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Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 2
  • Last Updated : 08 May, 2021

Prove the following identities: 

Question 16. cos2 (π/4 – x) – sin2 (π/4 – x) = sin 2x

Solution:

Let us solve LHS,

= cos2 (π/4 – x) – sin2(π/4 – x)

As we know that,

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                       [As we know that, cos (π/2 – A) = sin A]

LHS = RHS

Hence Proved.

Question 17. cos 4x = 1 – 8 cos2x + 8 cos4 x

Solution:

Let us solve LHS,

= cos 4x

As we know that,

cos 2x = 2 cos2x – 1

So,

cos 4x = 2 cos2 2x – 1

= 2(2 cos2 2x – 1)2 – 1 

= 2[(2 cos2 2x)2 + 12 – 2 × 2 cos2x] – 1

= 2(4 cos4 2x + 1 – 4 cos2x) – 1

= 8 cos4 2x + 2 – 8 cos2 x – 1



= 8 cos4 2x + 1 – 8 cos2x

LHS = RHS

Hence Proved.

Question 18. sin 4x = 4 sin x cos3 x – 4 cos x sin3 x

Solution:

Let us solve LHS,

= sin 4x

As we know that,

sin 2x = 2 sin x cos x

cos 2x = cos2x – sin2x

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

LHS = RHS

Hence proved.

Question 19. 3(sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) = 13

Solution:

Let us solve LHS,

= 3(sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) 

As we know that,



(a + b)2 = a2 + b2 + 2ab

(a – b)2 = a2 + b2 – 2ab

a3 + b3 = (a+b)(a2 + b2 – ab)

So,

= 3{(sinx – cosx)2}2 + 6 {(sinx)2 + (cosx)2 + 2 sinx cosx} + 4 {(sin2x)3 + (cos2x)3}

= 3{(sinx)2 + (cosx)2 – 2 sinx cosx}2 + 6(sin2x + cos2x + 2 sinx cosx) + 4{(sin2x + cos2x)(sin4x + cos4x – sin2x cos2x)}

= 3(1 – 2 sinx cosx)2 + 6(1 + 2 sinx cosx) + 4{(1)(sin4x + cos4x – sini2x cos2x)}

Since, 

sin2x + cos2x = 1

So,

= 3{12 + (2 sinx cosx)2 – 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin2x)2 + (cos2x)2 + 2 sin2x cos2x – 3 sin2x cos2x}

= 3{1 + 4 sin2x cos2x – 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin2x + cos2x)2 – 3 sin2x cos2x}

= 3 + 12 sin2x cos2x – 12 sinx cosx + 6 + 12 sinx cosx + 4{(1)2 – 3 sin2x cos2x}

= 9 + 12 sin2x cos2x + 4(1 – 3 sin2x cos2x)

= 9 + 12 sin2x cos2x + 4 – 12 sin2x cos2x

= 13

LHS = RHS

Hence proved.

Question 20. 2(sin6x + cos6x) – 3(sin4x + cos4x) + 1 = 0

Solution:

Let us solve LHS,

= 2(sin6 x + cos6 x) – 3(sin4 x + cos4 x) + 1 

As we know that,

(a + b)2 = a2 + b2 + 2ab

a3 + b3 = (a + b) (a2 + b2 – ab) 

So,

 = 2(sin6x + cos6x) – 3(sin4x + cos4x) + 1 

= 2{(sin2x)3 + (cos2x)3} – 3{(sin2x)2 + (cos2x)2} + 1

= 2((sin2x + cos2x)(sin4x + cos4x – sin2x cos2x) – 3{(sin2x)2 + (cos2x)2 + 2 sin2x cos2x – 2sin2x cos2x} + 1

= 2{(1)(sin4x + cos4x + 2 sin2x cos2x – 3 sin2x cos2x) – 3((sin2x + cos2x)2 – 2sin2x cos2x) + 1

Since

sin2x + cos2x = 1

So, 

= 2{(sin2x + cos2x)2 – 3 sin2x cos2x} – 3{(1)2 – 2 sin2x cos2x} + 1

= 2{(1)2 – 3 sin2x cos2x} – 3(1 – 2 sin2x cos2x) + 1

= 2(1 – 3 sin2x cos2x) – 3 + 6 sin2x cos2x + 1

= 2 – 6 sin2x cos2x – 2 + 6 sin2x cos2x

= 0 

LHS = RHS

Hence Proved.

Question 21. cos6 x – sin6 x = cos 2x (1 – 1/4 sin2 2x)

Solution:

Let us solve LHS,

= cos6 x – sin6 x

As we know that,

(a + b)2 = a2 + b2 + 2ab

a3 – b3 = (a – b) (a2 + b2 + ab)

So,

cos6 x – sin6 x = (cos2 x)3 – (sin2 x)3

 = (cos2x – sin2x) (cos4x + sin4x + cos2x sin2x)

As we know that,

cos 2x = cos2x – sin2x

So,

= cos 2x [(cos2x)2 + (sin2x)2 + 2 cos2x sin2x – cos2x sin2x]

= cos 2x [(cos2x)2 + (sin2x)2 – 1/4 × 4 cos2x sin2x]

As we know that,

sin2x + cos2x = 1

So,

= cos2x [(1)2 – 1/4 × (2 cosx sinx)2]

As we know that,

sin2x = 2 sinx cosx

So,

= cos 2x [1 – 1/4 × (sin 2x)2]

= cos 2x [1 – 1/4 × sin22x]

LHS = RHS

Hence proved.

Question 22. tan (π/4 + x) + tan (π/4 – x) = 2 sec2x

Solution:

Let us solve LHS,

= tan (π/4 + x) + tan (π/4 – x)

As we know that,

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, 

\frac{tan(\frac{π}{4})+tanx}{1-tan(\frac{π}{4})tanx}+\frac{tan(\frac{π}{4})-tanx}{1+tan(\frac{π}{4})tanx}

Since, tan π/4 = 1

So,

\frac{1+tanx}{1-tanx}+\frac{1-tanx}{1+tanx}

\frac{(1+tanx)^2+(1-tanx)^2}{(1-tanx)(1+tanx)}

By using formulas, we get

(a – b)(a + b) = a2 – b2

(a + b)2 = a2 + b2 + 2ab &  (a – b)2 = a2+ b2 – 2ab

So,

\frac{1^2+tan^2x+2tanx+1^2+tan^2x-2tanx}{1^2-tan^2x}

\frac{1+tan^2x+1+tan^2x}{1-tan^2x}

\frac{2(1+tan^2x)}{1-tan^2x}

As we know that,

tan x = sin x/cos x

So,

\frac{2(1+(sinx/cosx)^2)}{1-(sinx/cosx)^2}

\frac{2(1+sin^2x/cos^2x)}{1-sin^2x/cos^2x}

\frac{2(\frac{cos^2x+sin^2x}{cos^2x})}{\frac{cos^2x-sin^2x}{cos^2x}}

By using the formulas, we get

cos2x+ sin2x = 1 & cos 2x = cos2x – sin2x

So,

\frac{2(\frac{1}{cos^2x})}{\frac{cos2x}{cos^2x}}

= 2/cos2x

= 2 sec2x

LHS = RHS

Hence Proved.

Question 23. cot2x  – tan2x = 4cot2x cosec2x

Solution:

Let us solve LHS,

= cot2x  – tan2x

= cos2x/sin2x – sin2x/cos2x

= [(cos2x)2 – (sin2x)2] / sin2xcos2x

= [(cos2x + sin2x)(cos2x – sin2x)] / sin2xcos2x

= (1 × cos2x) / sin2xcos2x

= 4cos2x / 4sin2xcos2x

= 4(cos2x) / (sin2x)2

= 4(cos2x) / (sin2x) × 1 / (sin2x)

= 4 cot2x cosex2x 

LHS = RHS

Hence Proved

Question 24. cos4x – cos4α = 8(cosx – cosα)(cosx + cosα)(cosx – sinα)(cosx + sinα)

Solution:

Let us solve RHS,

= 8(cosx – cosα)(cosx + cosα)(cosx – sinα)(cosx + sinα)

= 8(cos2x – cos2α)(cos2x – sin2α)

= 8(cos4x – cos2x × sin2α – cos2α × cos2x + cos2α × sin2α)

= 8{cos4x – cos2x(sin2α + cos2α) + cos2α × sin2α}

= 8{cos4x – cos2x + cos2α × (1 – cos2α)}

= 8{cos4x – cos2x + cos2α – cos4α)}

= 8{cos2x(cos2x – 1) + cos2α × (1 – cos2α)}

= 8{1/2 cos2x (2cos2x – 1 – 1) – 1/2 cos2α (2cos2α – 1 -1)}

= 8{1/2 cos2x (cos2x – 1) – 1/2cos2α (cos2α – 1)}

= 8[1/4 {2cos2x (cos2x – 1) – 2cos2x (cos2α – 1)}]

= 8[1/4 {(1 + cos2x)(cos2x – 1) – (1 + cos2α)(cos2α – 1)}]

= 8[1/4 { cos22x – 1 – cos22α + 1}]

= 8[1/8 {2cos22x – 2cos22α}]

= [{(1 + cos4x) – (1 + cos4α)}]

= [1 + cos4x – 1 – cos4α]

= cos4x – cos4α 

LHS = RHS

Hence proved

Question 25. sin3x + sin2x – sinx = 4 sinx cos(x/2) cos(3x/2)

Solution:

Let us solve LHS,

= sin3x + sin2x – sinx

= sin3x + 2sin(2x – x)/2 cos(2x + x)/2

= sin3x + 2sin(x/2) cos(3x/2)

= 2sin(3x/2) cos(3x/2) + 2sin(3x/2) cos(x/2)

= 2cos(3x/2)[sin(3x/2) cos(x/2)]

= 2cos(3x/2)[2sin(3x/2+x/2)/2 cos(3x/2 – x/2)/2]

= 2cos(3x/2)[2sinx cos(x/2)]

= 4 sinx cos(x/2) cos(3x/2)

LHS = RHS

Hence proved.

Question 26. tan82\frac{1}{2}° = (√3 + √2)(√2 + 1) = √2 + √3 + √4 + √6

Solution:

Let us solve LHS,

tan(82.5)° = tan(90 – 7.5)° = cot(7.5)° = 1/ tan(7.5)°

We have,

tan(x/2) = sinx/(1 + cosx)

Now on putting x = 15°, we get

tan(15/2) = sin15°/(1 + cos15°)

= sin(45-30)°/{1 + cos(45-30)°}

 = (sin45°cos30° – sin30°cos45°) / (1 + cos45° sin30°)

\frac{(\frac{1}{√2})(\frac{√3}{2})-(\frac{1}{2})(\frac{1}{√2})}{1+(\frac{1}{√2})(\frac{√3}{2})+(\frac{1}{√2})(\frac{1}{2})}

\frac{\frac{√3}{2√2}-\frac{1}{2√2}}{1+\frac{√3}{2√3}+\frac{1}{2√2}}

\frac{√3-1}{2√2+√3+1}

Now,

tan(82.5)° = 1/tan(7.5)°

 = (2√2 + √3 + 1)/(√3 – 1)

= (2√2 + √3 + 1)/(√3 – 1) × (√3 + 1)/(√3 + 1)

= [√3 + 1(2√2 + √3 + 1)] / [(√3)2 – 12]

= (2√6 + 3 + √3 + 2√2 + √3 + 1) / (3 – 1)

= (2√6 + 2√3 + 2√2 + 4) / (2)

= √6 + √3 + √2 + 2

= √2 + √3 + √4 + √6        …..(i)

= √6 + √3 + 2 + √2

= √3(√2 + 1) + √2(√2 + 1)

= (√3 + √2)(√2 + 1)    …..(ii)

From equ (i) and (ii), we get 

tan(82.5)° = (√3 + √2)(√2 + 1) = √2 + √3 + √4 + √6

LHS = RHS

Hence proved

Question 27. cot22\frac{1}{2}° = √2 + 1

Solution:

As we know that, π/8 = 22\frac{1}{2}° = 45°

Let A = 22\frac{1}{2}°

By using the identity cot2A = (cot2A – 1)/2cotA, we get

cot45° = {cot2(22\frac{1}{2})° – 1} / 2cot(22\frac{1}{2}°

⇒ 1 = {cot2(22\frac{1}{2})° – 1} / 2cot(22\frac{1}{2}

⇒ 2cot(22\frac{1}{2})° – cot2(22\frac{1}{2})° + 1 = 0

⇒ cot2(22\frac{1}{2})° – 2cot(22\frac{1}{2})° – 1 = 0

⇒ { cot2(22\frac{1}{2}) – 2cot(22\frac{1}{2})° + 1} – 2 = 0

⇒ { cot(22\frac{1}{2})° – 1}2 = 2

⇒ cot(22\frac{1}{2})° – 1 = √2

⇒ cot(22\frac{1}{2})° = √2 + 1

LHS = RHS

Hence proved

Question 28 (i). If cosx = (-3/5) and x lies in the 3rd quadrant, find the values of cos(x/2), sin(x/2), sin2x.

Solution:

Given that, 

cosx = (-3/5)

⇒  cosx = cos2(x/2) – sin2(x/2)

⇒  -3/5 = 2cos2(x/2) – 1

⇒ 1 – 3/5 = 2cos2(x/2)

⇒ 2/5 = 2cos2(x/2)

⇒ 1/5 = cos2(x/2)

⇒ cos(x/2) = ± √(1/5)

Also, given that x lies in 3rd quardant, so x/2 lies in 2nd quadrant.

cos(x/2) = – √(1/5)

 Again,

cosx = cos2(x/2) – sin2(x/2)

⇒ -3/5 = (- √(1/5))2 – sin2(x/2)

⇒ – 3/5 = 1/5 – sin2(x/2)

⇒ -1/5 -3/5 = -sin2(x/2)

⇒ 4/5 = sin2(x/2)

⇒ sin(x/2) = ± 2/√5

It is given x lies in 3rd quardant, so x/2 lies in 2nd quadrant.

sin(x/2) = 2/√5

Now,

sinx = √(1 – cos2x)

= √(1 – (-3/5))2

= √(1 – 9/25)

=  ± 4/5

It is given x llies in 3rd quardant, so sinx is negative.

sinx = – 4/5

sin2x = 2 sinx cosx

= 2 (-4/5) (-3/5)

= 24/25

Hence, the value of cos(x/2) = – √(1/5), sin(x/2) = 2/√5, and sin2x = 24/25.

Question 28 (ii). If cosx = (-3/5) and x lies in the 3rd quadrant, find the values of sin2x and sin(x/2).

Solution:

Given that, 

 cosx = (-3/5)

 sinx = \sqrt{1-cos^2x}=\sqrt{1-(-3/5)^2}

⇒ sinx = ± 4/5

Here, x lies in the second quadant

So, sinx = 4/5

As we know that, 

sin2x = 2 sinx cosx

sin2x = 2 × 4/5 × (-3/5) = (-24/25)

Now,

cosx = 1 – 2 sin2(x/2)

⇒ 2sin2(x/2) = 1 – (-3/5) = 8/5

⇒ sinx2(x/2) = 4/5

⇒ sin(x/2) = ± 2/√5

Since x lies in the second quadrant,

x/2 lies in the first quadrant

So, sin(x/2) = 2/√5

Hence, the value of sin2x = (-24/25) and sin(x/2) = 2/√5

Question 29. If sinx = √5/3 and x lies in 2rd quadrant, find the values of cos(x/2), sin(x/2) and tan(x/2).

Solution:

Given that, sinx = √5/3

As we know that sinx = P/H 

So, P = √5, H = 3 and B = 2

Now, cosx = B/H = -2/3

So, 

cos(x/2) = √{(1 + cosx)/2} = √{(1 – 2/3)/2} = 1/√6

sin(x/2) = √{(1 – cosx)/2} = √{(1 + 2/3)/2} = √(5/6)

tan(x/2) = sin(x/2)/cos(x/2) = {√(5/6)} / (1/√6) = √5

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