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

### Question 16. cos2 (Ï€/4 – x) – sin2 (Ï€/4 – x) = sin 2x

Solution:

Let us solve LHS,

= cos2 (Ï€/4 – x) – sin2(Ï€/4 – x)

As we know that,

cos2 A – sin2 A = cos 2A

So,

= cos2 (Ï€/4 – x) sin2 (Ï€/4 – x)

= cos 2 (Ï€/4 – x)

= cos (Ï€/2 – 2x)

= sin 2x                       [As we know that, cos (Ï€/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 Ã— 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 Ã— 4 cos2x sin2x]

As we know that,

sin2x + cos2x = 1

So,

= cos2x [(1)2 – 1/4 Ã— (2 cosx sinx)2]

As we know that,

sin2x = 2 sinx cosx

So,

= cos 2x [1 – 1/4 Ã— (sin 2x)2]

= cos 2x [1 – 1/4 Ã— sin22x]

LHS = RHS

Hence proved.

### Question 22. tan (Ï€/4 + x) + tan (Ï€/4 – x) = 2 sec2x

Solution:

Let us solve LHS,

= tan (Ï€/4 + x) + tan (Ï€/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 Ï€/4 = 1

So,

By using formulas, we get

(a – b)(a + b) = a2 – b2

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

So,

As we know that,

tan x = sin x/cos x

So,

By using the formulas, we get

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

So,

= 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 Ã— cos2x) / sin2xcos2x

= 4cos2x / 4sin2xcos2x

= 4(cos2x) / (sin2x)2

= 4(cos2x) / (sin2x) Ã— 1 / (sin2x)

= 4 cot2x cosex2x

LHS = RHS

Hence Proved

### Question 24. cos4x – cos4Î± = 8(cosx – cosÎ±)(cosx + cosÎ±)(cosx – sinÎ±)(cosx + sinÎ±)

Solution:

Let us solve RHS,

= 8(cosx – cosÎ±)(cosx + cosÎ±)(cosx – sinÎ±)(cosx + sinÎ±)

= 8(cos2x – cos2Î±)(cos2x – sin2Î±)

= 8(cos4x – cos2x Ã— sin2Î± – cos2Î± Ã— cos2x + cos2Î± Ã— sin2Î±)

= 8{cos4x – cos2x(sin2Î± + cos2Î±) + cos2Î± Ã— sin2Î±}

= 8{cos4x – cos2x + cos2Î± Ã— (1 – cos2Î±)}

= 8{cos4x – cos2x + cos2Î± – cos4Î±)}

= 8{cos2x(cos2x – 1) + cos2Î± Ã— (1 – cos2Î±)}

= 8{1/2 cos2x (2cos2x – 1 – 1) – 1/2 cos2Î± (2cos2Î± – 1 -1)}

= 8{1/2 cos2x (cos2x – 1) – 1/2cos2Î± (cos2Î± – 1)}

= 8[1/4 {2cos2x (cos2x – 1) – 2cos2x (cos2Î± – 1)}]

= 8[1/4 {(1 + cos2x)(cos2x – 1) – (1 + cos2Î±)(cos2Î± – 1)}]

= 8[1/4 { cos22x – 1 – cos22Î± + 1}]

= 8[1/8 {2cos22x – 2cos22Î±}]

= [{(1 + cos4x) – (1 + cos4Î±)}]

= [1 + cos4x – 1 – cos4Î±]

= cos4x – cos4Î±

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.  = (âˆš3 + âˆš2)(âˆš2 + 1) = âˆš2 + âˆš3 + âˆš4 + âˆš6

Solution:

Let us solve LHS,

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

We have,

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

Now on putting x = 15Â°, we get

tan(15/2) = sin15Â°/(1 + cos15Â°)

= sin(45-30)Â°/{1 + cos(45-30)Â°}

= (sin45Â°cos30Â° – sin30Â°cos45Â°) / (1 + cos45Â° sin30Â°)

Now,

tan(82.5)Â° = 1/tan(7.5)Â°

= (2âˆš2 + âˆš3 + 1)/(âˆš3 – 1)

= (2âˆš2 + âˆš3 + 1)/(âˆš3 – 1) Ã— (âˆš3 + 1)/(âˆš3 + 1)

= [âˆš3 + 1(2âˆš2 + âˆš3 + 1)] / [(âˆš3)2 – 12]

= (2âˆš6 + 3 + âˆš3 + 2âˆš2 + âˆš3 + 1) / (3 – 1)

= (2âˆš6 + 2âˆš3 + 2âˆš2 + 4) / (2)

= âˆš6 + âˆš3 + âˆš2 + 2

= âˆš2 + âˆš3 + âˆš4 + âˆš6        …..(i)

= âˆš6 + âˆš3 + 2 + âˆš2

= âˆš3(âˆš2 + 1) + âˆš2(âˆš2 + 1)

= (âˆš3 + âˆš2)(âˆš2 + 1)    …..(ii)

From equ (i) and (ii), we get

tan(82.5)Â° = (âˆš3 + âˆš2)(âˆš2 + 1) = âˆš2 + âˆš3 + âˆš4 + âˆš6

LHS = RHS

Hence proved

### Question 27.  = âˆš2 + 1

Solution:

As we know that, Ï€/8 =  = 45Â°

Let A =

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

cot45Â° = {cot2()Â° – 1} / 2cot()Â°

â‡’ 1 = {cot2()Â° – 1} / 2cot()Â°

â‡’ 2cot()Â° – cot2()Â° + 1 = 0

â‡’ cot2()Â° – 2cot()Â° – 1 = 0

â‡’ { cot2() – 2cot()Â° + 1} – 2 = 0

â‡’ { cot()Â° – 1}2 = 2

â‡’ cot()Â° – 1 = âˆš2

â‡’ cot()Â° = âˆš2 + 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)

â‡’  cosx = cos2(x/2) – sin2(x/2)

â‡’  -3/5 = 2cos2(x/2) – 1

â‡’ 1 – 3/5 = 2cos2(x/2)

â‡’ 2/5 = 2cos2(x/2)

â‡’ 1/5 = cos2(x/2)

â‡’ cos(x/2) = Â± âˆš(1/5)

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

cos(x/2) = – âˆš(1/5)

Again,

cosx = cos2(x/2) – sin2(x/2)

â‡’ -3/5 = (- âˆš(1/5))2 – sin2(x/2)

â‡’ – 3/5 = 1/5 – sin2(x/2)

â‡’ -1/5 -3/5 = -sin2(x/2)

â‡’ 4/5 = sin2(x/2)

â‡’ sin(x/2) = Â± 2/âˆš5

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

sin(x/2) = 2/âˆš5

Now,

sinx = âˆš(1 – cos2x)

= âˆš(1 – (-3/5))2

= âˆš(1 – 9/25)

=  Â± 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) = – âˆš(1/5), sin(x/2) = 2/âˆš5, 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 =

â‡’ sinx = Â± 4/5

Here, x lies in the second quadrant

So, sinx = 4/5

As we know that,

sin2x = 2 sinx cosx

sin2x = 2 Ã— 4/5 Ã— (-3/5) = (-24/25)

Now,

cosx = 1 – 2 sin2(x/2)

â‡’ 2sin2(x/2) = 1 – (-3/5) = 8/5

â‡’ sinx2(x/2) = 4/5

â‡’ sin(x/2) = Â± 2/âˆš5

Since x lies in the second quadrant,

x/2 lies in the first quadrant

So, sin(x/2) = 2/âˆš5

Hence, the value of sin2x = (-24/25) and sin(x/2) = 2/âˆš5

### Question 29. If sinx = âˆš5/3 and x lies in 2nd quadrant, find the values of cos(x/2), sin(x/2) and tan(x/2).

Solution:

Given that, sinx = âˆš5/3

As we know that sinx = P/H

So, P = âˆš5, H = 3 and B = 2

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

So,

cos(x/2) = âˆš{(1 + cosx)/2} = âˆš{(1 – 2/3)/2} = 1/âˆš6

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

tan(x/2) = sin(x/2)/cos(x/2) = {âˆš(5/6)} / (1/âˆš6) = âˆš5

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Previous
Next