# Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 3

### Question 30(i). If 0 â‰¤ xâ‰¤ Ï€ and x lies in the 2nd quadrant such that sinx = 1/4, Find the values of cos(x/2), sin(x/2), and tan(x/2).

Solution:

Given that,

sinx = 1/4

As we know that, sinx = âˆš(1 – cos2x)

So,

â‡’ (1/4)2 = (1 – cos2x)

â‡’ (1/16) – 1 = – cos2x

cosx = Â± âˆš15/4

It is given that x is in 2nd quadrant, so cosx is negative.

cosx = – âˆš15/4

Now,

As we know that, cosx = 2 cos2(x/2) – 1

So,

â‡’ – âˆš15/4 = 2cos2(x/2) – 1

â‡’ cos2(x/2) = – âˆš15/8 + 1/2

cos(x/2) = Â± (4-âˆš15)/8

It is given that, x is in 2nd quadrant, so cos(x/2) is positive.

cos(x/2) = (4 – âˆš15)/8

Again,

cosx = cos2(x/2) – sin2(x/2)

â‡’ – âˆš15/4 = {(4 – âˆš15)/8}2 – sin2(x/2)

â‡’ sin2(x/2) = (4 + âˆš15)/8

â‡’ sin(x/2) = Â± âˆš{(4 + âˆš15)/8} = âˆš{(4 + âˆš15)/8}

Now,

tan(x/2) = sin(x/2) / cos(x/2)

= 4 + âˆš15

Hence, the value of cos(x/2) = (4 – âˆš15)/8, sin(x/2) = âˆš{(4 + âˆš15)/8}, and tan(x/2) = 4 + âˆš15 .

### Question 30(ii). If cosx = 4/5 and x is acute, find tan2x.

Solution:

Given that,

cosx = 4/5

As we know that, sinx = âˆš(1 – cos2x)

So,

= âˆš(1 – (4/5)2)

= âˆš(1 – 16/25)

= âˆš{(25 – 16)/25}

= âˆš(9/25)

= 3/5

Since, tanx = sinx/cosx, so

= (3/5) / (4/5)

= 3/4

As we know that,

tan2x = 2tanx / (1 – tan2x)

= 2(3/4) / {1 – (3/4)2}

= 2(3/4) / (1 – 9/16)

= (3/2) / (7/16)

Hence, the value of tan2x is 24/7

### Question 30(iii). If sinx = 4/5 and 0 < x < Ï€/2, then find the value of sin4x.

Solution:

Given that,

sinx = 4/5

As we know that, sinx = âˆš(1 – cos2x)

So,

â‡’ (4/5)2 = 1 – cos2x

â‡’ 16/25 – 1 = -cos2x

â‡’ 9/25 = cos2x

â‡’ cosx = Â±3/5

It is given that, x is ln the 1st quadrant

So, cosx = 3/5

Now,

sin4x = 2 sin2x cos2x

= 2 (2 sinx cosx)(1 – 2sin2x)

= 2(2 Ã— 4/5 Ã— 3/5)(1 – 2(4/5)2)

= 2(24/25)(1-32/25)

= 2(24/25)((25-32)/25)

= 2(24/25)(-7/25)

= -336/625

Hence, the value of sin4x is (- 336/625)

### Question 31. If tanx = b/a, then find the value of

Solution:

We have to find the value of

So,

It is given that tanx = b/a, so

Hence, the value of  is

### Question 32. If tanA = 1/7 and tanB = 1/3, show that cos2A = sin4B

Solution:

Given that, tanA = 1/7 and tanB = 1/3

Show: cos2A = sin4B

As we know that, tan2B = 2tanB / (1 – tan2B)

= (2 Ã— 1/3)(1 – 1/9) = 3/4

So, cos2A = (1 – tan2A)/(1 + tan2A)

= {1-(1/7)2}/{1+(1/7)2}

= 48/50

= 24/25

And sin4B = 2tan2B / (1 + tan22B)

= {2 Ã— 3/4}{1 + (3/4)2}

= 24/25

Hence, cos2A = sin4B

### Question 33. cos7Â° cos14Â° cos28Â° cos56Â° = sin68Â°/16cos83Â°

Solution:

Lets solve LHS

= cos7Â° cos14Â° cos28Â° cos56Â°

On dividing and multiplying by 2sin7Â°, we get

Ã— 2sin7Â° Ã— cos7Â° Ã— cos14Â° Ã— cos28Â° Ã— cos56Â°

Ã— cos28Â° Ã— cos56Â°

Ã— cos56Â°

LHS = RHS

Hence proved.

### Question 34. Proved that, cos(2Ï€/15)cos(4Ï€/15)cos(8Ï€/15)cos(16Ï€/15) = 1/16

Solution:

Let’s solve LHS

= cos(2Ï€/15)cos(4Ï€/15)cos(8Ï€/15)cos(16Ï€/15)

On dividing and multiplying by 2sin(2Ï€/15), we get

= 1/16

LHS = RHS

Hence proved.

### Question 35. Proved that, cos(Ï€/5)cos(2Ï€/5)cos(4Ï€/5)cos(8Ï€/5) = -1/16

Solution:

Lets solve LHS

= cos(Ï€/5)cos(2Ï€/5)cos(4Ï€/5)cos(8Ï€/5)

On dividing and multiplying by 2sin(2Ï€/5), we get

Ã— 2sin(Ï€/5)cos(Ï€/5)cos(2Ï€/5)cos(4Ï€/5)cos(8Ï€/5)

(sin(2Ï€/5)cos(2Ï€/5)cos(4Ï€/5)cos(8Ï€/5))

[2sin(2Ï€/5)cos(2Ï€/5)cos(4Ï€/5)cos(8Ï€/5)]

[sin(4Ï€/5)cos(4Ï€/5)cos(8Ï€/5)]

[2sin(4Ï€/5)cos(4Ï€/5)cos(8Ï€/5)]

= [sin(8Ï€/5)cos(8Ï€/5)]

=[2sin(8Ï€/5)cos(8Ï€/5)]

=

= -1/16

LHS = RHS

Hence proved.

### Question 36. Proved that, cos(Ï€/65)cos(2Ï€/65)cos(4Ï€/65)cos(8Ï€/65)cos(16Ï€/65)cos(32Ï€/65) = 1/64

Solution:

Lets solve LHS

= cos(Ï€/65)cos(2Ï€/65)cos(4Ï€/65)cos(8Ï€/65)cos(16Ï€/65)cos(32Ï€/65)

Now on dividing and multiplying by 2sin(Ï€/65), we get

Ã— 2sin(Ï€/65)cos(Ï€/65)cos(2Ï€/65)cos(4Ï€/65)cos(8Ï€/65)cos(16Ï€/65)cos(32Ï€/65)

Ã— [cos(2Ï€/65) Ã— cos(4Ï€/65) Ã— cos(8Ï€/65) Ã— cos(16Ï€/65) Ã— cos(32Ï€/65)]

Ã— cos(4Ï€/65) Ã— cos(8Ï€/65) Ã— cos(16Ï€/65) Ã— cos(32Ï€/65)

Ã— cos(8Ï€/65) Ã— cos(16Ï€/65) Ã— cos(32Ï€/65)

Ã— cos(16Ï€/65) Ã— cos(32Ï€/65)

Ã— cos(32Ï€/65)

= 1/64

LHS = RHS

Hence proved

### Question 37. If 2tanÎ± = 3tanÎ², prove that tan(Î± – Î²) = sin2Î² / (5 – cos2Î²)

Solution:

Given that,

2tanÎ± = 3tanÎ²

Prove: tan(Î± – Î²) = sin2Î² / (5 – cos2Î²)

Proof:

Lets solve LHS

LHS = RHS

Hence proved.

### Question 38(i). If sinÎ± + sinÎ² = a and cosÎ± + cosÎ² = b, prove that sin(Î± + Î²) = 2ab/(a2 + b2)

Solution:

Given that,

sinÎ± + sinÎ² = a and cosÎ± + cosÎ² = b

Prove: sin(Î± + Î²) = 2ab/(a2 + b2)

Proof:

As we know that,

So       ……(i)

Now, using the identity

…..(ii)

Now on dividing  eq(i) and (ii), we get

tan(Î± + Î²)/2 = a/b

As we know that,

sin2x = 2tanx/(1 + tan2x)

= 2ab/(a2 + b2)

LHS = RHS

Hence proved

### Question 38(ii). If sinÎ± + sinÎ² = a and cosÎ± + cosÎ² = b, prove that cos(Î± – Î²) = (a2 + b2 – 2)/2

Solution:

Given that,

sinÎ± + sinÎ² = a ……(i)

cosÎ± + cosÎ² = b …….(ii)

Now on squaring eq(i) and (ii) and then adding them, we get

sin2Î± + sin2Î² + 2sinÎ±sinÎ² + cos2Î± + cos2Î² + 2cosÎ±cosÎ² = a2 + b2

â‡’ 1 + 1 + 2(sinÎ±sinÎ² + cosÎ±cosÎ²) = a2 + b2

â‡’ 2(sinÎ±sinÎ² + cosÎ±cosÎ²) = a2 + b2 – 2

â‡’ 2cos(Î± – Î²) = a2 + b2 – 2

â‡’ cos(Î± – Î²) = (a2 + b2 – 2)/2

Hence proved.

### Question 39. If 2tan(Î±/2) = tan(Î²/2), prove that cosÎ± =

Solution:

Given that,

2tan(Î±/2) = tan(Î²/2)

Prove: cosÎ± =

Proof:

Let us solve RHS

= cosÎ±

RHS = LHS

Hence proved.

### Question 40. If cosx = , prove that tan(x/2) = Â± tan(Î±/2)tan(Î²/2).

Solution:

Given that,

…..(i)

â‡’

Now, by componendo and dividendo, we get

â‡’

â‡’

â‡’

â‡’

â‡’ tan2(x/2) = tan2(Î±/2)tan2(Î²/2)

â‡’ tan(x/2) = Â±tan(Î±/2)tan(Î²/2)

Hence Proved.

### Question 41. If sec(x + Î±) + sec(x – Î±) = 2secx, prove that cosx = Â± âˆš2 cos(Î±/2).

Solution:

Given that,

sec(x + Î±) + sec(x – Î±) = 2secx

So,

â‡’

â‡’

â‡’

â‡’ cos2xcosÎ± = cos2x(cos2Î± + sin2Î±) – sin2Î±

â‡’ cos2x(1 – cosÎ±) = sin2Î±

â‡’

â‡’ cosx = Â± âˆš2 cos(Î±/2)

Hence Proved

### Question 42. If cosÎ± + cosÎ² = 1/3 and sinÎ± + sinÎ² = 1/4, prove that cos(Î± – Î²)/2 = Â±5/24.

Solution:

Given that,

cosÎ± + cosÎ² = 1/3

sinÎ± + sinÎ² = 1/4, we get

Prove:  cos(Î± – Î²)/2 = Â±5/24

Proof:

(cos2Î± + cos2Î² + cosÎ±cosÎ²) + (sin2Î± + sin2Î² + 2sinÎ±sinÎ²) = 1/9 + 1/16

1 + 1 + 2(cosÎ±cosÎ² + sinÎ±sinÎ²) = 25/144

2 + 2cos(Î± – Î²) = -263/288  …..(i)

Now,

[From (i)]

= 25/576

= Â± 5/24

Hence proved.

### Question 43. If sinÎ± = 4/5 and cosÎ² = 5/13, prove that cos{(Î± – Î²)/2} = 8/âˆš65.

Solution:

Given that,

sinÎ± = 4/5 and cosÎ² = 5/13

As we know that.

cosÎ± = âˆš(1 – sin2Î±)

So,

= âˆš{1 – (4/5)2}

= 3/5

Also, sinÎ² = âˆš(1 – cos2Î²)

= âˆš{1 – (5/13)2}

= 12/13

Now,

cos(Î± – Î²) = cosÎ± cosÎ² + sinÎ± sinÎ²

= (3/5)(5/13)(4/5)(12/13)

= 63/65

Thus,

cos{(Î± – Î²)/2} =

= 8/âˆš65

Hence Proved.

### Question 44. If acos2Î¸ + bsin2Î¸ = c has Î± and Î² as its roots prove that,

(i) tanÎ± + tanÎ² = 2b/(a + c)

(ii) tanÎ± tanÎ² = (c – a)/(c + a)

(iii) tan(Î± + Î²) = b/a

Solution:

As we know that

Now substitute these values in the given equation, we get

a(1 – tan2Î¸) + b(2tanÎ¸) = c(1 + tan2Î¸)

(c + a)tan2Î¸ + 2btanÎ¸ + c – a = 0

(i) As Î± and Î² are roots

So, sum of the roots:

tanÎ± + tanÎ² = 2b / (c + a)

(ii) As Î± and Î² are roots

So, product of roots:

tanÎ± tanÎ² = (c – a) / (c + a)

(iii) tan(Î± + Î²)=

= b/a

Hence proved.

### Question 45. If cosÎ± + cosÎ² = 0 = sinÎ± + sinÎ², then prove that cos2Î± + cos2Î² = -2cos(Î± + Î²).

Solution:

Given that,

cosÎ± + cosÎ² = 0 = sinÎ± + sinÎ²

Prove: cos2Î± + cos2Î² = -2cos(Î± + Î²)

Proof:

cosÎ± + cosÎ² = 0

On squaring on both sides, we get

cos2Î± + cos2Î² + 2 cosÎ± cosÎ² = 0     ….(i)

Similarly

sinÎ± + sinÎ² = 0

On squaring on both sides, we get

sin2Î± + sin2Î² + 2 sinÎ± sinÎ² = 0       …..(ii)

Now, subtract eq (ii) from (i), we get

â‡’ (cos2Î± + cos2Î² + 2 cosÎ± cosÎ²) – (sin2Î± + sin2Î² + 2 sinÎ± sinÎ²) = 0

â‡’ cos2Î± – sin2Î± + cos2Î² – sin2Î² + 2(cosÎ± cosÎ² – sinÎ± sinÎ²) = 0

â‡’ cos2Î± + cos2Î² + 2cos(Î± + Î²) = 0

â‡’ cos2Î± + cos2Î² = -2cos(Î± + Î²)

Hence proved.

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