Question 1(i)
Convert 25° to radians.
Steps only
25π/180
Answer
5π/36 radians.
CLASS 11 · MATHEMATICS · CHAPTER 3
Trigonometric Functions
Work through angle conversions, trigonometric functions, identities and half-angle values with concise calculations and step-by-step explanations. One exact-value statement in Exercise 3.3 appears misprinted; its printed left side evaluates to 8, not 6.
61 worked answersAll 3 exercise setsAll 1 end-of-chapter questions
61 worked answers · short steps first, explanations when you need them
Exercise 3.1
Textbook pages 48–49
Question 1(ii)
Convert −47°30′ to radians.
Steps only
−47.5π/180
Answer
−19π/72 radians.
Question 1(iii)
Convert 240° to radians.
Steps only
240π/180
Answer
4π/3 radians.
Question 1(iv)
Convert 520° to radians.
Steps only
520π/180
Answer
26π/9 radians.
Question 2(i)
Convert 11/16 radians to degrees, using π=22/7.
Steps only
(11/16)×(180/π)
Answer
315/8° (39.375°).
Question 2(ii)
Convert −4 radians to degrees, using π=22/7.
Steps only
−4×180/π
Answer
−2520/11° (about −229.09°).
Question 2(iii)
Convert 5π/3 radians to degrees.
Steps only
(5π/3)×(180/π)
Answer
300°.
Question 2(iv)
Convert 7π/6 radians to degrees.
Steps only
(7π/6)×(180/π)
Answer
210°.
Question 3
A wheel makes 360 revolutions per minute. Through how many radians does it turn in one second?
Steps only
360/60=6 revolutions
6×2π
Answer
12π radians.
Question 4
Find the central angle for an arc of 22 cm in a circle of radius 100 cm. Use π=22/7.
Steps only
θ=s/r=22/100
Answer
11/50 radians (about 12.6°).
Question 5
A circle has diameter 40 cm and a chord of length 20 cm. Find the minor arc length.
Steps only
r=20, chord=2r sin(θ/2)
θ=60°=π/3
s=rθ
Answer
20π/3 cm.
Question 6
Equal arcs subtend central angles 60° and 75° in two circles. Find the ratio of their radii.
Steps only
s=r₁θ₁=r₂θ₂
r₁/r₂=θ₂/θ₁
Answer
r₁:r₂=75:60=5:4.
Question 7(i)
A 75 cm pendulum tip traces an arc of 10 cm. Find the angle in radians.
Steps only
θ=s/r=10/75
Answer
2/15 radians.
Question 7(ii)
A 75 cm pendulum tip traces an arc of 15 cm. Find the angle in radians.
Steps only
θ=15/75
Answer
1/5 radian.
Question 7(iii)
A 75 cm pendulum tip traces an arc of 21 cm. Find the angle in radians.
Steps only
θ=21/75
Answer
7/25 radians.
Exercise 3.2
Textbook pages 57
Question 1
Given cos x=−1/2 and x is in quadrant III, find the other five trigonometric functions.
Steps only
sin x=−√3/2
tan x=√3
cot x=√3/3
sec x=−2; cosec x=−2√3/3
Answer
sin x=−√3/2; tan x=√3; cot x=√3/3; sec x=−2; cosec x=−2√3/3.
Question 2
Given sin x=3/5 and x is in quadrant II, find the other five trigonometric functions.
Steps only
cos x=−4/5
tan x=−3/4
cot x=−4/3
sec x=−5/4; cosec x=5/3
Answer
cos x=−4/5; tan x=−3/4; cot x=−4/3; sec x=−5/4; cosec x=5/3.
Question 3
Given cot x=3/4 and x is in quadrant III, find the other five trigonometric functions.
Steps only
Take adjacent=−3, opposite=−4, hypotenuse=5
sin x=−4/5; cos x=−3/5
tan x=4/3
sec x=−5/3; cosec x=−5/4
Answer
sin x=−4/5; cos x=−3/5; tan x=4/3; sec x=−5/3; cosec x=−5/4.
Question 4
Given sec x=13/5 and x is in quadrant IV, find the other five trigonometric functions.
Steps only
cos x=5/13
sin x=−12/13
tan x=−12/5
cot x=−5/12; cosec x=−13/12
Answer
cos x=5/13; sin x=−12/13; tan x=−12/5; cot x=−5/12; cosec x=−13/12.
Question 5
Given tan x=−5/12 and x is in quadrant II, find the other five trigonometric functions.
Steps only
Take opposite=12, adjacent=−5, hypotenuse=13
sin x=12/13; cos x=−5/13
cot x=−12/5
sec x=−13/5; cosec x=13/12
Answer
sin x=12/13; cos x=−5/13; cot x=−12/5; sec x=−13/5; cosec x=13/12.
Question 6
Find sin 765°.
Steps only
765°−720°=45°
sin45°=√2/2
Answer
√2/2.
Question 7
Find cosec(−1410°).
Steps only
−1410°+4×360°=30°
cosec30°=2
Answer
2.
Question 8
Find tan(19π/3).
Steps only
19π/3−6π=π/3
tan(π/3)=√3
Answer
√3.
Question 9
Find sin(−11π/3).
Steps only
−11π/3+4π=π/3
sin(π/3)=√3/2
Answer
√3/2.
Question 10
Find cot(−15π/4).
Steps only
−15π/4+4π=π/4
cot(π/4)=1
Answer
1.
Exercise 3.3
Textbook pages 67–68
Question 1
Verify sin²(π/6)+cos²(π/3)−tan²(π/4)=−1/2.
Steps only
1/4+1/4−1
=−1/2
Answer
The left side is −1/2.
Question 2
Verify 2sin²(π/6)+cosec²(7π/6)cos²(π/3)=3/2.
Steps only
2(1/4)+4(1/4)
=3/2
Answer
The left side is 3/2.
Question 3
Check the printed statement cot²(π/6)+cosec²(5π/6)+3tan²(π/6)=6.
Steps only
3+4+3(1/3)=8
The printed right side appears inconsistent
Answer
As printed, the left side equals 8, not 6. The right side likely has a printing error.
Question 4
Verify 2sin²(3π/4)+2cos²(π/4)+2sec²(π/3)=10.
Steps only
2(1/2)+2(1/2)+2(4)
=10
Answer
The left side is 10.
Question 5(i)
Find sin75°.
Steps only
sin(45°+30°)
=(√6+√2)/4
Answer
(√6+√2)/4.
Question 5(ii)
Find tan15°.
Steps only
tan(45°−30°)
=(1−1/√3)/(1+1/√3)
Answer
2−√3.
Question 6
Prove cos(π/4−x)cos(π/4−y)−sin(π/4−x)sin(π/4−y)=sin(x+y).
Steps only
Use cosAcosB−sinAsinB=cos(A+B)
cos(π/2−x−y)=sin(x+y)
Answer
The left side equals sin(x+y).
Question 7
Prove tan(π/4+x)/tan(π/4−x)=((1+tanx)/(1−tanx))².
Steps only
tan(π/4+x)=(1+t)/(1−t)
tan(π/4−x)=(1−t)/(1+t)
Divide the expressions
Answer
The stated ratio identity holds wherever both sides are defined.
Question 8
Prove cos(π+x)cos(−x)/[sin(π−x)cos(π/2+x)]=cot²x.
Steps only
cos(π+x)=−cosx
cos(−x)=cosx
sin(π−x)=sinx; cos(π/2+x)=−sinx
Answer
The quotient simplifies to cot²x.
Question 9
Prove cos(3π/2+x)cos(2π+x)[cot(3π/2−x)+cot(2π+x)]=1.
Steps only
cos(3π/2+x)=sinx
cos(2π+x)=cosx
cot(3π/2−x)=tanx; cot(2π+x)=cotx
Answer
The product is 1.
Question 10
Prove sin((n+1)x)sin((n+2)x)+cos((n+1)x)cos((n+2)x)=cosx.
Steps only
Use cos(A−B)=cosAcosB+sinAsinB
A−B=−x
Answer
The left side is cosx.
Question 11
Prove cos(3π/4+x)−cos(3π/4−x)=−√2 sinx.
Steps only
cosA−cosB=−2sin((A+B)/2)sin((A−B)/2)
=−2sin(3π/4)sinx
Answer
The difference is −√2 sinx.
Question 12
Prove sin²6x−sin²4x=sin2x sin10x.
Steps only
sin²A−sin²B=sin(A+B)sin(A−B)
A=6x, B=4x
Answer
The difference is sin10x sin2x.
Question 13
Prove cos²2x−cos²6x=sin4x sin8x.
Steps only
cos²A−cos²B=sin(A+B)sin(B−A)
A=2x, B=6x
Answer
The difference is sin8x sin4x.
Question 14
Prove sin2x+2sin4x+sin6x=4cos²x sin4x.
Steps only
sin2x+sin6x=2sin4x cos2x
2sin4x(1+cos2x)
=4cos²x sin4x
Answer
The sum is 4cos²x sin4x.
Question 15
Prove cot4x(sin5x+sin3x)=cotx(sin5x−sin3x).
Steps only
sin5x+sin3x=2sin4x cosx
sin5x−sin3x=2cos4x sinx
Answer
Both sides equal 2cos4x cosx.
Question 16
Prove (cos9x−cos5x)/(sin17x−sin3x)=−sin2x/cos10x.
Steps only
cos9x−cos5x=−2sin7x sin2x
sin17x−sin3x=2cos10x sin7x
Answer
The quotient is −sin2x/cos10x.
Question 17
Prove (sin5x+sin3x)/(cos5x+cos3x)=tan4x.
Steps only
Numerator=2sin4x cosx
Denominator=2cos4x cosx
Answer
The quotient is tan4x.
Question 18
Prove (sinx−siny)/(cosx+cosy)=tan((x−y)/2).
Steps only
Numerator=2cos((x+y)/2)sin((x−y)/2)
Denominator=2cos((x+y)/2)cos((x−y)/2)
Answer
The quotient is tan((x−y)/2).
Question 19
Prove (sinx+sin3x)/(cosx+cos3x)=tan2x.
Steps only
Numerator=2sin2x cosx
Denominator=2cos2x cosx
Answer
The quotient is tan2x.
Question 20
Prove (sinx−sin3x)/(sin²x−cos²x)=2sinx.
Steps only
sinx−sin3x=−2cos2x sinx
sin²x−cos²x=−cos2x
Answer
The quotient is 2sinx.
Question 21
Prove (cos4x+cos3x+cos2x)/(sin4x+sin3x+sin2x)=cot3x.
Steps only
cos4x+cos2x=2cos3x cosx
sin4x+sin2x=2sin3x cosx
Answer
The quotient is cot3x.
Question 22
Prove cotx cot2x−cot2x cot3x−cot3x cotx=1.
Steps only
Group as cot2x(cotx−cot3x)−cot3x cotx
Use cot(A−B)=(cotA cotB+1)/(cotB−cotA)
Answer
The left side is 1.
Question 23
Prove tan4x=4tanx(1−tan²x)/(1−6tan²x+tan⁴x).
Steps only
Let t=tanx
tan2x=2t/(1−t²)
tan4x=2tan2x/(1−tan²2x)
Answer
The stated formula follows.
Question 24
Prove cos4x=1−8sin²x cos²x.
Steps only
cos4x=1−2sin²2x
sin2x=2sinx cosx
Answer
The expression simplifies to cos4x.
Question 25
Prove cos6x=32cos⁶x−48cos⁴x+18cos²x−1.
Steps only
Let c=cosx
cos3x=4c³−3c
cos6x=2cos²3x−1
Answer
The expansion is 32c⁶−48c⁴+18c²−1.
Miscellaneous Exercise on Chapter 3
Textbook pages 71–72
Question 1
Prove 2cos(π/13)cos(9π/13)+cos(3π/13)+cos(5π/13)=0.
Steps only
2cosAcosB=cos(A+B)+cos(A−B)
cos(10π/13)=−cos(3π/13)
cos(8π/13)=−cos(5π/13)
Answer
The terms cancel in pairs, giving 0.
Question 2
Prove (sin3x+sinx)sinx+(cos3x−cosx)cosx=0.
Steps only
sin3x sinx+cos3x cosx=cos2x
sin²x−cos²x=−cos2x
Answer
0.
Question 3
Prove (cosx+cosy)²+(sinx−siny)²=4cos²((x+y)/2).
Steps only
Expand and use sin²+cos²=1
=2+2cos(x+y)
=4cos²((x+y)/2)
Answer
The left side equals 4cos²((x+y)/2).
Question 4
Prove (cosx−cosy)²+(sinx−siny)²=4sin²((x−y)/2).
Steps only
Expand and use sin²+cos²=1
=2−2cos(x−y)
=4sin²((x−y)/2)
Answer
The left side equals 4sin²((x−y)/2).
Question 5
Prove sinx+sin3x+sin5x+sin7x=4cosx cos2x sin4x.
Steps only
sinx+sin7x=2sin4x cos3x
sin3x+sin5x=2sin4x cosx
Answer
The sum is 4cosx cos2x sin4x.
Question 6
Prove [(sin7x+sin5x)+(sin9x+sin3x)]/[(cos7x+cos5x)+(cos9x+cos3x)]=tan6x.
Steps only
Numerator=2sin6x(cosx+cos3x)
Denominator=2cos6x(cosx+cos3x)
Answer
The quotient is tan6x.
Question 7
Prove sin3x+sin2x−sinx=4sinx cos(x/2)cos(3x/2).
Steps only
sin3x−sinx=2cos2x sinx
sin2x=2sinx cosx
2sinx(cos2x+cosx)
Answer
The sum is 4sinx cos(x/2)cos(3x/2).
Question 8
Given tanx=−4/3 and x is in quadrant II, find sin(x/2), cos(x/2) and tan(x/2).
Steps only
sinx=4/5; cosx=−3/5
x/2 lies in quadrant I
sin(x/2)=2/√5; cos(x/2)=1/√5; tan(x/2)=2
Answer
sin(x/2)=2/√5; cos(x/2)=1/√5; tan(x/2)=2.
Question 9
Given cosx=−1/3 and x is in quadrant III, find sin(x/2), cos(x/2) and tan(x/2).
Steps only
x/2 lies in quadrant II
sin(x/2)=√((1−cosx)/2)=√6/3
cos(x/2)=−√((1+cosx)/2)=−√3/3
Answer
sin(x/2)=√6/3; cos(x/2)=−√3/3; tan(x/2)=−√2.
Question 10
Given sin x=4/5 and x is in quadrant II, find sin(x/2), cos(x/2) and tan(x/2).
Steps only
cosx=−√(1−16/25)=−3/5
x/2 lies in quadrant I
sin(x/2)=√((1−cosx)/2)=2/√5; cos(x/2)=√((1+cosx)/2)=1/√5
Answer
sin(x/2)=2/√5; cos(x/2)=1/√5; tan(x/2)=2.