â 2(1 â€“ 3 sin2Â x cos2Â x) â€“ 3 + 6 sin2Â x cos2Â x + 1

=Â 2 â€“ 6 sin2Â x cos2Â x â€“ 2 + 6 sin2Â x cos2Â x

= 0

= RHS

Hence proved.

21. cos6 x â€“ sin6 x = cos 2x (1 â€“ 1/4 sin2 2x)

Solution:

Let us consider LHS:

cos6 x â€“ sin6 x

We know, (a + b) 2Â = a2Â + b2Â + 2ab

a3Â â€“ b3Â = (a â€“ b) (a2Â + b2Â + ab)

So,

cos6 x â€“ sin6 x = (cos2 x)3 â€“ (sin2 x)3

= (cos2 x â€“ sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)

We know, cos 2x = cos2 x â€“ sin2 x

So,

= cos 2x [(cos2 x) 2 + (sin2 x) 2 + 2 cos2 x sin2 x â€“ cos2 x sin2 x]

= cos 2x [(cos2 x) 2 + (sin2 x) 2 â€“ 1/4 Ã— 4 cos2 x sin2 x]

We know, sin2Â x + cos2Â x = 1

So,

= cos 2x [(1)2 â€“ 1/4 Ã— (2 cos x sin x) 2]

We know, sin 2x = 2 sin x cos x

So,

= cos 2x [1 â€“ 1/4 Ã— (sin 2x) 2]

= cos 2x [1 â€“ 1/4 Ã— sin2 2x]

= RHS

Hence proved.

22. tan (Ï€/4 + x) + tan (Ï€/4 – x) = 2 sec 2x

Solution:

Let us consider LHS:

tan (Ï€/4 + x) + tan (Ï€/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 â€“ 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 yâ€¦..(i)

So,

sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin (2x + x) cos 2x + cos (2x + x) sin 2xâ€¦â€¦..(ii)

And

cos (x + y) = cos x cos y â€“ sin x sin yâ€¦â€¦(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 â€“ sin 2x sin x) sin 2x

= sin 2x cos 2x cos x + cos2Â 2x sin x + (sin 2x cos 2x cos x â€“ sin2Â 2x sin x)

= 2sin 2x cos 2x cos x + cos2Â 2x sin x â€“ sin2Â 2x sin x â€¦â€¦.(iv)

Now sin 2x = 2sin x cos xâ€¦â€¦â€¦(v)

And cos 2x = cos2x â€“ sin2xâ€¦â€¦â€¦(vi)

Substituting equation (v) and (vi) in equation (iv), we get

sin 5x = 2(2sin x cos x) (cos2x â€“sin2x) cos x + (cos2x â€“ sin2x)2 sin x â€“ (2sin x cos x)2 sin x

= 4(sin x cos2Â x) ([1â€“ sin2x] â€“ sin2x) + ([1â€“sin2x] â€“ sin2x)2 sin x â€“ (4sin2Â x cos2Â x)sin x

(as cos2x + sin2x = 1Â â‡’Â cos2x = 1â€“ sin2x)

sin 5x = 4(sin x [1 â€“ sin2x]) (1 â€“ 2sin2x) + (1 â€“ 2sin2x)2 sin x â€“ 4sin3Â x [1 â€“ sin2x]

= 4sin x (1 â€“ sin2x) (1 â€“ 2sin2 x) + (1 â€“ 4sin2x + 4sin4x) sin x â€“ 4sin3Â x + 4sin5x

= (4sin x â€“ 4sin3x) (1 â€“ 2sin2x) + sin x â€“ 4sin3x + 4sin5x â€“ 4sin3Â x + 4sin5x

= 4sin x â€“ 8sin3x â€“ 4sin3x + 8sin5x + sin x â€“ 8sin3x + 8sin5x

= 5sin x â€“ 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 = âˆš3/2 = cos 30o

Sin 30o = cos 60o = 1/2

So,

Sin (3Ã—20o) = cos (3Ã—10o)

3sin 20Â°â€“ 4sin320Â° = 4cos310Â° â€“ 3cos 10Â°

(we know, sin 3Î¸ = 3sin Î¸ â€“ 4sin3Â Î¸ and cos 3Î¸ = 4cos3Î¸ â€“ 3cosÎ¸)

So,

4(cos310Â°+sin320Â°) = 3(sin 20Â°+cos 10Â°)

= RHS

Hence proved.

3. cos3 x sin 3x + sin3 x cos 3x = 3/4 sin 4x

Solution:

We know that,

cos 3Î¸ = 4cos3Î¸ â€“ 3cosÎ¸

So, 4 cos3Î¸ = cos3Î¸ + 3cosÎ¸

cos3 Î¸ = [cos3Î¸ + 3cosÎ¸]/4 â€¦â€¦ (i)

Similarly,

sin 3Î¸ = 3sin Î¸ â€“ 4sin3Â Î¸

4 sin3Î¸ = 3sinÎ¸ â€“ sinÂ 3Î¸

sin3Î¸ = [3sinÎ¸ â€“ sinÂ 3Î¸]/4 â€¦â€¦.. (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 â€“ sin 3x)/4 cos 3x

= 1/4 (sin 3x cos 3x + 3 sin 3x cox x + 3sin x cos 3x â€“ 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 â€“ 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 yâ€¦..(i)

So,

sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin (2x + x) cos 2x + cos (2x + x) sin 2xâ€¦â€¦..(ii)

And

cos (x + y) = cos x cos y â€“ sin x sin yâ€¦â€¦(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 â€“ sin 2x sin x) sin 2x â€¦ (iv)

Now sin 2x = 2sin x cos xâ€¦â€¦â€¦(v)

And cos 2x = cos2x â€“ sin2xâ€¦â€¦â€¦(vi)

Substituting equation (v) and (vi) in equation (iv), we get

sin 5x = [(2 sin x cos x) cos x + (cos2x â€“ sin2x) sin x] (cos2x â€“ sin2x) + [(cos2x â€“ sin2x) cos x â€“ (2 sin x cos x) sin x)] (2 sin x cos x)

= [2 sin x cos2Â x + sin x cos2x â€“ sin3x] (cos2x â€“ sin2x) + [cos3x â€“ sin2x cos x â€“ 2 sin2Â x cos x] (2 sin x cos x)

= cos2x [3 sin x cos2Â x â€“ sin3x] â€“ sin2x [3 sin x cos2Â x â€“ sin3x] + 2 sin x cos4x â€“ 2 sin3Â x cos2Â x â€“ 4 sin3Â x cos2Â x

= 3 sin x cos4Â x â€“ sin3x cos2x â€“ 3 sin3Â x cos2Â x â€“ sin5x + 2 sin x cos4x â€“ 2 sin3Â x cos2Â x â€“ 4 sin3Â x cos2Â x

= 5 sin x cos4Â x â€“10sin3xcos2x +sin5x

= RHS

Hence proved.

5. sin 5x = 5 sin x â€“ 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 yâ€¦..(i)

So,

sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin (2x + x) cos 2x + cos (2x + x) sin 2xâ€¦â€¦..(ii)

And

cos (x + y) = cos x cos y â€“ sin x sin yâ€¦â€¦(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 â€“ sin 2x sin x) sin 2x

= sin 2x cos 2x cos x + cos2Â 2x sin x + (sin 2x cos 2x cos x â€“ sin2Â 2x sin x)

= 2sin 2x cos 2x cos x + cos2Â 2x sin x â€“ sin2Â 2x sin x â€¦â€¦.(iv)

Now sin 2x = 2sin x cos xâ€¦â€¦â€¦(v)

And cos 2x = cos2x â€“ sin2xâ€¦â€¦â€¦(vi)

Substituting equation (v) and (vi) in equation (iv), we get

sin 5x = 2(2sin x cos x) (cos2x â€“sin2x) cos x + (cos2x â€“ sin2x)2 sin x â€“ (2sin x cos x)2 sin x

= 4(sin x cos2Â x) ([1â€“ sin2x] â€“ sin2x) + ([1â€“sin2x] â€“ sin2x)2 sin x â€“ (4sin2Â x cos2Â x)sin x

(as cos2x + sin2x = 1Â â‡’Â cos2x = 1â€“ sin2x)

sin 5x = 4(sin x [1 â€“ sin2x]) (1 â€“ 2sin2x) + (1 â€“ 2sin2x)2 sin x â€“ 4sin3Â x [1 â€“ sin2x]

= 4sin x (1 â€“ sin2x) (1 â€“ 2sin2 x) + (1 â€“ 4sin2x + 4sin4x) sin x â€“ 4sin3Â x + 4sin5x

= (4sin x â€“ 4sin3x) (1 â€“ 2sin2x) + sin x â€“ 4sin3x + 4sin5x â€“ 4sin3Â x + 4sin5x

= 4sin x â€“ 8sin3x â€“ 4sin3x + 8sin5x + sin x â€“ 8sin3x + 8sin5x

= 5sin x â€“ 20sin3x + 16sin5x

= RHS

Hence proved.

EXERCISE 9.3 PAGE NO: 9.42

Prove that:

1. sin2 2Ï€/5 â€“ sin2 Ï€/3 = (âˆš5 – 1)/8

Solution:

Let us consider LHS:

sin2 2Ï€/5 â€“ sin2 Ï€/3 = sin2 (Ï€/2 – Ï€/10) â€“ sin2 Ï€/3

we know, sin (90Â°â€“ A) = cos A

So, sin2 (Ï€/2 – Ï€/10) = cos2 Ï€/10

Sin Ï€/3 = âˆš3/2

Then the above equation becomes,

= Cos2 Ï€/10 â€“ (âˆš3/2)2

We know, cos Ï€/10 = âˆš(10+2âˆš5)/4

the above equation becomes,

= [âˆš(10+2âˆš5)/4]2 â€“ 3/4

= [10 + 2âˆš5]/16 â€“ 3/4

= [10 + 2âˆš5 – 12]/16

= [2âˆš5 â€“ 2]/16

= [âˆš5 â€“ 1]/8

= RHS

Hence proved.

2. sin2 24o â€“ sin2 6o = (âˆš5 – 1)/8

Solution:

Let us consider LHS:

sin2 24o â€“ sin2 6o

we know, sin (A + B) sin (A â€“ B) = sin2A â€“ sin2B

Then the above equation becomes,

sin2 24o â€“ sin2 6o = sin (24o + 6o) â€“ sin (24o â€“ 6o)

= sin 30o â€“ sin 18o

= sin 30o â€“ (âˆš5 – 1)/4 [since, sin 18o = (âˆš5 – 1)/4]

= 1/2 Ã— (âˆš5 – 1)/4

= (âˆš5 – 1)/8

= RHS

Hence proved.

3. sin2 42o â€“ cos2 78o = (âˆš5 + 1)/8

Solution:

Let us consider LHS:

sin2 42o â€“ cos2 78o = sin2 (90o â€“ 48o) â€“ cos2 (90o â€“ 12o)

= cos2 48o â€“ sin2 12o [since, sin (90 – A) = cos A and cos (90 – A) = sin A]

We know, cos (A + B) cos (A â€“ B) = cos2A â€“ sin2B

Then the above equation becomes,

= cos2 (48o + 12o) cos (48o â€“ 12o)

= cos 60o cos 36o [since, cos 36o = (âˆš5 + 1)/4]

= 1/2 Ã— (âˆš5 + 1)/4

= (âˆš5 + 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 â€“ B)

Then the above equation becomes,

= 1/2 (cos (78o + 42o) + cos (78o â€“ 42o)) Ã— cos 36o

= 1/2 (cos 120o + cos 36o) Ã— cos 36o

= 1/2 (cos (180o â€“ 60o) + cos 36o) Ã— cos 36o

= 1/2 (-cos (60o) + cos 36o) Ã— cos 36o [since, cos(180Â° â€“ A) = â€“ A]

= 1/2 (-1/2 + (âˆš5 + 1)/4) ((âˆš5 + 1)/4) [since, cos 36o = (âˆš5 + 1)/4]

= 1/2 (âˆš5 + 1 – 2)/4 ((âˆš5 + 1)/4)

= 1/2 (âˆš5 – 1)/4) ((âˆš5 + 1)/4)

= 1/2 ((âˆš5)2 – 12)/16

= 1/2 (5-1)/16

= 1/2 (4/16)

= 1/8

= RHS

Hence proved.

5. cos Ï€/15 cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15 = 1/16

Solution:

Let us consider LHS:

cos Ï€/15 cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15

Let us multiply and divide by 2 sin Ï€/15, we get,

= [2 sin Ï€/15 cos Ï€/15] cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15] / 2 sin Ï€/15

We know, 2sin A cos A = sin 2A

Then the above equation becomes,

= [(sin 2Ï€/15) cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15] / 2 sin Ï€/15

Now, multiply and divide by 2 we get,

= [(2 sin 2Ï€/15 cos 2Ï€/15) cos 4Ï€/15 cos 7Ï€/15] / 2 Ã— 2 sin Ï€/15

We know, 2sin A cos A = sin 2A

Then the above equation becomes,

= [(sin 4Ï€/15) cos 4Ï€/15 cos 7Ï€/15] / 4 sin Ï€/15

Now, multiply and divide by 2 we get,

= [(2 sin 4Ï€/15 cos 4Ï€/15) cos 7Ï€/15] / 2 Ã— 4 sin Ï€/15

We know, 2sin A cos A = sin 2A

Then the above equation becomes,

= [(sin 8Ï€/15) cos 7Ï€/15] / 8 sin Ï€/15

Now, multiply and divide by 2 we get,

= [2 sin 8Ï€/15 cos 7Ï€/15] / 2 Ã— 8 sin Ï€/15

We know, 2sin A cos B = sin (A+B) + sin (Aâ€“B)

Then the above equation becomes,

= [sin (8Ï€/15 + 7Ï€/15) + sin (8Ï€/15 – 7Ï€/15)] / 16 sin Ï€/15

= [sin (Ï€) + sin (Ï€/15)] / 16 sin Ï€/15

= [0 + sin (Ï€/15)] / 16 sin Ï€/15

= sin (Ï€/15) / 16 sin Ï€/15

= 1/16

= RHS

Hence proved.

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