# Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 2

### Prove the following identities:

### Question 16. cos^{2} (π/4 – x) – sin^{2} (π/4 – x) = sin 2x

**Solution:**

Let us solve LHS,

= cos

^{2}(π/4 – x) – sin^{2}(π/4 – x)As we know that,

cos

^{2}A – sin^{2}A = cos 2ASo,

= cos

^{2}(π/4 – x) sin^{2}(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 cos^{2}x + 8 cos^{4} x

**Solution:**

Let us solve LHS,

= cos 4x

As we know that,

cos 2x = 2 cos

^{2}x – 1So,

cos 4x = 2 cos

^{2}2x – 1= 2(2 cos

^{2}2x – 1)^{2 }– 1= 2[(2 cos

^{2}2x)^{2}+ 1^{2}– 2 × 2 cos^{2}x] – 1= 2(4 cos

^{4}2x + 1 – 4 cos^{2}x) – 1= 8 cos

^{4}2x + 2 – 8 cos^{2}x – 1= 8 cos

^{4}2x + 1 – 8 cos^{2}xLHS = RHS

Hence Proved.

### Question 18. sin 4x = 4 sin x cos^{3} x – 4 cos x sin^{3} x

**Solution:**

Let us solve LHS,

= sin 4x

As we know that,

sin 2x = 2 sin x cos x

cos 2x = cos

^{2}x – sin^{2}xSo,

sin 4x = 2 sin 2x cos 2x

= 2 (2 sin x cos x) (cos

^{2 }x – sin^{2}x)= 4 sin x cos x (cos

^{2}x – sin^{2}x)= 4 sin x cos

^{3}x – 4 sin^{3}x cos xLHS = RHS

Hence proved.

### Question 19. 3(sin x – cos x)^{4} + 6 (sin x + cos x)^{2} + 4 (sin^{6} x + cos^{6} x) = 13

**Solution:**

Let us solve LHS,

= 3(sin x – cos x)

^{4}+ 6 (sin x + cos x)^{2}+ 4 (sin^{6}x + cos^{6}x)As we know that,

(a + b)

^{2}= a^{2}+ b^{2}+ 2ab(a – b)

^{2}= a^{2}+ b^{2}– 2aba

^{3}+ b^{3}= (a+b)(a^{2}+ b^{2}– ab)So,

= 3{(sinx – cosx)

^{2}}^{2}+ 6 {(sinx)^{2}+ (cosx)^{2}+ 2 sinx cosx} + 4 {(sin^{2}x)^{3}+ (cos^{2}x)^{3}}= 3{(sinx)

^{2}+ (cosx)^{2}– 2 sinx cosx}^{2}+ 6(sin^{2}x + cos^{2}x + 2 sinx cosx) + 4{(sin^{2}x + cos^{2}x)(sin^{4}x + cos^{4}x – sin^{2}x cos^{2}x)}= 3(1 – 2 sinx cosx)

^{2}+ 6(1 + 2 sinx cosx) + 4{(1)(sin^{4}x + cos^{4}x – sini^{2}x cos^{2}x)}Since,

sin

^{2}x + cos^{2}x = 1So,

= 3{1

^{2}+ (2 sinx cosx)^{2}– 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin^{2}x)^{2}+ (cos^{2}x)^{2}+ 2 sin^{2}x cos^{2}x – 3 sin^{2}x cos^{2}x}= 3{1 + 4 sin

^{2}x cos^{2}x – 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin^{2}x + cos^{2}x)^{2}– 3 sin^{2}x cos^{2}x}= 3 + 12 sin

^{2}x cos^{2}x – 12 sinx cosx + 6 + 12 sinx cosx + 4{(1)^{2}– 3 sin^{2}x cos^{2}x}= 9 + 12 sin

^{2}x cos^{2}x + 4(1 – 3 sin^{2}x cos^{2}x)= 9 + 12 sin

^{2}x cos^{2}x + 4 – 12 sin^{2}x cos^{2}x= 13

LHS = RHS

Hence proved.

### Question 20. 2(sin^{6}x + cos^{6}x) – 3(sin^{4}x + cos^{4}x) + 1 = 0

**Solution:**

Let us solve LHS,

= 2(sin

^{6}x + cos^{6}x) – 3(sin^{4}x + cos^{4}x) + 1As we know that,

(a + b)

^{2}= a^{2}+ b^{2}+ 2aba

^{3}+ b^{3}= (a + b) (a^{2}+ b^{2}– ab)So,

= 2(sin

^{6}x + cos^{6}x) – 3(sin^{4}x + cos^{4}x) + 1= 2{(sin

^{2}x)^{3}+ (cos^{2}x)^{3}} – 3{(sin^{2}x)^{2}+ (cos^{2}x)^{2}} + 1= 2((sin

^{2}x + cos^{2}x)(sin^{4}x + cos^{4}x – sin^{2}x cos^{2}x) – 3{(sin^{2}x)^{2}+ (cos^{2}x)^{2}+ 2 sin^{2}x cos^{2}x – 2sin^{2}x cos^{2}x} + 1= 2{(1)(sin

^{4}x + cos^{4}x + 2 sin^{2}x cos^{2}x – 3 sin^{2}x cos^{2}x) – 3((sin^{2}x + cos^{2}x)^{2}– 2sin^{2}x cos^{2}x) + 1Since

sin

^{2}x + cos^{2}x = 1So,

= 2{(sin

^{2}x + cos^{2}x)^{2}– 3 sin^{2}x cos^{2}x} – 3{(1)^{2}– 2 sin^{2}x cos^{2}x} + 1= 2{(1)

^{2}– 3 sin^{2}x cos^{2}x} – 3(1 – 2 sin^{2}x cos^{2}x) + 1= 2(1 – 3 sin

^{2}x cos^{2}x) – 3 + 6 sin^{2}x cos^{2}x + 1= 2 – 6 sin

^{2}x cos^{2}x – 2 + 6 sin^{2}x cos^{2}x= 0

LHS = RHS

Hence Proved.

### Question 21. cos^{6} x – sin^{6} x = cos 2x (1 – 1/4 sin^{2} 2x)

**Solution:**

Let us solve LHS,

= cos

^{6}x – sin^{6}xAs we know that,

(a + b)

^{2}= a^{2}+ b^{2}+ 2aba

^{3}– b^{3}= (a – b) (a^{2}+ b^{2}+ ab)So,

cos

^{6}x – sin^{6}x = (cos^{2}x)^{3}– (sin^{2}x)^{3}= (cos

^{2}x – sin^{2}x) (cos^{4}x + sin^{4}x + cos^{2}x sin^{2}x)As we know that,

cos 2x = cos

^{2}x – sin^{2}xSo,

= cos 2x [(cos

^{2}x)^{2}+ (sin^{2}x)^{2}+ 2 cos^{2}x sin^{2}x – cos^{2}x sin^{2}x]= cos 2x [(cos

^{2}x)^{2}+ (sin^{2}x)^{2}– 1/4 × 4 cos^{2}x sin^{2}x]As we know that,

sin

^{2}x + cos^{2}x = 1So,

= 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 × sin

^{2}2x]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,

=

Since, tan π/4 = 1

So,

=

=

By using formulas, we get

(a – b)(a + b) = a

^{2}– b^{2}(a + b)

^{2}= a^{2}+ b^{2}+ 2ab & (a – b)^{2}= a^{2}+ b^{2}– 2abSo,

=

=

=

As we know that,

tan x = sin x/cos x

So,

=

=

=

By using the formulas, we get

cos

^{2}x+ sin^{2}x = 1 & cos 2x = cos^{2}x – sin^{2}xSo,

=

= 2/cos2x

= 2 sec2x

LHS = RHS

Hence Proved.

### Question 23. cot^{2}x – tan^{2}x = 4cot2x cosec2x

**Solution:**

Let us solve LHS,

= cot

^{2}x – tan^{2}x= cos

^{2}x/sin^{2}x – sin^{2}x/cos^{2}x= [(cos

^{2}x)^{2}– (sin^{2}x)^{2}] / sin^{2}xcos^{2}x= [(cos

^{2}x + sin^{2}x)(cos^{2}x – sin^{2}x)] / sin^{2}xcos^{2}x= (1 × cos2x) / sin

^{2}xcos^{2}x= 4cos2x / 4sin

^{2}xcos^{2}x= 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(cos

^{2}x – cos^{2}α)(cos^{2}x – sin^{2}α)= 8(cos

^{4}x – cos^{2}x × sin^{2}α – cos^{2}α × cos^{2}x + cos^{2}α × sin^{2}α)= 8{cos

^{4}x – cos^{2}x(sin^{2}α + cos^{2}α) + cos^{2}α × sin^{2}α}= 8{cos

^{4}x – cos^{2}x + cos^{2}α × (1 – cos^{2}α)}= 8{cos

^{4}x – cos^{2}x + cos^{2}α – cos^{4}α)}= 8{cos

^{2}x(cos^{2}x – 1) + cos^{2}α × (1 – cos^{2}α)}= 8{1/2 cos

^{2}x (2cos^{2}x – 1 – 1) – 1/2 cos^{2}α (2cos^{2}α – 1 -1)}= 8{1/2 cos

^{2}x (cos2x – 1) – 1/2cos^{2}α (cos^{2}α – 1)}= 8[1/4 {2cos

^{2}x (cos2x – 1) – 2cos^{2}x (cos2α – 1)}]= 8[1/4 {(1 + cos2x)(cos2x – 1) – (1 + cos2α)(cos2α – 1)}]

= 8[1/4 { cos

^{2}2x – 1 – cos^{2}2α + 1}]= 8[1/8 {2cos

^{2}2x – 2cos^{2}2α}]= [{(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. = (√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°)

=

=

=

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}– 1^{2}]= (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. = √2 + 1

**Solution:**

As we know that, π/8 = = 45°

Let A =

By using the identity cot2A = (cot

^{2}A – 1)/2cotA, we getcot45° = {cot

^{2}()° – 1} / 2cot()°⇒ 1 = {cot

^{2}()° – 1} / 2cot()°⇒ 2cot()° – cot

^{2}()° + 1 = 0⇒ cot

^{2}()° – 2cot()° – 1 = 0⇒ { cot

^{2}() – 2cot()° + 1} – 2 = 0⇒ { cot()° – 1}

^{2}= 2⇒ cot()° – 1 = √2

⇒ cot()° = √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 = cos

^{2}(x/2) – sin^{2}(x/2)⇒ -3/5 = 2cos

^{2}(x/2) – 1⇒ 1 – 3/5 = 2cos

^{2}(x/2)⇒ 2/5 = 2cos

^{2}(x/2)⇒ 1/5 = cos

^{2}(x/2)⇒ cos(x/2) = ± √(1/5)

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

cos(x/2) = – √(1/5)

Again,

cosx = cos

^{2}(x/2) – sin^{2}(x/2)⇒ -3/5 = (- √(1/5))

^{2}– sin^{2}(x/2)⇒ – 3/5 = 1/5 – sin

^{2}(x/2)⇒ -1/5 -3/5 = -sin

^{2}(x/2)⇒ 4/5 = sin

^{2}(x/2)⇒ sin(x/2) = ± 2/√5

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

sin(x/2) = 2/√5

Now,

sinx = √(1 – cos

^{2}x)= √(1 – (-3/5))

^{2}= √(1 – 9/25)

= ± 4/5

It is given x lies in 3rd quadrant, 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 =

⇒ sinx = ± 4/5

Here, x lies in the second quadrant

So, sinx = 4/5

As we know that,

sin2x = 2 sinx cosx

sin2x = 2 × 4/5 × (-3/5) = (-24/25)

Now,

cosx = 1 – 2 sin

^{2}(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 2nd 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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