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# Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Miscellaneous Exercise on Chapter 3

• Last Updated : 22 Mar, 2021

### Question 1. 2cos(Ï€/13)cos(9Ï€/13) + cos(3Ï€/13) + cos(5Ï€/13) = 0

Solution:

Let’s take L.H.S = 2cos(Ï€/13)cos(9Ï€/13) + cos(3Ï€/13) + cos(5Ï€/13)

Rearranging terms, we get

= 2cos(Ï€/13)cos(9Ï€/13)+ cos(5Ï€/13) + cos(3Ï€/13)

By using factorization formula,

cos A + cos B = 2 cos((A + B) / 2) cos((A – B) / 2)

We get

= 2cos(Ï€/13)cos(9Ï€/13)+ 2cos((5Ï€ + 3Ï€)/(2 x13))cos((5Ï€ – 3Ï€)/(2×13))

= 2cos(Ï€/13)cos(9Ï€/13)+ 2cos((8Ï€)/(2×13))cos((2Ï€)/(2×13))

= 2cos(Ï€/13)cos(9Ï€/13)+ 2cos(4Ï€/13)cos(Ï€/13)

= 2cos(Ï€/13)(cos(9Ï€/13) + cos(4Ï€/13))

= 2cos(Ï€/13)(2cos((9Ï€ + 4Ï€)/(2 x13))cos((9Ï€ – 4Ï€)/(2×13)))

= 2cos(Ï€/13)(2cos(Ï€/2)cos(5Ï€ /(2×13)))

As we know that,

cos (Ï€ /2) = 0

= 2cos(Ï€/13)(0 x cos(5Ï€ /(2×13)))

= 0

L.H.S = R.H.S

Hence Proved.

### Question 2. (sin 3x + sin x)sin x + (cos 3x – cos x)cos x = 0

Solution:

Let’s take L.H.S = (sin 3x + sin x)sin x + (cos 3x – cos x)cos x

By using factorization formula,

sin A + sin B = 2 sin((A + B) / 2) cos((A – B) / 2)

and

cos A – cos B = -2 sin((A + B) / 2) sin((A – B) / 2)

We get

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

= (2 sin((4x) / 2) cos((2x) / 2))sin x + (-2 sin((4x) / 2) sin((2x) / 2))cos x

= 2 sin2x cos x sin x – 2 sin 2x sin x cos x

= 0

L.H.S = R.H.S

Hence Proved.

### Question 3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2((x + y)/2)

Solution:

Let’s take L.H.S = (cos x + cos y)2 + (sin x – sin y)2

By using factorization formula,

cos A + cos B = 2 cos((A + B) / 2) cos((A – B) / 2)

and

sin A – sin B = 2 cos((A + B) / 2) sin((A – B) / 2)

We get

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

= 4cos2((x + y) / 2) cos2((x – y) / 2)) + 4cos2((x + y) / 2) sin2((x – y) / 2))

= 4cos2((x + y) / 2) (cos2((x – y) / 2) + sin2((x – y) / 2))

As we know that,

sin2 X + cos2 X = 1

Therefore,

= 4cos2((x + y) / 2) (1)

= 4cos2((x + y) / 2)

L.H.S = R.H.S

Hence Proved.

### Question 4. (cos x – cos y)2 + (sin x – sin y)2 = 4 sin2((x – y)/2)

Solution:

Let’s take L.H.S = (cos x – cos y)2 + (sin x – sin y)2

By using factorization formula,

cos A – cos B = 2 sin((A + B) / 2) sin((A – B) / 2)

and

sin A – sin B = 2 cos((A + B) / 2) sin((A – B) / 2)

We get

= (-2 sin((x + y) / 2) sin((x – y) / 2)))2 + (2 cos((x + y) / 2) sin((x – y) / 2)))2

= 4sin2((x + y) / 2) sin2((x – y) / 2)) + 4cos2((x + y) / 2) sin2((x – y) / 2))

= 4sin2((x – y) / 2) (sin2((x + y) / 2) + cos2((x + y) / 2))

As we know that,

sin2 X + cos2 X = 1

Therefore,

= 4sin2((x – y) / 2) (1)

= 4sin2((x – y) / 2)

L.H.S = R.H.S

Hence Proved.

### Question 5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

Let’s take L.H.S = sin x + sin 3x + sin 5x + sin 7x

= sin 7x + sin x + sin 5x + sin 3x

By using factorization formula,

sin A + sin B = 2 sin((A + B) / 2) cos((A – B) / 2)

We get

L.H.S = 2 sin((7x + x) / 2) cos((7x – x) / 2) + 2 sin((5x + 3x) / 2) cos((5x – 3x) / 2)

= 2 sin((8x) / 2) cos((6x) / 2) + 2 sin((8x) / 2) cos((2x) / 2)

= 2 sin 4x (cos 3x + cos x)

Again by using factorization formula,

cos A + cos B = 2 cos((A + B)/2) cos((A – B)/2)

We get

= 2 sin 4x(2 cos((3x + x)/2) cos((3x – x)/2))

= 4 sin 4x cos (4x/2) cos (2x/2)

= 4 cos x cos 2x sin 4x

L.H.S = R.H.S

Hence Proved.

### Question 6.

Solution:

Let’s take L.H.S =

By using factorization formula.

and

We get

= sin 6x/cos 6x (âˆµ sinÎ¸â€‹/cosÎ¸ = tanÎ¸â€‹)

= tan 6x

L.H.S = R.H.S

Hence Proved.

### Question 7. sin 3x + sin 2x – sin x = 4 sin x cos (x/2) cos (3x/2)

Solution:

Lets take L.H.S = sin 3x + sin 2x – sin x

By using factorization formula,

sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

we get

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

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

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

We know that,

sin 2a = 2 sin a cos a

Therefore,

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

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

Again by using factorization formula,

sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)

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

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

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

L.H.S = R.H.S

Hence Proved.

### Question 8. tan x = -4/3, x is in Quadrant II

Solution:

Given: x is in 2nd Quadrant

i.e

90Â° < x < 180Â°

On dividing by 2 throughout we get,

90Â°/2 < x/2 < 180Â°/2

45Â° < x/2 < 90Â°

Thus x/2 lies in 1st Quadrant.

Since Sine, cosine and tangent of any angle is positive in1st Quadrant.

Therefore, sin x/2, cos x/2, tan x/2 are positive.

Given:

tan x = – 4/3        ..(1)

By using double angle formula,

tan 2x = 2 tan(x) / (1 – tan2(x))

i.e tan (2x/2) = 2 tan(x/2) / (1 – tan2(x/2))

i.e tan (x) = 2 tan(x/2) / (1 – tan2(x/2))         ..(2)

From eq(1) and (2) we have

– 4/3 = 2 tan(x/2) / (1 – tan2(x/2))

-4 + 4tan2(x/2) = 6 tan x/2

2tan2(x/2) – 3 tan(x/2) – 2 = 0

2tan2(x/2) – 4tan(x.2) + tan(x/2) – 2 = 0

2tan(x/2)(tan(x/2) – 2) + 1(tan(x/2) – 2) = 0

(2tan(x/2) + 1) (tan(x/2) – 2) = 0

2tanx(x/2) + 1 = 0 or tan(x/2) – 2 = 0

tan x/2 = -1/2 or tan x/2 = 2

But tan x/2 is Positive,

Therefore

tan (x/2) = 2           ..(3)

By identity,

1 + tan2a = sec2a

Therefore,

1 + tan2x/2 = sec2x/2

sec2x/2 = 1 + 22

sec2x/2 = 1 + 4

sec x/2 = Â± (5)1/2

Since sec x/2 lies in 1st Quadrant.

Therefore, sec x/2 = (5)1/2

We know that

cos a = 1 / sec a

Therefore,

cos x/2 = 1 / sec (x/2)

cos x/2 = 1/âˆš5 = âˆš5/5

Also we have,

sin2a + cos2a = 1

sin2x/2 = (1 – cos2x/2)

sin2x/2 = (1 – 1/5)

sin x/2 = Â±(4/5)1/2

Since sin x/2 is in 1st Quadrant,

Therefore, sin x/2 = 2/âˆš5 = 2âˆš5/5

Hence, the values are,

sin x/2 = 2âˆš5/5, cos x/2 = âˆš5/5, tan x/2 = 2

### Question 9. cos x = -1/3, x is in Quadrant III

Solution:

Given: x is in 3rd Quadrant.

i.e 180Â° < x < 270Â°

i.e 180Â°/2 < x/2 < 270Â°.2

i.e 90Â° < x/2 < 135Â°

Thus, x/2 lies in 2nd Quadrant.

Therefore, sin x/2 is positive while cos x/2, tan x/2 are negative.

Here,

cos x = -1/3          ..(1)

By using double angle formula

cos 2x = 2cos2x – 1

cos 2(x/2) = 2cos2(x/2) – 1

cos x = 2cos2(x/2) – 1

2cos2(x/2) = 1 + cos x

2cos2(x/2) = 1 – 1/3          (From eq(1))

2cos2(x/2) = (3 – 1)/3

cos2(x/2) = 1/3

cos x/2 = Â±(1/3)1/2

Since cos x/2 is negative

Therefore

cos x/2 = -1/âˆš3 = -âˆš3/3

We know that,

sin2a + cos2a = 1

Therefore,

sin2(x/2) + cos2(x/2) = 1

sin2(x/2) = 1 – cos2(x/2)

sin2(x/2) = 1 – (-1/âˆš3)2

sin2(x/2) = (3 -1)/3

sin x/2 = Â±(2/3)1/2

But sin x/2 is positive.

Therefore, sin x/2 = (2/3)1/2 = âˆš6/3

Since tan a = sin a / cos a

Therefore,

tan x/2 = sin x/2 / cos x/2

tan x/2 = (2/3)1/2/ -(1/3)1/2

tan x/2 = -âˆš2

Therefore, the values are,

sin x/2 = âˆš6/3, cos x/2 = -âˆš3/3, tan x/2 = -âˆš2

### Question 10. sin x = 1/3, x is in Quadrant II

Solution:

Given: x is in 2nd Quadrant

i.e

90Â° < x < 180Â°

On dividing by 2 throughout we get,

90Â°/2 < x/2 < 180Â°/2

45Â° < x/2 < 90Â°

Thus x/2 lies in 1st Quadrant.

Since Sine, cosine and tangent of any angle is positive in1st Quadrant.

Therefore, sin x/2, cos x/2, tan x/2 are positive.

We know that,

cos2x + sin2x = 1

cos2x = 1 – sin2x

cos2x = 1 – (1/4)2          (Given sin x = 1/4)

cos2x = 1 – 1/16

cos2x = (16 – 1)/16

cos2x = 15/16

cos x = Â± (âˆš15/4)

Since x is in 2nd Quadrant,

Therefore, cos x = – (âˆš15/4)

By using double angle formula,

cos 2x = 1 – 2sin2x

cos 2(x/2) = 1 – 2sin2(x/2)

cos x = 1 – 2sin2(x/2)

sin2(x/2) = (1 – cos x)/2

sin2(x/2) = (1 – (-âˆš15/4)/2

sin2(x/2) = (4 + âˆš15)/8

sin x/2 = Â±((4 + âˆš15)/8)1/2

But sin x/2 is Positive in 1st Quadrant.

Therefore, sin x/2 = ((4 + âˆš15)/8)1/2

Also,

sin2x/2 + cos2x/2 = 1

cos2x/2 = 1 – sin2x/2

cos2x/2 = 1 – (4 + âˆš15)/8

cos2x/2 = (8 – (4 + âˆš15))/8

cos2x/2 = (8 – 4 – âˆš15)/8

cos2x/2 = (8 – 4 – âˆš15)/8

cos x/2 = Â±((4 – âˆš15)/8)1/2

But cos x/2 is negative in 1st Quadrant.

Therefore, cos x/2 = ((4 – âˆš15)/8)1/2

Since,

tan x/2 = sin x/2 / cos x/2

Therefore,

tan x/2 = ((4 + âˆš15)/8)1/2/-((4 – âˆš15)/8)1/2

tan x/2 = ((4 + âˆš15) / (4 – âˆš15))1/2

tan x/2 = (((4 + âˆš15)(4 + âˆš15)) / ((4 – âˆš15)(4 + âˆš15)))1/2

tan x/2 = ((4 + âˆš15)2/(42 – 15))1/2

tan x/2 = (4 + âˆš15)

Therefore, the values are,

sin x/2 = ((4 + âˆš15)/8)1/2, cos x/2 = ((4 – âˆš15)/8)1/2, tan x/2 = (4 + âˆš15)

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