MATHEMATICS XI

Solutions Chapter 3 – Trigonometric Functions

Page No 54:

Question 1:

Find the radian measures corresponding to the following degree measures:

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

Answer:

(i) 25°

We know that 180° = π radian

(ii) –47° 30′

–47° 30′ = degree [1° = 60′]

 degree

Since 180° = π radian

(iii) 240°

We know that 180° = π radian

(iv) 520°

We know that 180° = π radian

Page No 55:

Question 2:

Find the degree measures corresponding to the following radian measures

.

(i)  (ii) – 4 (iii)  (iv) 

Answer:

(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?

Answer:

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.,

12 π radian

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.

Answer:

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.

Answer:

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.

Answer:

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 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

Answer:

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, = 15 cm

(iii) Here, = 21 cm

Page No 63:

Question 1:

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

Answer:

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.

Answer:

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.

Answer:

Since 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.

Answer:

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.

Answer:

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°

Answer:

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

Question 7:

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

Answer:

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

Question 8:

Find the value of the trigonometric function 

Answer:

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

Question 9:

Find the value of the trigonometric function

Answer:

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

Question 10:

Find the value of the trigonometric function

Answer:

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

Page No 73:

Question 1:

Answer:

L.H.S. = 

Question 2:

Prove that 

Answer:

L.H.S. = 

Question 3:

Prove that 

Answer:

L.H.S. =

Question 4:

Prove that 

Answer:

L.H.S =

Question 5:

Find the value of:

(i) sin 75°

(ii) tan 15°

Answer:

(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°)

Question 6:

Prove that:

Answer:

Question 7:

Prove that: 

Answer:

It is known that 

∴L.H.S. =

Question 8:

Prove that

Answer:

Question 9:

Answer:

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

Answer:

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

Question 11:

Prove that

Answer:

It is known that.

∴L.H.S. = 

Question 12:

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

Answer:

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 4sin 8x

Answer:

It is known that 

∴L.H.S. = cos2 2x – cos2 6x

= (cos 2x + cos 6x) (cos 2– 6x)

= [2 cos 4x cos 2x] [–2 sin 4(–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

Answer:

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (– 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)

Answer:

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.

Question 16:

Prove that 

Answer:

It is known that

∴L.H.S =

Question 17:

Prove that 

Answer:

It is known that

∴L.H.S. =

Question 18:

Prove that 

Answer:

It is known that

∴L.H.S. =

Question 19:

Prove that 

Answer:

It is known that

∴L.H.S. =

Question 20:

Prove that 

Answer:

It is known that

∴L.H.S. = 

Question 21:

Prove that 

Answer:

L.H.S. =

Page No 74:

Question 22:

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

Answer:

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) (cot 2x + cot x)

= cot x cot 2– (cot 2cot x – 1)

= 1 = R.H.S.

Question 23:

Prove that 

Answer:

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

Question 24:

Prove that cos 4x = 1 – 8sincosx

Answer:

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 – 1

Answer:

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 – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 – 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 cos6– 48 cos4x + 18 cos2x – 1

= R.H.S.

Page No 78:

Question 1:

Find the principal and general solutions of the equation 

Answer:

Therefore, the principal solutions are x =and.

Therefore, the general solution is

Question 2:

Find the principal and general solutions of the equation 

Answer:

Therefore, the principal solutions are x =and.

Therefore, the general solution is, where n ∈ Z

Question 3:

Find the principal and general solutions of the equation 

Answer:

Therefore, the principal solutions are x = and.

Therefore, the general solution is 

Question 4:

Find the general solution of cosec x = –2

Answer:

cosec x = –2

Therefore, the principal solutions are x =.

Therefore, the general solution is

Question 5:

Find the general solution of the equation 

Answer:

Question 6:

Find the general solution of the equation 

Answer:

Question 7:

Find the general solution of the equation 

Answer:

Therefore, the general solution is.

Question 8:

Find the general solution of the equation 

Answer:

Therefore, the general solution is.

Question 9:

Find the general solution of the equation 

Answer:

Therefore, the general solution is

Page No 81:

Question 1:

Prove that: 

Answer:

L.H.S.

= 0 = R.H.S

Question 2:

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

Answer:

L.H.S.

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

= RH.S.

Page No 82:

Question 3:

Prove that: 

Answer:

L.H.S. = 

Question 4:

Prove that: 

Answer:

L.H.S. = 

Question 5:

Prove that: 

Answer:

It is known that.

∴L.H.S. = 

Question 6:

Prove that: 

Answer:

It is known that

.

L.H.S. = 

= tan 6x

= R.H.S.

Question 7:

Prove that: 

Answer:

L.H.S. = 

Question 8:

x in quadrant II

Answer:

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

Answer:

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

Answer:

Here, x is in quadrant II.

Therefore,, and  are all positive.

 [cosx is negative in quadrant II]

Thus, the respective values of are .

Solutions Chapter 2 – Relations and Functions (Prev Lesson)
(Next Lesson) Solutions Chapter 4 – Principle of Mathematical Induction
Back to MATHEMATICS XI

No Comments

Post a Reply