# 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 – cos

^{2}x)So,

â‡’ (1/4)

^{2}= (1 – cos^{2}x)â‡’ (1/16) – 1 = – cos

^{2}xcosx = 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 cos

^{2}(x/2) – 1So,

â‡’ – âˆš15/4 = 2cos

^{2}(x/2) – 1â‡’ cos

^{2}(x/2) = – âˆš15/8 + 1/2cos(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 = cos

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

^{2}– sin^{2}(x/2)â‡’ sin

^{2}(x/2) = (4 + âˆš15)/8â‡’ 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 – cos

^{2}x)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 – tan

^{2}x)= 2(3/4) / {1 – (3/4)

^{2}}= 2(3/4) / (1 – 9/16)

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

= 24/7

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 – cos

^{2}x)So,

â‡’ (4/5)

^{2}= 1 – cos^{2}xâ‡’ 16/25 – 1 = -cos

^{2}xâ‡’ 9/25 = cos

^{2}xâ‡’ 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 – 2sin

^{2}x)= 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 – tan

^{2}B)= (2 Ã— 1/3)(1 – 1/9) = 3/4

So, cos2A = (1 – tan

^{2}A)/(1 + tan^{2}A)= {1-(1/7)

^{2}}/{1+(1/7)^{2}}= 48/50

= 24/25

And sin4B = 2tan2B / (1 + tan

^{2}2B)= {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/(a^{2 }+ b^{2})

**Solution:**

Given that,

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

Prove: sin(Î± + Î²) = 2ab/(a

^{2 }+ b^{2})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 + tan

^{2}x)=

= 2ab/(a

^{2 }+ b^{2})LHS = RHS

Hence proved

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

sin

^{2}Î± + sin^{2}Î² + 2sinÎ±sinÎ² + cos^{2}Î± + cos^{2}Î² + 2cosÎ±cosÎ² = a^{2 }+ b^{2}â‡’ 1 + 1 + 2(sinÎ±sinÎ² + cosÎ±cosÎ²) = a

^{2 }+ b^{2}â‡’ 2(sinÎ±sinÎ² + cosÎ±cosÎ²) = a

^{2 }+ b^{2 }– 2â‡’ 2cos(Î± – Î²) = a

^{2 }+ b^{2 }– 2â‡’ cos(Î± – Î²) = (a

^{2 }+ b^{2 }– 2)/2Hence 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‡’ tan

^{2}(x/2) = tan^{2}(Î±/2)tan^{2}(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‡’ cos

^{2}xcosÎ± = cos^{2}x(cos^{2}Î± + sin^{2}Î±) – sin^{2}Î±â‡’ cos

^{2}x(1 – cosÎ±) = sin^{2}Î±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:

(cos

^{2}Î± + cos^{2}Î² + cosÎ±cosÎ²) + (sin^{2}Î± + sin^{2}Î² + 2sinÎ±sinÎ²) = 1/9 + 1/161 + 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 – sin

^{2}Î±)So,

= âˆš{1 – (4/5)

^{2}}= 3/5

Also, sinÎ² = âˆš(1 – cos

^{2}Î²)= 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 – tan

^{2}Î¸) + b(2tanÎ¸) = c(1 + tan^{2}Î¸)(c + a)tan

^{2}Î¸ + 2btanÎ¸ + c – a = 0

(i)As Î± and Î² are rootsSo, sum of the roots:

tanÎ± + tanÎ² = 2b / (c + a)

(ii)As Î± and Î² are rootsSo, 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

cos

^{2}Î± + cos^{2}Î² + 2 cosÎ± cosÎ² = 0 ….(i)Similarly

sinÎ± + sinÎ² = 0

On squaring on both sides, we get

sin

^{2}Î± + sin^{2}Î² + 2 sinÎ± sinÎ² = 0 …..(ii)Now, subtract eq (ii) from (i), we get

â‡’ (cos

^{2}Î± + cos^{2}Î² + 2 cosÎ± cosÎ²) – (sin^{2}Î± + sin^{2}Î² + 2 sinÎ± sinÎ²) = 0â‡’ cos

^{2}Î± – sin^{2}Î± + cos^{2}Î² – sin^{2}Î² + 2(cosÎ± cosÎ² – sinÎ± sinÎ²) = 0â‡’ cos2Î± + cos2Î² + 2cos(Î± + Î²) = 0

â‡’ cos2Î± + cos2Î² = -2cos(Î± + Î²)

Hence proved.

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