1st PUC Maths Question Bank Chapter 3 Trigonometric Functions

Students can Download Maths Chapter 3 Trigonometric Functions Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 3 Trigonometric Functions

Measurement of Angles:

Question 1.
What is the meaning of trigonometry?
Answer :
The literal meaning of the word trigonometry is the science of measuring the sides and the angles of triangles.

Question 2.
Define an angle.
Answer :
Angle is a measure of rotation of a given ray about its initial point.
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 1

Question 3.
Define a degree measure.
Answer :
If a rotation from the initial side to terminal side is \( \left(\frac{1}{360}\right)^{t h} \) of a revolution, the angle is said to have a measure of one degree, written as 1°.

Keen eye:
A degree is divided into 60 minutes and a minute is divided into 60 seconds. One sixtieth of a degree is called a minute, written as 1′ and one sixtieth of a minute is called second, written as 1″
∴ 1 ° – 60′
1’= 60″

KSEEB Solutions

Question 4.
Define a radian measure.
Answer :
Angle subtended at the centre by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian.

Keen eye:
\(1 \mathrm{rad}=\frac{180^{\circ}}{\pi}=57^{\circ} 16^{\prime} \)

Degree 30° 45° 60° 90° 180° 270° 360°
Radian \(\frac{\pi}{6}\) \(\frac{\pi}{4}\) \(\frac{\pi}{3}\) \(\frac{\pi}{2}\) π \(\frac{3 \pi}{2}\)

Question 5.
Convert the following into radian measure,
(i) 25°
(ii) 240°
(iii) 520°
(iv) 225°
(v) 150°
(vi) 330°
(vii) – 47° 30′
(viii) 40° 20′
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 2

Question 6.
Convert the following into degree measure.
(i) 6
(ii) \( \frac{11}{16}\)
(iii) -4
(iv) \( \frac{5 \pi}{3} \)
(v) \( \frac{7 \pi}{6} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 3
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 4
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 5

Question 7.
A wheel makes 360° revolutions in one minute. Through how many radians does it turn in one second?
Answer :
Number of revolutions in one minute = 360
Number of revolutions in one second = \( \frac{360}{60}=6  \)
But, in one revolutions angle traced = 360°
= 2π rad
⇒ Angle traced in 6 revolutions
= 6 x 2π= 12π radians
Angle traced in one second = 12π radians

KSEEB Solutions

Question 8.
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 \(\left(\text { use } \pi=\frac{22}{7}\right)\)
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 6

Question 9.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length \(37.4 \mathrm{cm}\left(\text { use } \pi=\frac{22}{7}\right)\)
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 7

Question 10.
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 :
Given, diameter AB = 40 cm
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 8
∴ Radius OB = 20 cm
Chord BC = 20 cm
∴  Δ OBC is an equilateral triangle
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 9

Question 11.
The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minute?
(use π = 3.14).
Answer :
In 60 minutes, the minute hand of a watch completes one revolution.
∴ In 40 minutes, the minute hand turns through  of a revolution
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 10

Question 12.
If the arcs of the same lengths in two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 11

Question 13.
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 r1 and r2 be the radii of two circles
\( \text { Given } \quad \theta_{1}=60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 12

Question 14.
Find the angle in radian through 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 :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 13

Trigonometric Functions:

Recall:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 14

1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 15

1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 16

1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 18

Graph of trigonometric function:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 19
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 20
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 21

Question 1.
If \(\cos x=-\frac{3}{5}, x\) lies in the third quadrant, find the values of other five trigonometric functions.
Answer
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 22

Question 2.
If \( \cos x=-\frac{1}{2}, x \) lies in the third quadrant, find the values of other five trigonometric functions.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 23

Question 3.
If \( \sin x=\frac{3}{5}, x \) lies in the second quadrant, find the values of other five trigonometric functions.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 24
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 25

Question 4.
If \( \tan x=-\frac{5}{12}, x \)  lies in the second quadrant, find the values of other five trigonometric functions.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 26

Question 5.
If \(\cot x=\frac{3}{4}, x \) lies in third quadrant, find the values of other five trigonometric functions.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 27
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 28

Question 6.
If \(\\cot x=-\frac{5}{12}, x lies\) in second quadrant, find the values of other five trigonometric functions.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 29
\(\cot x=-\frac{5}{12}\)

Question 7.
If \(\sec x=\frac{13}{5}, x\) lies in fourth quadrant, find the values of other five trigonometric functions.
Answer:

Measure of angle sin cos tan cosec sec cot
– sinθ cosθ -tanθ – cosecθ secθ – cotθ
\(\frac{\pi}{2}-\theta \) cosθ -sinθ cot θ secθ cosecθ tanθ
\(\frac{\pi}{2}+\theta \) cosθ – sinθ -cotθ – secθ cosecθ -tanθ
π-θ sinθ -cosθ -tanθ cosec θ -secθ -cotθ
π+θ – sinθ -cosθ tanθ -cosccθ – secθ cotθ
\(\frac{3 \pi}{2}-\theta\) – cosθ -sinθ cotθ – secθ – cosecθ tan θ
\(\frac{3 \pi}{2}+\theta\) -cosθ sinθ -cotθ – secθ cosecθ tanθ
2π-θ -sinθ cosθ -tanθ – cosccθ secθ -cotθ
2π+θ sinθ cos θ tanθ cosecθ secθ cosθ

Question 8.
Find the value of
\( \sin \frac{31 \pi}{3} \)
Answer:
We know that values of sinx repeats after an interval of 2π.
\( \sin \frac{31 \pi}{3}=\sin \left(10 \pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

KSEEB Solutions

Question 9.
Find the value of sin 765°.
Answer:
\( \sin 765^{\circ}=\sin \left(720^{\circ}+45^{\circ}\right)=\sin 45^{\circ}=\frac{1}{\sqrt{2}}\)

Question 10.
Find the value of
\( \sin \left(-\frac{11 \pi}{3}\right) \)
Answer:
\(\begin{aligned} \sin \left(-\frac{11 \pi}{3}\right) &=-\sin \left(\frac{11 \pi}{3}\right)=-\sin \left(4 \pi-\frac{\pi}{3}\right) \\ &=\sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2} \end{aligned}\)

Question 11.
Find the value of cos (-1710°)
Answer :
cos (1710°) = cos (1710°)
\( \begin{array}{l}{=\cos \left(1800-90^{\circ}\right)=\cos \left(10 \pi-\frac{\pi}{2}\right)} \\ {=\cos \frac{\pi}{2}=0}\end{array}\)

Question 12.
Find the value of cosec (-1410°)
Answer :
cosec (- 1410°)
= – cosec (1410°)
= – cosec[4(360°) – 30°]
= – cosec (- 30°)
= cosec 30° = 2

Question 13.
Find the value of
\(  \tan \frac{19 \pi}{3}\)
Answer:
\(\tan \frac{19 \pi}{3}=\tan \left(6 \pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3} \)

Question 14.
Find the value of
\( \cot \left(-\frac{15 \pi}{4}\right)\)
Answer:
\(\begin{aligned} \cot \left(-\frac{15 \pi}{4}\right) &=-\cot \left(\frac{15 \pi}{4}\right)=-\cot \left(4 \pi-\frac{\pi}{4}\right) \\ &=\cot \frac{\pi}{4}=1 \end{aligned}\)

Question 15.
Derive cos(x + y) = cos x cos y – sin x sin y , using unit circle concept.
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 30
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 31
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 32

Question 16.
Prove that:
cos(x-y) = cosxcosy+ sinxsiny.
Answer :
We have cos(x + y) = cos xcos y – sin xsin y
Replace y by – y, we get
cos[x + (-y)] = cos xcos(-y) – sin xsin(-y)
cos(x – y) = cos x cos y + sin x sin y

Question 17.
Prove that
\( \cos \left(\frac{\pi}{2}-x\right)=\sin x \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 33

Question 18.
Prove that
\( \sin \left(\frac{\pi}{2}-x\right)=\cos x\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 34

Question 19.
Prove that sin(x + y) = sinx cosy + cosx siny .
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 35
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 36

Question 20.
Prove that sin( x – y) = sin x cos y – cos x sin y.
Answer :
sin(x – y) = sin[x + (-y)]
= sin x cos(-y) + cos x  sin(-y)
= sin x cos y-cos x sin y .

KSEEB Solutions

Question 21.
Prove the following
(i) \(\cos \left(\frac{\pi}{2}+x\right)=-\sin x \)
(ii) \( \sin \left(\frac{\pi}{2}+x\right)=\cos x \)
(iii) cos(π – x) = – cosx
(iv) sin(π – x) =sinx
(v) cos(π + x) = – cosx
(vi) sin(π + x) = – sinx
(vii) cos(2π – x) =cosx
(viii) sin(2π – x) = – sin x.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 37

Question 22.
If none of the angles ac, y and x + y is an odd multiple of \(\frac{\pi}{2} \)then prove that
\(\tan (x \pm y)=\frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 38
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 39

Question 23.
If none of the angles x,y and (x+y) is a multiple of π, then prove that
\( \cot (x \pm y)=\frac{\cot x \cot y \pm 1}{\cot y \pm \cot x}\)
Answer :
As x, y and x+y are not the multiple of π, so Sinx, sin y and sin(x+ y) are non zero.
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 40
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 41

Question 24.
Prove that
\(\cos 2 x=\left\{\begin{array}{l}{\cos ^{2} x-\sin ^{2} x} \\ {2 \cos ^{2} x-1} \\ {1-2 \sin ^{2} x} \\ {\frac{1-\tan ^{2} x}{1+\tan ^{2} x}}\end{array}\right.\)
Answer:
We have cos(x + y) = cos xcos y – sin xsin y
Replacing y by x, we get
cos(x + x) = cosx cosx-sinx sinx
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 42

Question 25.
Prove that
\( \sin 2 x=\left\{\begin{array}{c}{2 \sin x \cos x} \\ {\frac{2 \tan x}{1+\tan ^{2} x}}\end{array}\right.\)
Answer:
We have sin(x + y) = sin xcos y + cos xsin y
Replacing y by x we get
sin 2x = 2sinx cosx

1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 43

Question 26.
Prove that
\(\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 44

KSEEB Solutions

Question 27.
Prove that sin3x = 3sinx – 4sin3 x .
Answer :
sin 3x = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= 2sin xcos x cos x + (1 – 2sin2 x)sin x
= 2 sin x(1- sin2 x) + sin x – 2 sin3 x
= 3sinx-4sin3 x

Question 28.
Prove that cos3x = 4cos3 x -3cosx .
Answer :
cos3x = cos(2x + x)
= cos 2x cos x – sin 2x sin x
= (2 cos2 x -1) cos x – 2 sin x cos x sin x
= 2cos3x-cosx-2cosx(1-cos2x)
= 4cos3 x-3cosx

Question 29.
Prove that
\(\tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 45
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 46

Question 30.
Prove the following
(i) \(\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\)
(ii) \( \cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \)
(iii) \( \sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} \)
(iv) \( \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \)
Answer:
We have
cos(A + B) = cosAcosB-sinAsinB …(1)
cos(A – B) = cos AcosB + sin Asin B … (2)
Adding and subtracting (1) and (2) we get
cos( A + B) + cos( A -B) = 2 cos A cos B … (3)
cos( A + B) – cos (A -B) = – 2 sin A sin B … (4)
Further
sin(A + B) = sinAcosB + cosAsinB         …(5)
sin(A-B) = sinAcosB-cosAsinB …(6)
Adding and subtracting (5) and (6) we get
sin( A + B) + sin( A – B) = 2 sin A cos B … (7)
sin( A + B) – sin(A – B) = 2cosA sinB … (8)
Let A + B = x and A-B = y
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 47
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 48
Keen eye;

  • 2 cos xcos y = cos(x + y) + cos(x – y)
  • -2 sin x sin y = cos(x + y) – cos(x – y)
  • 2sin x cos y = sin(x + y) 4-sin(x- y)
  • 2cosx siny = sin(x+ y)-sin(x-y)

KSEEB Solutions

Question 31.
Prove the following
(i) \( 3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}=1\)
(ii) \( \sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}\)
(iii) \( 2 \sin ^{2} \frac{\pi}{6}+\csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2} \)
(iv) \( 2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10 \)
(v) \( \cot ^{2} \frac{\pi}{6}+\csc \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6 \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 49
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 50

Question 32.
Find the Value of
(i) sin 75°
(ii) cos 75°
(iii) tan 75°
(iv) sin 15°
(v) cos 15°
(vi) tan 15°
(vii) sin 105°
(viii) cos 105°.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 51
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 52

Question 33.
Find the value of \( \tan \frac{13 \pi}{12} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 53

Question 34.
Prove that
\( \frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 54

Question 35.
Prove that
\(\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)} \\ {=\sin (x+y)}\end{array} \)
Answer:
\( \begin{aligned} L H S &=\cos \left[\frac{\pi}{4}-x+\frac{\pi}{4}-y\right] \\ &=\cos \left[\frac{\pi}{2}-(x+y)\right] \\ &=\sin (x+y)=R H S \end{aligned} \)

Question 36.
Prove that
sin[(n + l)x]sin[(n + 2)x] +cos[(n+1)x]cos[(n + 2)x] = cosx
Answer :
LHS = cos[(n + 2)x – (n + 1)x]
= cos x = RHS .

Question 37.
Prove that
(sin3x + sinx)sinx + (cos3x -cosx)cosx = 0
Answer :
LHS = sin 3x  sin x+sin2 x + cos 3xcos x – cos2 x
= [cos 3x cos x + sin 3x sin x] – [cos2 x – sin2 x]
= cos(3x – x) – cos 2x
= 0
= RHS

Question 38.
Prove that
\(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4} x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2} \)
Answer:
\( L H S=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}=R H S\)

KSEEB Solutions

Question 39.
Prove that
\( \frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x\)
Answer:
\(L H S=\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}=\cot ^{2} x=R H S \)

Question 40.
Prove that
\( \begin{array}{l}{\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)} \\ {\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1}\end{array}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 55

Question 41.
Prove that
\(\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x\)
Answer:

1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 56

Question 42.
Prove that sin26x-sin24x=sin2xsinl0x.
Answer :
LHS = sin(6x + 4x)sin(6x – 4x)
∵ sin( A + B) sin(A – B)  = sin2 A – sin2 B
LHS = sin 10x . sin2x
= RHS.

Question 43.
Prove that cos2 2x-cos26x =sin4x sin8x .
Answer :
LHS = (1 – sin2 2x) – (1 – sin2 6x)
= sin2 6x – sin2 2x = sin(6x + 2x) sin(6x – 2x)
= sin8x sin4x
= RHS

Question 44.
Prove that
\( \cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x \)
Answer:
\(\begin{aligned} L H S &=\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x+\sin \frac{\pi}{4} \sin x \\ &=2 \cos \frac{\pi}{4} \cos x=2\left(\frac{1}{\sqrt{2}}\right) \cos x=\sqrt{2} \cos x \\ &=R H S \end{aligned} \)

Question 45.
Prove that
tan3x tan2x tanx = tan3x-tan2x-tanx.
Answer :
Consider: tan 3x = tan(2x + x) tan 2x + tan x
\( \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x} \)
⇒ tan 3x(1 – tan 2x tan x) = tan 2x + tan x
⇒ tan3x-tan3x tan2x tanx = tan2x + tanx
∴ tan 3x – tan 2x – tan x = tan 3x tan 2x tan x

Question 46.
Prove that
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1.
Answer :
Consider cot 3x = cot(2x + x)
\( =\frac{\cot 2 x \cot x-1}{\cot 2 x+\cot x} \)
⇒ cot 3x(cot 2x+cot x) – cot 2x cot x -1
⇒ 1 = cot 2x cot x – cot 3x cot 2x – cot 3x cot x

Question 47.
Prove that
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 57

Question 48.
Prove that
\(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 58

Question 49.
Prove that
\(\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 59

Question 50.
Prove that
\(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 60

Question 51.
Prove that
\(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right) \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 61

Question 52.
Prove that
\(\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 62

Question 53.
Prove that
\(\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 63

Question 54.
Prove that
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 64

Question 55.
Prove that
\(\frac{\sin 9 x+\sin 7 x+\sin 5 x+\sin 3 x}{\cos 9 x+\cos 7 x+\cos 5 x+\cos 3 x}=\tan 6 x\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 65

Question 56.
Prove that
\((\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2}\left(\frac{x+y}{2}\right)\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 66

Question 57.
Prove that
\((\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 67
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 68

Question 58.
Prove that
sinx + sin3x + sin5x + sin7x
= 4 cos x cos 2x sin 4x
Answer :
LHS = (sin 7x + sin x) + (sin 5x + sin 3x)
= 2 sin 4x cos 3x + 2 sin 4x cos x
= 2 sin 4x(cos 3x + cos x)
= 2 sin 4x 2 cos 2x cos x
= 4cosx cos2x sin4x
= RHS.

KSEEB Solutions

Question 59.
Prove that
\(\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 69

Question 60.
Prove that
\(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 70

Question 61.
Prove that
\(\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 71
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 72

Question 62.
Prove that cos4x = 1 – 8sin2 x cos2 x.
Answer :
LHS = cos2(2x)
= 1-2sin2 2x = 1-2(2sinx cosx)2
= 1-8sin2xcos2x
= RHS

Question 63.
Prove that
cos6x = 32cos6 x – 48cos4 x + 18cos2 x -1.
Answer :
LHS = cos3(2x)
= 4cos3 2x-3cos2x
= 4(2 cos2 x -1)3 – 3(2 cos2 x -1)
= 4[8 cos6 x -1 – 3(2cos2 x)(2cos2 x -1)] -6cos2x + 3
= 32cos6x-4-24cos2x(2cos2x-1) -6cos2x + 3
= 32 cos6 x – 48 cos4 x +18 cos2 x -1
= RHS.

Question 64.
If \(\sin x=\frac{3}{5}, \cos y=-\frac{12}{13} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 73
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 74

 

Question 65.
Prove that
\(\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 75

Question 66.
Find the value of  \( \tan \frac{\pi}{8}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 76
Question 67.
Prove that
\( \cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 77

Question 68.
\(\begin{aligned} &\text { If } \tan x=\frac{3}{4}, \pi<x<\frac{3 \pi}{2}, \text { find the value of }\\ &\sin \frac{x}{2}, \cos \frac{x}{2} \text { and } \tan \frac{x}{2} \end{aligned}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 78

Question 69.
Find \(\sin \frac{x}{2}, \cos \frac{x}{2} \text { and } \tan \frac{x}{2} \)
(i) \(\tan x=-\frac{4}{3}, \) x in quadrant II
(ii) \( \cos x=-\frac{1}{3}, \) x in quadrant III
(iii) \( \sin x=\frac{1}{4}, x \)in quadrant II
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 79
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 80
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 81
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 82

Trigonometric Equations

Question 1.
Define trigonometric equations.
Answer :
Equations involving trigonometric functions of a variable are called trigonometric equations.

Question 2.
Define principal solutions of trigonometric equations.
Answer :
The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

KSEEB Solutions

Question 3.
Define general solution of trigonometric equation.
Answer :
The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution.

Question 4.
Find the principal solution of the following equations
(i) \(\sin x=\frac{\sqrt{3}}{2}\)
(ii) \(\tan x=-\frac{1}{\sqrt{3}}\)
(iii) \( \tan x=\sqrt{3}\)
(iv) sec x=2
(v) \( \cot x=\sqrt{3}\)
(vi) cosec x=-2
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 83
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 84
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 85

Question 5.
Prove that, for any real numbers x and y, sin x = sin y implies x =nπ + (-1)n y , where n∈Z
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 86

Question 6.
Prove that, for any real numbers x and y, cosx=cosy, implies x = 2nπ±y,where n∈Z
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 87

Question 7.
Prove that, if x and y are not odd multiple of \(\frac{\pi}{2} \)
Answer:
Given: tan x=tan y
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 88

Question 8.
Find the solution of  \(\sin x=-\frac{\sqrt{3}}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 89

Note:
\(\frac{4 \pi}{3}\) is one such value of x for which \(\sin x=-\frac{\sqrt{3}}{2} \)
One may take any other value of x for which \(\sin x=-\frac{\sqrt{3}}{2}\)
The solutions obtained will be the same although these may apparently look different.

Question 9.
Solve: \(\cos x=\frac{1}{2}\)
Answer:
\(\begin{array}{l}{\text { We have, } \cos x=\frac{1}{2}=\cos \frac{\pi}{3}} \\ {x=2 n \pi \pm \frac{\pi}{3}, n \in Z}\end{array}\)

Question 10.
Solve:
\(\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 90

Question 11.
Find the general solution for which cos4x = cos2x.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 91

Question 12.
Solve: sin 2x – sin4x + sin 6x = 0
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 92

Question 13.
Solve: cos3x + cosjc-cos2jc = 0.
Answer :
Given: (cos 3x + cosx) – cos 2x = 0
⇒ 2cos2x cosac-cos 2x = 0
cos 2x(2cos x -1) = 0
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 93

Question 14.
Solve: sin2x+cosx = 0.
Answer :
Given sin 2x + cosx = 0
⇒ 2sinxcosx + cosx = 0
⇒ cosx(2sin x+1)=0
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 94

Question 15.
Solve: sinx + sin3x+sin5x = 0.
Answer :
Given: (sin5x + sinx) + sin3x = 0
⇒2sin3xcos2x + sin3x = 0
∴ sin3x(2cos2x + 1) = 0
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 95

Question 16.
Solve: 2cos2x + 3sinx = 0
Answer :
Given: 2(1 – sin2 x) + 3sin x = 0
⇒ 2sin2 x-3sinx-2 = 0
⇒ (2sin x+1)(sin x-2)=0
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 96

Question 17.
Solve: sec22x = 1-tan2x.
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 97

Sine and cosine formulae

Question 1.
State and prove sine formula.
Answer :
Statement: In any triangle, sides are proportional to the sines of the opposite angles. That is, in a triangle ABC
\(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)
Proof : Let ABC be either acute angled triangle or obtuse angled triangle.
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 98
The altitude h is drawn from the vertex B to meet the side AC in point D.
From the right angled triangle ABD, we have

1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 99

Question 2.
State and prove cosine formula.
Answer :
Statement: Let A, B and C be angles of a triangle and a, b and c be lengths of sides opposite to angles A, B and C respectively, then
a2 =b2 + c2-2bc cos A
b2 =c2 +a2-2ca cos B
c2 =a2 +b2 – 2ab cos C
Proof:
Let ABC be triangle as given below
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 100
Referring to second triangle ,we have
BC2 = BD2 + DC2 = BD2 + (AC – AD)2
= BD2 + AD2 + AC2 – 2AC AD
= AB2 + AC2 -2AC AB cos A
= a2 =b2 +c2 -2bc cos A
Similarly we can obtain
b2 =c2 + a2 – 2ca cosB
and c2 = a2 +b2 – 2ab cos C

Keen eye:
\(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\)
\(\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 a c} \)
\(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

KSEEB Solutions

Question 3.
(Napier Analogies) In triangle ABC, .prove that
(i)\(\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}\)
(ii) \(\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}\)
(iii) \(\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 101

Question 4.
In any triangle ABC , if a = 18, b = 24, c = 30, find
(i) cosA, cosB, cosC
(ii) sinA, sinB, sinC
Answer :
(i) We have
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 102

Question 5.
For any triangle ABC,Prove that
\(\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 103

Question 6.
For any triangle ABC, Prove that
\(\frac{a-b}{c}=\frac{\sin \left(\frac{A-B}{2}\right)}{\cos \frac{C}{2}}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 104

Question 7.
For any triangle ABC, Prove that
\(\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a} \cos \frac{A}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 105

Question 8.
In any ΔABC, prove that a(bcosC-c cos B) = b2-c2.
Answer :
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 106

Question 9.
In any ΔABC, prove that
\( a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 107
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 108

Question 10.
In any ΔABC, prove that
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}} \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 109

Question 11.
In any ΔABC, prove that
\((b+c) \cos \frac{B+C}{2}=a \cos \frac{B-C}{2}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 110

Question 12.
In any ΔABC, prove that
acosA+bcosB + ccosC = 2asin BsinC.
Answer :
LHS – k sin A cos A + k sin B cos B + k sin C cos C
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 111

Question 13.
In any ΔABC, prove that
\(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 112

Question 14.
In any ΔABC, prove that
\( \left(b^{2}-c^{2}\right) \cot A+\left(c^{2}-a^{2}\right) \cot B+\left(a^{2}-b^{2}\right) \cot C=0\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 113

KSEEB Solutions

Question 15.
In any ΔABC, prove that
\( \frac{b^{2}-c^{2}}{a^{2}} \sin 2 A+\frac{c^{2}-a^{2}}{b^{2}} \sin 2 B+\frac{a^{2}-b^{2}}{c^{2}} \sin 2 C=0\)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 114

Question 16.
In any ΔABC, prove that
asin(B -C) + bsin(C – A) + csin(A-B) = 0
Answer :
LHS = k sin A sin(B -C) + ksinB sin(C – A) +k sin C sin( A – B)
= k[sin(B + C) sin(B – C)+sin(C + A) sin(C – A) +sin( A + B) sin( A – B)]
= K[sin2 B – sin2 C + sin2 C -sin2 A + sin2 A – sin2 B]
= k x 0
=0=RHS.

Question 17.
A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35 m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.
Answer :
In ΔOBT,
\( \frac{h}{\sin 45^{\circ}}=\frac{35}{\sin 30^{\circ}}\)
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 115

Question 18.
Two ships leave a port at the same time. One goes 24 km per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.
Answer :
Let C be the port A and B are the positions of the two ships after 3 hours.
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 116

Question 19.
Two trees, A and B are on the same side of a river. From a point C in the river the distance of the trees A and B is 250 m and 300 m, respectively. If the angle C is 45°, find the distance of the trees.
Answer :
Given AC = 250 m
BC = 300 m
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 117

Question 20.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance ‘d’ from the point A measured along the line AB which makes an angle 30° with AQ. Prove that \(d=h(\sqrt{3}-1) \)
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 118
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 119

Question 21.
A lamp past is situated at the middle point M of the side AC of a triangular plot ABC with BC = 7 m, CA = 8 m and AB = 9 m. Lamp post subtends an angle 15° at the point B. Determine the height of the lamp post.
Answer:
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 120
1st PUC Maths Question Bank Chapter 3 Trigonometric Functions 121