NCERT Solutions for Class 11 Science Math Chapter 3 Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Trigonometric Functions are extremely popular among Class 11 Science students for Math Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.

Question 1:

Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30' (iii) 240° (iv) 520°

(i) 25°

We know that 180° = π radian

(ii) –47° 30'

–47° 30' = degree [1° = 60']

degree

(iii) 240°

We know that 180° = π radian

(iv) 520°

We know that 180° = π radian

Question 2:

Find the degree measures corresponding to the following radian measures

.

(i) (ii) – 4 (iii) (iv)

(i)

We know that π radian = 180°

(ii) – 4

We know that π radian = 180°

(iii)

We know that π radian = 180°

(iv)

We know that π radian = 180°

Question 3:

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Number of revolutions made by the wheel in 1 minute = 360

∴Number of revolutions made by the wheel in 1 second =

In one complete revolution, the wheel turns an angle of 2π radian.

Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,

Thus, in one second, the wheel turns an angle of 12π radian.

Question 4:

Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm.

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then

Therefore, forr = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

Question 5:

In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Diameter of the circle = 40 cm

∴Radius (r) of the circle =

Let AB be a chord (length = 20 cm) of the circle.

In ΔOAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ΔOAB is an equilateral triangle.

∴θ = 60° =

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then.

Thus, the length of the minor arc of the chord is.

Question 6:

If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Let the radii of the two circles be and. Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r1, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r2.

Now, 60° =and 75° =

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then.

Thus, the ratio of the radii is 5:4.

Question 7:

Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then.

It is given that r = 75 cm

(i) Here, l = 10 cm

(ii) Here, l = 15 cm

(iii) Here, l = 21 cm

Question 1:

Find the values of other five trigonometric functions if , x lies in third quadrant.

Since x lies in the 3rd quadrant, the value of sin x will be negative.

Question 2:

Find the values of other five trigonometric functions if , x lies in second quadrant.

Since x lies in the 2nd quadrant, the value of cos x will be negative

Question 3:

Find the values of other five trigonometric functions if , x lies in third quadrant.

Since x lies in the 3rd quadrant, the value of sec x will be negative.

Question 4:

Find the values of other five trigonometric functions if , x lies in fourth quadrant.

Since x lies in the 4th quadrant, the value of sin x will be negative.

Question 5:

Find the values of other five trigonometric functions if , x lies in second quadrant.

Since x lies in the 2nd quadrant, the value of sec x will be negative.

∴sec x =

Question 6:

Find the value of the trigonometric function sin 765°

It is known that the values of sin x repeat after an interval of 2π or 360°.

Question 7:

Find the value of the trigonometric function cosec (–1410°)

It is known that the values of cosec x repeat after an interval of 2π or 360°.

Question 8:

Find the value of the trigonometric function

It is known that the values of tan x repeat after an interval of π or 180°.

Question 9:

Find the value of the trigonometric function

It is known that the values of sin x repeat after an interval of 2π or 360°.

Question 10:

Find the value of the trigonometric function

It is known that the values of cot x repeat after an interval of π or 180°.

L.H.S. =

Prove that

L.H.S. =

Prove that

L.H.S. =

Prove that

L.H.S =

Question 5:

Find the value of:

(i) sin 75°

(ii) tan 15°

(i) sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

[sin (x + y) = sin x cos y + cos x sin y]

(ii) tan 15° = tan (45° – 30°)

Prove that:

Prove that:

It is known that

∴L.H.S. =

Prove that

L.H.S. =

Question 10:

Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x

Question 11:

Prove that

It is known that.

∴L.H.S. =

Question 12:

Prove that sin2 6x – sin2 4x = sin 2x sin 10x

It is known that

∴L.H.S. = sin26x – sin24x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= (2 sin 5x cos x) (2 cos 5x sin x)

= (2 sin 5x cos 5x) (2 sin x cos x)

= sin 10x sin 2x

= R.H.S.

Question 13:

Prove that cos2 2x – cos2 6x = sin 4x sin 8x

It is known that

∴L.H.S. = cos2 2x – cos2 6x

= (cos 2x + cos 6x) (cos 2x – 6x)

= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]

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

= sin 8x sin 4x

= R.H.S.

Question 14:

Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (â€“ 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos2 x â€“ 1 + 1)

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

Question 15:

Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x â€“ sin 3x)

= 2 cos 4x. cos x

L.H.S. = R.H.S.

Prove that

It is known that

∴L.H.S =

Prove that

It is known that

∴L.H.S. =

Prove that

It is known that

∴L.H.S. =

Prove that

It is known that

∴L.H.S. =

Prove that

It is known that

∴L.H.S. =

Prove that

L.H.S. =

Question 22:

Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1)

= 1 = R.H.S.

Question 23:

Prove that

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

Question 24:

Prove that cos 4x = 1 – 8sin2 x cos2 x

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]

= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]

= 1 – 8 sin2x cos2x

= R.H.S.

Question 25:

Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]

= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]

= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3

= 32 cos6x – 48 cos4x + 18 cos2x – 1

= R.H.S.

Question 1:

Find the principal and general solutions of the equation

Therefore, the principal solutions are x =and.

Therefore, the general solution is

Question 2:

Find the principal and general solutions of the equation

Therefore, the principal solutions are x =and.

Therefore, the general solution is, where nZ

Question 3:

Find the principal and general solutions of the equation

Therefore, the principal solutions are x = and.

Therefore, the general solution is

Question 4:

Find the general solution of cosec x = –2

cosec x = –2

Therefore, the principal solutions are x =.

Therefore, the general solution is

Question 5:

Find the general solution of the equation

Question 6:

Find the general solution of the equation

Question 7:

Find the general solution of the equation

Therefore, the general solution is.

Question 8:

Find the general solution of the equation

Therefore, the general solution is.

Question 9:

Find the general solution of the equation

Therefore, the general solution is

Prove that:

L.H.S.

= 0 = R.H.S

Question 2:

Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

L.H.S.

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

= RH.S.

Prove that:

L.H.S. =

Prove that:

L.H.S. =

Question 5:

Prove that:

It is known that.

∴L.H.S. =

Prove that:

It is known that

.

L.H.S. =

= tan 6x

= R.H.S.

Prove that:

L.H.S. =

Question 8:

Here, x is in quadrant II.

i.e.,

Therefore, are all positive.

As x is in quadrant II, cosx is negative.

Thus, the respective values of are.

Question 9:

Find for , x in quadrant III

Here, x is in quadrant III.

Therefore, and are negative, whereasis positive.

Now,

Thus, the respective values of are.

Question 10:

Find for , x in quadrant II