1st PUC Maths Question Bank Chapter 3 Trigonometric Functions

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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.

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″

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	\(\frac{3 \pi}{2}\)	2π

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:

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:

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

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 :

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 :

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

∴ Radius OB = 20 cm
Chord BC = 20 cm
∴  Δ OBC is an equilateral triangle

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

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 :

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
\( \text { Given } \quad \theta_{1}=60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)

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 :

Trigonometric Functions:

Recall:

Graph of trigonometric function:

Question 1.
If \(\cos x=-\frac{3}{5}, x\) lies in the third quadrant, find the values of other five trigonometric functions.
Answer

Question 2.
If \( \cos x=-\frac{1}{2}, x \) lies in the third quadrant, find the values of other five trigonometric functions.
Answer:

Question 3.
If \( \sin x=\frac{3}{5}, x \) lies in the second quadrant, find the values of other five trigonometric functions.
Answer:

Question 4.
If \( \tan x=-\frac{5}{12}, x \)  lies in the second quadrant, find the values of other five trigonometric functions.
Answer:

Question 5.
If \(\cot x=\frac{3}{4}, x \) lies in third quadrant, find the values of other five trigonometric functions.
Answer:

Question 6.
If \(\\cot x=-\frac{5}{12}, x lies\) in second quadrant, find the values of other five trigonometric functions.
Answer:

\(\cot x=-\frac{5}{12}\)

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
-θ	– 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.
\( \sin \frac{31 \pi}{3}=\sin \left(10 \pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

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.

Answer:

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:

Question 18.
Prove that
\( \sin \left(\frac{\pi}{2}-x\right)=\cos x\)
Answer:

Question 19.
Prove that sin(x + y) = sinx cosy + cosx siny .
Answer :

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 .

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:

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:

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.

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

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

Question 26.
Prove that
\(\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x} \)
Answer:

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:

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


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)

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:

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:

Question 33.
Find the value of \( \tan \frac{13 \pi}{12} \)
Answer:

Question 34.
Prove that
\( \frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} \)
Answer:

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:
\( L H S=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}=R H S\)

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:

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:

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:

Question 48.
Prove that
\(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x \)
Answer:

Question 49.
Prove that
\(\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x\)
Answer:

Question 50.
Prove that
\(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x\)
Answer:

Question 51.
Prove that
\(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right) \)
Answer:

Question 52.
Prove that
\(\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x \)
Answer:

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:

Question 54.
Prove that
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Answer:

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:

Question 56.
Prove that
\((\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2}\left(\frac{x+y}{2}\right)\)
Answer:

Question 57.
Prove that
\((\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)\)
Answer:

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.

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:

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:

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:

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:

 

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:

Question 66.
Find the value of  \( \tan \frac{\pi}{8}\)
Answer:

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:

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:

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:

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.

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:

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:

Question 6.
Prove that, for any real numbers x and y, cosx=cosy, implies x = 2nπ±y,where n∈Z
Answer:

Question 7.
Prove that, if x and y are not odd multiple of \(\frac{\pi}{2} \)
Answer:
Given: tan x=tan y

Question 8.
Find the solution of  \(\sin x=-\frac{\sqrt{3}}{2}\)
Answer:

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:

Question 11.
Find the general solution for which cos4x = cos2x.
Answer:

Question 12.
Solve: sin 2x – sin4x + sin 6x = 0
Answer :

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

Question 14.
Solve: sin2x+cosx = 0.
Answer :
Given sin 2x + cosx = 0
⇒ 2sinxcosx + cosx = 0
⇒ cosx(2sin x+1)=0

Question 15.
Solve: sinx + sin3x+sin5x = 0.
Answer :
Given: (sin5x + sinx) + sin3x = 0
⇒2sin3xcos2x + sin3x = 0
∴ sin3x(2cos2x + 1) = 0

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

Question 17.
Solve: sec²2x = 1-tan2x.
Answer :

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.

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

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

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

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:

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

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:

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:

Question 7.
For any triangle ABC, Prove that
\(\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a} \cos \frac{A}{2}\)
Answer:

Question 8.
In any ΔABC, prove that a(bcosC-c cos B) = b²-c².
Answer :

Question 9.
In any ΔABC, prove that
\( a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\)
Answer:

Question 10.
In any ΔABC, prove that
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}} \)
Answer:

Question 11.
In any ΔABC, prove that
\((b+c) \cos \frac{B+C}{2}=a \cos \frac{B-C}{2}\)
Answer:

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

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:

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:

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:

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,
\( \frac{h}{\sin 45^{\circ}}=\frac{35}{\sin 30^{\circ}}\)

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.

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

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:

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: