Class 11 - Maths - Trigonometric Functions

Exercise 3.1

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

So, 25° = (π/180) * 25 radian = 25π/180 radian = 5π/36 radian

(ii) -47° 30'

-47° 30' = -47½ = -95/2 degree

Since 180° = π radian

So, -95/2 degree = (π/180) * (-95/2) radian

= -19π/(36 * 2)

= -19π/72 radian

So, -47° 30' = -19π/72 radian

(iii) 240°

We know that 180° = π radian

So, 240° = (π/180) * 240 radian = 4π/3 radian

(iv) 520°

We know that 180° = π radian

So, 520° = (π/180) * 520 radian = 26π/9 radian

Question 2:

Find the degree measures corresponding to the following radian measures. (Use π = 22/7)       (i) 11/16      (ii) -4      (iii) 5π/3  (iv) 7π/6

Answer:

(i) 11/16

We know that π radian = 180°

So, 11/16 radian = (180/π) * (11/16) degree

= (45 * 11)/(π * 4) degree

= (45 * 11 * 7)/(22 * 4) degree

= 315/8 degree

= 393/8 degree

= 39° + (3 * 60)/8 minutes             [Since 1° = 60’]

= 39° + 22’ + 1/2 minutes

= 39° 22’ 30’’                                    [Since 1’ = 60’’]

(ii) -4

We know that π radian = 180°

So, -4 radian = (180/π) * (-4) degree

= {180 * (-4) * 7}/22 degree

= -2520/11 degree

= -229 1/11 degree

= -229° + (1 * 60)/11 minutes             [Since 1° = 60’]

= -229° + 5’ + 5/11 minutes

= -229° 5’ 27’’                                         [Since 1’ = 60’’]

(iii) 5π/3

We know that π radian = 180°

So, 5π/3 = (180/ π) * (5π/3) = 300°

(iv) 7π/6

We know that π radian = 180°

So, 7π/6 = (180/ π) * (7π/6) = 210°

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 (60 seconds) = 360

So, the number of revolutions made by the wheel in 1 second = 360/60 = 6

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., 12π 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.   (Use π = 22/7)

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

θ = l/r

Therefore, for r = 100 cm, l = 22 cm, we have

θ = 22/100 radian = (180/π) * (22/100) degree

= (180 * 22 * 7)/(22 * 100) degree

= 126/10 degree

= 12 3/5 degree

= 12° 36’                      [Since 1° = 60’]

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 = 40/2 = 20 cm

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

In ∆OAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ∆OAB is an equilateral triangle.

θ = 60° = π/3 radian

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

θ = l/r

=> π/3 = AB/20

=> AB = 20π/3

Thus, the length of the minor arc of the chord is 20π/3 cm.

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 r1 and r2. 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°= π/3 radian and 75° = 5π/12 radian

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

So, l = (r1 * π)/3 and I = (r2 * 5π)/12

=> (r1 * π)/3 = (r2 * 5π)/12

=> r1 = (r2 * 5)/4

=> r1 / r2 = 5/4

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 θ = l/r

It is given that r = 75 cm

(i) Here, l = 10 cm

Now, θ = 10/75 = 2/15 radian

(ii) Here, l = 15 cm

Now, θ = 15/75 = 1/5 radian

(iii) Here, l = 21 cm

Now, θ = 21/75 = 7/25 radian

Exercise 3.2

Question 1:

Find the values of other five trigonometric functions if cos x = -1/2, x lies in third quadrant.

Answer:

Given, cos x = -1/2

So, sec x = 1/cos x = 1/(-1/2) = -2

We know that

sin2 x + cos2 x = 1

=> sin2 x =1 - cos2 x

=> sin2 x =1 – (-1/2)2

=> sin2 x = 1 – 1/4

=> sin2 x = 3/4

=> sin x = ±√3/2

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

So, sin x = -√3/2

cosec x = 1/sin x = 1/(-√3/2) = -2/√3

tan x = sin x/cos x = (-√3/2)/(-1/2) = √3

cot x = 1/tan x = 1/√3

Question 2:

Find the values of other five trigonometric functions if sin x = 3/5, x lies in second quadrant.

Answer:

Given, sin x = 3/5

cosec x = 1/sin x = 1/(3/5) = 5/3

We know that

sin2 x + cos2 x = 1

=> cos2 x =1 - sin 2 x

=> cos2 x =1 – (3/5)2

=> cos2 x = 1 – 9/25

=> cos2 x = 16/25

=> cos x = ±4/5

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

So, cos x = -4/5

sec x = 1/cos x = 1/(-4/5) = -5/4

tan x = sin x/cos x = (3/5)/(-4/5) = -3/4

cot x = 1/tan x = 1/(-3/4) = -4/3

Question 3:

Find the values of other five trigonometric functions if cot x = 3/4, x lies in third quadrant.

Answer:

Given, cot x = 3/4

tan x = 1/cot x = 1/(3/4) = 4/3

We know that

1 + tan2 x = sec2 x

=> 1 + (4/3)2 = sec2 x

=> sec2 x = 1 + 16/9

=> sec2 x = 25/9

=> sec x = ±5/3

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

So, sec x = -5/3

cos x = 1/sec x = 1/(-5/3) = -3/5

Again, tan x = sin x/cos x

=> 4/3 = sin x/(-3/5)

=> sin x = (4/3) * (-3/5)

=> sin x = -4/5

and cosec x = 1/sin x = 1/(-4/5) = -5/4

Question 4:

Find the values of other five trigonometric functions if sec x = 13/5, x lies in fourth quadrant.

Answer:

Given, sec x = 13/5

So, cos x = 1/sec x = 1/(13/5) = 5/13

We know that

sin2 x + cos2 x = 1

=> sin2 x =1 - cos2 x

=> sin2 x =1 – (5/13)2

=> sin2 x = 1 – 25/169

=> sin2 x = 144/169

=> sin x = ±12/13

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

So, sin x = -12/13

cosec x = 1/sin x = 1/(-12/13) = -13/12

Now, tan x = sin x/cos x = (-12/13)/(5/13) = -12/5

cot x = 1/tan x = 1/(-12/5) = -5/12

Question 5:

Find the values of other five trigonometric functions if tan x = -5/12, x lies in second quadrant.

Answer:

Given, tan x = -5/12

So, cot x = 1/tan x = 1/(-5/12) = -12/5

We know that

1 + tan2 x = sec2 x

=> 1 + (-5/12)2 = sec2 x

=> sec2 x = 1 + 25/144

=> sec2 x = 169/12

=> sec x = ±13/12

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

So, sec x = -13/12

cos x = 1/sec x = 1/(-13/12) = -12/13

Again, tan x = sin x/cos x

=> -5/12 = sin x/(-12/13)

=> sin x = (-5/12) * (-12/13)

=> sin x = 5/13

and cosec x = 1/sin x = 1/(5/13) = 13/5

Question 6:

Find the value of the trigonometric function sin 765°

Answer:

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

Now, sin 765° = sin (2 * 360° + 45°) = sin 45° = 1/√2

Question 7:

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

Answer:

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

Now, cosec(-1410°) = cosec(-1410° + 4 * 360°)

= cosec(-1410° + 1440°)

= cosec 30°

= 2

Question 8:

Find the value of the trigonometric function tan 19π/3

Answer:

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

Now, tan 19π/3 = tan 6 1/3 π = tan(6π + π/3) = tan π/3 = √3

Question 9:

Find the value of the trigonometric function sin(-11π/3)

Answer:

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

Now, sin (-11π/3) = sin (-11π/3 + 2 * 2π) = sin (-11π/3 + 4π) = sin π/3 = √3/2

Question 10:

Find the value of the trigonometric function cot(-15π/4)

Answer:

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

Now, cot(-15π/4) = cot(-15π/4 + 4π) = cot π/4 = 1

Exercise 3.3

Question 1:

sin2 π/6 + cos2 π/3 - tan2 π/4 = -1/2

Answer:

LHS:

sin2 π/6 + cos2 π/3 - tan2 π/4

= (1/2)2 + (1/2)2 – 12

= 1/4 + 1/4 - 1

= 1/2 - 1

= -1/2

= RHS

Question 2:

2sin2 π/6 + cosec2 7π/6 * cos2 π/3 = 3/2

Answer:

LHS:

2sin2 π/6 + cosec2 7π/6 * cos2 π/3

= 2(1/2)2 + cosec2 (π + π/6) * (1/2)2

= 2/4 + (-cosec2 π/6) * 1/4

= 2/4 + (-cosec π/6)2 * 1/4

= 1/2 + (-2)2 * 1/4

= 1/2 + 4 * 1/4

= 1/2 + 1

= 3/2

= RHS

Question 3:

Prove that: cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6

Answer:

LHS:

cot2 π/6 + cosec 5π/6 + 3 tan2 π/6

= cot2 π/6 + cosec 5π/6 + 3 (tan π/6)2

= (√3)2 + cosec (π - π/6) + 3(1/√3)2

= 3 + cosec π/6 + 3 * 1/3

= 3 + 2 + 1

= 6

= RHS

Question 4:

Prove that: 2sin2 3π/4 + 2cos2 π/4 + 2sec2 π/3 = 6

Answer:

LHS:

2sin2 3π/4 + 2cos2 π/4 + 2sec2 π/3

= 2(sin 3π/4)2 + 2(cos π/4)2 + 2(sec π/3)2

= 2[sin (π - π/4)]2 + 2(1/√2)2 + 2 * 22

= 2[sin π/4]2 + 2 * 1/2 + 2 * 4

= 2(1/√2)2 + 1 + 8

= 2 * 1/2 + 1 + 8

= 1 + 1 + 8

= 10

= RHS

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]

= (1/√2) * (√3/2) + (1/√2) * (1/2)

= √3/2√2 + 1/2√2

= (√3 + 1)/2√2

(ii) tan 15° = tan(45° – 30°)

= (tan 45° – tan 30°)/(1 + tan 45° * tan 30°)

= (1 – 1/√3)/(1 + 1 * 1/√3)

= (1 – 1/√3)/(1 + 1/√3)

= {(√3 – 1)/√3}/{(√3 + 1)/√3}

= (√3 – 1)/(√3 + 1)

= {(√3 – 1)/(√3 + 1)} * {(√3 – 1)/(√3 – 1)}

= (√3 – 1)2/{(√3)2 - 1)}

= (3 + 1 - 2√3)/(3 - 1)

= (4 - 2√3)/2

= 2 - √3

Question 6:

Prove that: cos(π/4 - x) * cos(π/4 - y) - sin(π/4 - x) * sin(π/4 - y) = sin(x + y)

Answer:

LHS:

cos(π/4 - x) * cos(π/4 - y) - sin(π/4 - x) * sin(π/4 - y)

= (1/2)[2 cos(π/4 - x) * cos(π/4 - y)] + (1/2)[-2 sin(π/4 - x) * sin(π/4 - y)]

= (1/2)[cos{(π/4 - x) + (π/4 - y)} + cos{(π/4 - x) - (π/4 - y)}]

+ (1/2)[cos{(π/4 - x) + (π/4 - y)}] - cos{(π/4 - x) - (π/4 - y)}]

= 2 * (1/2)[cos{(π/4 - x) + (π/4 - y)}

= cos[π/2 – (x + y)]

= sin(x + y)

= RHS

Question 7:

Prove that: tan(π/4 + x)/ tan(π/4 - x) = {(1 + tan x)/(1 + tan x)}2

Answer:

LHS:

tan(π/4 + x)/tan(π/4 - x) = {(tan π/4 + tan x)/(1 – tan π/4 * tan x)}

{(tan π/4 - tan x)/(1 + tan π/4 * tan x)}

= {(1 + tan x)/(1 – tan x)}/{(1 - tan x)/(1 + tan x)}

= {(1 + tan x)/(1 – tan x)}2

= RHS

Question 8:

Prove that: {cos(π + x) cos(-x)}/{sin(π - x) cos(π/2 + x)} = cot2 x

Answer:

LHS:

{cos(π + x) cos(-x)}/{sin(π - x) cos(π/2 + x)}

= {- cos x * cos x}/{sin x * (-sin x)}

= (-cos2 x)/(-sin2 x)

= cot2 x

= RHS

Question 9:

cos(3π/2 + x) cos(2π + x)[cot(3π/2 - x) + cot(2π + x)] = 1

Answer:

LHS:

cos(3π/2 + x) cos(2π + x)[cot(3π/2 - x) + cot(2π + x)]

= sin x * cos x[tan x + cot x]

= sin x * cos x[sin x/cos x + cos x/sin x]

= sin x * cos x[(sin2 x + cos2 x)/(cos x * sin x)]

= 1

= RHS

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(n + 2)x

= (1/2)[2sin(n + 1)x sin (n + 2)x + 2cos (n + 1)x cos (n + 2)x]

= (1/2)[cos{(n + 1)x - (n + 2)x} - cos{(n + 1)x + (n + 2)x}

+ (1/2)[cos{(n + 1)x + (n + 2)x}] + cos{(n + 1)x  - (n + 2)x}]

= (1/2) * 2 cos{(n + 1)x - (n + 2)x}

= cos(-x)

= cos x

= RHS

Question 11:

Prove that: cos(3π/4 + x) - cos(3π/4 - x) = -√2sin x

Answer:

LHS:

cos(3π/4 + x) - cos(3π/4 - x)

= -2sin[{(3π/4 + x) + (3π/4 - x)}/2] * sin[{(3π/4 + x) - (3π/4 - x)}/2]

= -2sin 3π/4 * sin x

= -2sin (π - π/4) * sin x

= -2sin π/4 * sin x

= -2 * 1/√2 * sin x

= -√2 sin x

= RHS

Question 12:

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

Answer:

L.H.S.

sin2 6x – sin2 4x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= [2 sin(6x + 4x)/2 * cos (6x - 4x)/2][2 cos(6x + 4x)/2 * sin (6x - 4x)/2]

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

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

= sin 10x sin 2x                                                  [Since sin 2x = 2 * sin x * cos x]

= R.H.S.

Question 13:

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

Answer:

L.H.S.

cos2 2x – cos2 6x

= (cos 2x + cos 6x) (cos 2x – 6x)

= [2 cos(2x + 6x)/2 * cos (2x - 6x)/2][-2 sin(2x + 6x)/2 * sin (2x - 6x)/2]

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

= [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

Answer 14:

L.H.S.

sin 2x + 2 sin 4x + sin 6x

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

= 2 sin(2x + 6x)/2 * cos (2x - 6x)/2 + 2sin 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 – 1 + 1)

= 2 sin 4x (2 cos2 x)

= 4 cos2 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)

= (cos 4x/sin 4x)[2sin (5x + 3x)/2 * cos (5x - 3x)/2]

= (cos 4x/sin 4x)[2sin 4x * cos x]

= 2 cos 4x cos x

RHS:

cot x (sin 5x – sin 3x)

= (cos x/sin x)[2cos (5x + 3x)/2 * sin (5x - 3x)/2]

= (cos x/sin x)[2cos 4x * sin x]

= 2 cos 4x cos x

So, L.H.S. = R.H.S.

Question 16:

Prove that: (cos 9x – cos 5x)/(sin 17x – sin 3x) = -sin 2x/cos 10x

Answer:

LHS:

(cos 9x – cos 5x)/(sin 17x – sin 3x)

= [-2 sin(9x + 5x)/2 * sin(9x – 5x)/2]/ [2 cos(17x + 3x)/2 * sin(17x – 3x)/2]

= (-2 sin 7x * sin 2x)/(2 cos 10x * sin 7x)

= -sin 2x/cos 10x

= RHS

Question 17:

Prove that: (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x

Answer:

LHS:

(sin 5x + sin 3x)/(cos 5x + cos 3x)

= {2 * sin (5x + 3x)/2 * cos (5x - 3x)/2}/{2 * cos (5x + 3x)/2 * cos (5x - 3x)/2}

= {2*sin 4x * cos x}/{2*cos 4x * cos x}

= sin 4x/ cos 4x

= tan 4x

= RHS

Question 18:

Prove that: (sin x – sin y)/(cos x + cos y) = tan (x - y)/2

Answer:

LHS:

(sin x – sin y)/(cos x + cos y)

= {2 * cos (x + y)/2 * sin (x - y)/2}/{2 * cos (x + y)/2 * cos (x - y)/2}

= {sin (x - y)/2}/{cos (x - y)/2}

= tan (x – y)/2

= RHS

Question 19:

Prove that: (sin x + sin 3x)/(cos x + cos 3x) = tan 2x

Answer:

LHS:

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

= {2 * sin (x + 3x)/2 * cos (x – 3x)/2}/{2 * cos (x + 3x)/2 * cos (x – 3x)/2}

= {2 * sin 2x * cos (-x)}/{2 * cos 2x * cos (-x)}

= sin 2x/cos 2x

= tan 2x

= RHS

Question 20:

Prove that: (sin x – sin 3x)/(sin2 x – cos2 x) = 2 sin x

Answer:

LHS:

(sin x – sin 3x)/(sin2 x – cos2 x)

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

= {2 * cos 2x * sin(-x)}/ (-cos 2x)

= -2 * sin(-x)

= 2 sin x

= RHS

Question 21:

Prove that: (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x

Answer:

LHS:

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

= {(cos 4x + cos 2x) + cos 3x}/{(sin 4x + sin 2x) + sin 3x}

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

= {2 * cos 3x * cos x + cos 3x}/{2 * sin 3x * cos x + sin 3x}

= {cos 3x (2 cos x + 1)}/{sin 3x (2 cos x + 1)}

= cos 3x/ sin 3x

= cot 3x

= RHS

Question 22:

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

Answer:

LHS:

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)/(cot x + cot 2x)}(cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1)

= cot x cot 2x – cot 2x cot x + 1

= 1

= R.H.S.

Question 23:

Prove that: tan 4x = {4 tan x (1 – tan2 x)}/( 1 – 6tan2 x + tan4 x)

Answer:

LHS:

tan 4x

= tan 2(2x)

= 2 tan 2x/(1 – tan2 2x)

= [2{2tan x/(1 – tan2 x)}]/[1 – {2tan x/(1 – tan2 x)}2]

= [4tan x/(1 – tan2 x)]/[1 – 4tan2 x/(1 – tan2 x)2]

= [4tan x/(1 – tan2 x)]/[{(1 – tan2 x)2 – 4tan2 x}/(1 – tan2 x)2]

= [4tan x(1 – tan2 x)]/[{(1 + tan4 x - 2tan2 x - 4tan2 x}]

= {4 tan x(1 – tan2 x)}/( 1 – 6tan2 x + tan4 x)

= RHS

Question 24:

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

Answer:

LHS:

cos 4x

= cos 2(2x)

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

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

= 1 – 8 sin2 x cos2 x

= RHS

Question 25:

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

Answer:

LHS:

cos 6x

= cos 3(2x)

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

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

= 4 [(2 cos2 x)3 – 13 – 3(2 cos2 x)2 + 3(2 cos2 x)] – 6cos2 x + 3

= 4 [8 cos6 x – 1 – 12 cos4 x + 6 cos2 x] – 6 cos2 x + 3

= 32 cos6 x – 4 – 48 cos4 x + 24 cos2 x – 6 cos2 x + 3

= 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

= R.H.S.

Exercise 3.4

Question 1:

Find the principal and general solutions of the equation tan x = √3

Answer:

Given, tan x = √3

We know that tan π/3 = √3

and tan 4π/3 = tan (π + π/3) = tan π/3 = √3

Therefore, the principal solutions are: x = π/3 and x = 4π/3

Therefore, the general solution is

Now, tan x = tan π/3

=> x = nπ + π/3, where n є Z

Hence, the general solution is x = nπ + π/3, where n є Z

Question 2:

Find the principal and general solutions of the equation sec x = 2

Answer:

Given, sec x = 2

We know that sec π/3 = 2

and sec 5π/3 = tan (2π - π/3) = sec π/3 = 2

Therefore, the principal solutions are: x = π/3 and x = 5π/3

Now, sec x = sec π/3

=> cos x = cos π/3

=> x = 2nπ ± π/3, where n є Z

Hence, the general solution is x = 2nπ ± π/3, where n є Z

Question 3:

Find the principal and general solutions of the equation cot x = -√3

Answer:

Given, cot x = -√3

We know that cot π/6 = √3

Now, cot (π - π/6) = -cot π/6 = -√3

and cot (2π - π/6) = -cot π/6 = -√3

So, cot 5π/6 = -√3 and cot 11π/6 = -√3

Therefore, the principal solutions are: x = 5π/3 and x = 11π/3

Now, cot x = cot 5π/6

=> tan x = tan 5π/6

=> x = nπ + 5π/6, n є Z

Therefore, the general solution is x = nπ + 5π/6, n є Z

Question 4:

Find the general solution of cosec x = –2

Answer:

Given, cosec x = –2

We know that cosec π/6 = 2

Now, cosec (π + π/6) = - cosec π/6 = -2

and cosec (2π - π/6) = - cosec π/6 = -2

So, cosec 7π/6 = -2 and cosec 11π/6 = -2

Therefore, the principal solutions are: x = 7π/6 and x = 11π/6

Now, cosec x = cosec 7π/6

=> sin x = sin 7π/6

=> x = nπ + (-1)n 7π/6, n є Z

Therefore, the general solution is x = nπ + (-1)n 7π/6, n є Z

Question 5:

Find the general solution of the equation cos 4x = cos 2x

Answer:

Given, cos 4x = cos 2x

=> cos 4x - cos 2x = 0

=> -2 * sin {(4x + 2x)/2} * sin {(4x - 2x)/2} = 0       [Apply cos A – cos B formula]

=> sin 3x * sin x = 0

=> sin 3x = 0 or sin x = 0

=> 3x = nπ or x = nπ, where n є Z

=> x = nπ/3 or x = nπ, where n є Z

Question 6:

Find the general solution of the equation cos 3x + cos x – cos 2x = 0

Answer:

Given, cos 3x + cos x – cos 2x = 0

=> 2 * cos {(3x + x)/2} * cos {(3x - x)/2} – cos 2x = 0       [Apply cos A + cos B formula]

=> 2 * cos 2x * cos x – cos 2x = 0

=> cos 2x(2cos x - 1) = 0

=> cos 2x = 0 or 2cos x – 1 = 0

=> cos 2x = 0 or cos x = 1/2

=> 2x = (2n + 1)π/2 or x = 2nπ ± π/3, where n є Z

=> x = (2n + 1)π/4 or x = 2nπ ± π/3, where n є Z

Question 7:

Find the general solution of the equation sin 2x + cos x = 0

Answer:

Given, sin 2x + cos x = 0

=> 2 * sin x * cos x + cos x = 0

=> cos x(2sin x + 1) = 0

=> cos x = 0 or 2sin x + 1 = 0

Now, cos x = 0 = cos π/2

=> x = (2n + 1)π/2, where n є Z

Again 2sin x + 1 = 0

=> sin x = -1/2 = -sin π/6

=> sin x = sin (π + π/6)

=> sin x = sin 7π/6

=> x = nπ + (-1)n 7π/6, where n є Z

Thus, the general solution is: = (2n + 1)π/2 or nπ + (-1)n 7π/6, where n є Z

Question 8:

Find the general solution of the equation sec2 2x = 1 – tan 2x

Answer:

Given, sec2 2x = 1 – tan 2x

=> 1 + tan2 2x = 1 – tan 2x

=> tan2 2x + tan 2x = 0

=> tan 2x(tan 2x + 1) = 0

=> tan 2x = 0 or tan 2x + 1 = 0

=> tan 2x = 0 or tan 2x = -1

=> tan 2x = tan 0 or tan 2x = -tan π/4

=> tan 2x = tan 0 or tan 2x = tan (π - π/4)

=> tan 2x = tan 0 or tan 2x = tan 3π/4

=> 2x = nπ + 0 or 2x = nπ + 3π/4, n є Z

=> x = nπ/2 or x = nπ/2 + 3π/8, n є Z

Thus, the general solution is nπ/2 or nπ/2 + 3π/8, n є Z

Question 9:

Find the general solution of the equation sin x + sin 3x + sin 5x = 0

Answer:

Given, sin x + sin 3x + sin 5x = 0

=> sin x + sin 5x + sin 3x = 0

=> 2*sin(x + 5x)/2 * cos(x - 5x)/2 + sin 3x = 0

=> 2*sin 3x * cos(-2x) + sin 3x = 0

=> 2*sin 3x * cos 2x + sin 3x = 0

=> sin 3x(2 cos 2x + 1) = 0

=> sin 3x = 0 or 2 cos 2x + 1 = 0

=> sin 3x = 0 or cos 2x = -1/2

=> sin 3x = 0 or cos 2x = -1/2

=> sin 3x = 0 or cos 2x = -cos π/3

=> sin 3x = 0 or cos 2x = cos (π - π/3)

=> sin 3x = 0 or cos 2x = cos 2π/3

=> 3x = nπ or 2x = 2nπ ± 2π/3, n є Z

=> x = nπ/3 or x = nπ ± π/3

Therefore, the general solution is: nπ/3 or nπ ± π/3, n є Z

Miscellaneous Exercise on chapter 3

Question 1:

Prove that: 2cos π/13 cos 9π/13 + cos 3π/13 cos 5π/13 = 0

Answer:

LHS:

2cos π/13 * cos 9π/13 + cos 3π/13 * cos 5π/13

= 2cos π/13 * cos 9π/13 + 2 cos{(3π/13 + 5π/13)/2} * cos{(3π/13 - 5π/13)/2}

= 2cos π/13 * cos 9π/13 + 2 cos 4π/13 * cos(-π/13)

= 2cos π/13 * cos 9π/13 + 2 cos 4π/13 * cos π/13

= 2cos π/13(cos 9π/13 + cos 4π/13)

= 2cos π/13[2 cos{(9π/13 + 4π/13)/2} * cos{(9π/13 - 4π/13)/2}]

= 2cos π/13[2 cos π/2 * cos 5π/26]

= 2cos π/13 * 2 * 0 * cos 5π/26

= 0

= RHS

Question 2:

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

Answer:

LHS:

(sin 3x + sin x) sin x + (cos 3x - cos x) cos x

= {2 * sin(3x + x)/2 * cos (3x - x)/2} * sin x + {-2 * sin (3x + x)/2 * sin(3x - x)/2} * cos x

= 2 * sin 2x *cos x * sin x – 2 * sin 2x * sin x * cos x

= 0 = RHS

Question 3:

Prove that: (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2 (x + y)/2

Answer:

LHS:

(cos x + cos y)2 + (sin x – sin y)2

= cos2 x + cos2 y + 2 * cos x * cos y + sin2 x + sin2 y – 2 * sin x * sin y

= cos2 x + sin2 x + cos2 y + sin2 y + 2 * cos x * cos y – 2 * sin x * sin y

= 1 + 1 + 2 * cos x * cos y – 2 * sin x * sin y

= 2 + 2 * cos (x + y)

= 2[1 + cos(x + y)]

= 2[1 + 2cos2 (x + y)/2 - 1]

= 4 cos2 (x + y)/2

= RHS

Question 4:

Prove that: (cos x - cos y)2 + (sin x – sin y)2 = 4 sin2 (x - y)/2

Answer:

LHS:

(cos x - cos y)2 + (sin x – sin y)2

= cos2 x + cos2 y - 2 * cos x * cos y + sin2 x + sin2 y – 2 * sin x * sin y

= cos2 x + sin2 x + cos2 y + sin2 y - 2 * cos x * cos y – 2 * sin x * sin y

= 1 + 1 - 2 * cos x * cos y – 2 * sin x * sin y

= 2 - 2 * cos (x - y)

= 2[1 - cos(x - y)]

= 2[1 – {1 – 2 sin2 (x - y)/2}]

= 2[1 – 1 + 2 sin2 (x - y)/2]

= 4 sin2 (x - y)/2

= RHS

Question 5:

Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 * cos x * cos 2x * sin 4x

Answer:

LHS:

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

= 2 * sin(x + 3x)/2 * cos(x - 3x)/2 + 2 * sin(5x + 7x)/2 * cos(5x - 7x)/2

= 2 * sin(4x/2) * cos(-2x/2) + 2 * sin(12x/2) * cos(-2x/2)

= 2 * sin 2x * cos(-x) + 2 * sin 6x * cos(-x)

= 2 * sin 2x * cos x + 2 * sin 6x * cos x

= 2 * cos x * (sin 2x + sin 6x)

= 2 * cos x * {sin(2x + 6x)/2 * cos(2x - 6x)/2}

= 2 * cos x * {2 * sin(8x/2) * cos(-4x/2)}

= 2 * cos x * {2 * sin 4x * cos(-2x)}

= 4 * cos x * sin 4x * cos 2x

= 4 * cos x * cos 2x * sin 4x

= RHS

Question 6:

Prove that: {(sin 7x + sin 5x) + (sin 9x + sin 3x)}/{(cos 7x + cos 5x) + (cos 9x + cos 3x)} = tan 6x

Answer:

We know that

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

and cos A + cos B = 2cos (A + B)/2  * cos (A - B)/2

LHS:

{(sin 7x + sin 5x) + (sin 9x + sin 3x)}/{(cos 7x + cos 5x) + (cos 9x + cos 3x)}

= [2sin (7x + 5x)/2 * cos (7x – 5x)/2 + 2sin (9x + 3x)/2 * cos (9x – 3x)/2]

[2cos (7x + 5x)/2 * cos (7x – 5x)/2 + 2cos (9x + 3x)/2 * cos (9x – 3x)/2]

= [2 * sin 6x * cos x + 2 * sin 6x * cos 3x]/[2 * cos 6x * cos x + 2 * cos 6x * cos 3x]

= [2 * sin 6x * (cos x + cos 3x)]/[2 * sin 6x * (cos x + cos 3x)]

= tan 6x

= RHS

Question 7:

Prove that: sin 3x + sin 2x – sin x = 4 * sin x * cos(x/2) * cos(3x/2)

Answer:

LHS:

sin 3x + sin 2x - sin x

= sin 3x - sin x + sin 2x

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

= 2 * cos 2x * sin x + 2 * sin x * cos x                      (sin 2x = 2 * sin x * cos x)

= 2 * sin x * (cos 2x + cos x)

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

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

= RHS

Question 8:

Find sin x/2, cos x/2 and tan x/2, if tan x = −4/3, x in quadrant II.

Answer:

Here, x is in quadrant II.

i.e π/2 < x < π

=> π/4 < x/2 < π/2

Therefore, sin x/2, cos x/2 and tan x/2 are lies in first quadrant.

It is given that

tan x = -4/3

Now, sec2 x = 1 + tan2 x = 1 + (-4/3)2 = 1 + 16/9 = 25/9

So, cos2 x = 9/25

=> cos x = ± 3/5

Since x is in quadrant II, cos x is negative.

So, cos x = -3/5

Now, cos x = 2cos2 x/2 – 1

=> -3/5 = 2cos2 x/2 – 1

=> 2cos2 x/2 = 1 – 3/5

=> 2cos2 x/2 = 2/5

=> cos2 x/2 = 1/5

=> cos x/2 = 1/√5             [Since cos x is positive]

Again sin2 x/2 + cos2 x/2 = 1

=> sin2 x/2 + (1/√5)2 = 1

=> sin2 x/2 + 1/5 = 1

=> sin2 x/2 = 1 – 1/5

=> sin2 x/2 = 4/5

=> sin x/2 = 2/√5              [Since sin x/2 is positive]

tan x/2 = (sin x/2)/(cos x/2) = (2/√5)/(1/√5) = 2

Thus, the respective values of sin x/2, cos x/2 and tan x/2 are 2√5/5, √5/5 and 2.

Question 9:

Find sin x/2, cos x/2 and tan x/2 for cos x = −1/3, x in quadrant III.

Answer:

Here, x is in quadrant III.

i.e π < x < 3π/2

=> π/2 < x/2 < 3π/4

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

Given, cos x = -1/3

Now, cos x = 1 – 2 sin2 x/2

=> sin2 x/2 = (1 – cos x)/2

= {1 – (-1/3)}/2

= (1 + 1/3)/2

= (4/3)/2

= 2/3

=> sin x/2 = √(2/3)

=> sin x/2 = (√2/√3) * (√3/√3)

=> sin x/2 = √6/3

cos x = 2 cos2 x/2 – 1

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

= {1 + (-1/3)}/2

= (1 - 1/3)/2

= (2/3)/2

= 1/3

=> cos x/2 = -1/√3                         [Since cos x/2 is negative]

=> cos x/2 = (-1/√3) * (√3/√3)

=> cos x/2 = -√3/3

tan x/2 = (sin x/2)/(cos x/2) = (√2/√3)/( -1/√3) = -√2

Thus, the values of sin x/2, cos x/2 and tan x/2 are: √6/3, -√3/3, -√2

Question 10:

Find sin x/2, cos x/2 and tan x/2 for sin x = 1/4, x in quadrant II

Answer:

Here, x is in quadrant II.

i.e π/2 < x < π

=> π/4 < x/2 < π/2

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

Given, sin x = 1/4

cos2 x = 1 – sin2 x

= 1 – (1/4)2

= 1 – 1/16

= 15/16

cos x = -√15/4                                          [since cos x is negative in quadrant II]

Now, sin2 x/2 = (1 – cos x)/2

= {1 – (-√15/4)}/2

= (1 + √15/4)/2

= (4 + √15)/8

=> sin x/2 = √[(4 + √15)/8]                     [Since sin x/2 is positive]

=> sin x/2 = √[{(4 + √15)/8} * (2/2)]

=> sin x/2 = √[(8 + 2√15)/16]

=> sin x/2 = √{(8 + 2√15)}/4

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

= {1 + (-√15/4)}/2

= (1 - √15/4)/2

= (4 - √15)/8

=> cos x/2 = √[(4 - √15)/8]                          [Since cos x/2 is positive]

=> cos x/2 = √[{(4 - √15)/8} * (2/2)]

=> cos x/2 = √[(8 - 2√15)/16]

=> cos x/2 = √{(8 - 2√15)}/4

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

= [√{(8 + 2√15)}/4]/[ √{(8 - 2√15)}/4]

= [√(8 + 2√15)]/[ √(8 - 2√15)]

= √[{(8 + 2√15)]/(8 - 2√15)} * {(8 + 2√15)]/(8 + 2√15)}]

= √[(8 + 2√15)2]/(64 - 60)]

= (8 + 2√15)/2

= 4 + √15

Therefore, value of sin x/2, cos x/2 and tan x/2 are: √{(8 + 2√15)}/4, √{(8 - 2√15)}/4 and 4 + √15

.