NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2 - Trigonometric Functions

Question:1 Find the values of other five trigonometric functions $\small \cos x = -\frac{1}{2}$ , x lies in the third quadrant.

Answer:
$\cos x = -\frac {1}{2}$
$\because \sec x = \frac{1}{\cos x} = \frac{1}{-\frac {1}{2}} = -2$
x lies in the III quadrant. Therefore, sec x is negative

$\sin ^{2}x +\cos^{2}x = 1 \\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 -\left ( -\frac{1}{2} \right )^{2}\\ \sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}\\ \sin x = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$
x lies in the III quadrant. Therefor, sin x is negative
$\therefore \sin x= - \frac{\sqrt{3}}{2}$

$\because cosec \ x = \frac {1}{\sin x}= \frac{1}{- \frac{\sqrt{3}}{2}} =- \frac{2}{\sqrt{3}}$

x lies in the III quadrant. Therefore, cosec x is negative

$\tan x = \frac{\sin x}{\cos x} = \frac {-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$
x lies in the III quadrant. Therefore, tan x is positive

$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$
x lies in the III quadrant. Therefore, cot x is positive.

Question:2 Find the values of the other five trigonometric functions $\small \sin x = \frac{3}{5}$ x lies in the second quadrant.

Answer:

$\sin x = \frac {3}{5}$

$cosec \ x = \frac{1}{\sin x}=\frac {1}{\frac {3}{5}} = \frac {5}{3}$
x lies in the second quadrant. Therefore, cosec x is positive

$\sin^{2}x + \cos ^{2}x = 1\\ \cos ^{2}x = 1 - \sin ^{2}x\\ \cos ^{2}x = 1 - \left ( \frac{3}{5} \right )^{2}\\ \cos ^{2}x = 1 - \frac {9}{25} = \frac {16}{25}\\ \cos x = \sqrt{\frac {16}{25}} = \pm \frac {4}{5}$
x lies in the second quadrant. Therefore, cos x is negative
$\therefore \cos x = - \frac {4}{5}$

$\sec x = \frac {1}{\cos x} = \frac{1}{- \frac {4}{5}} = -\frac {5}{4}$
x lies in the second quadrant. Therefore, sec x is negative

$\tan x = \frac {\sin x}{\cos x} = \frac {\frac{3}{5}}{-\frac{4}{5}} = -\frac {3}{4}$
x lies in the second quadrant. Therefore, tan x is negative

$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac {3}{4}} = -\frac{4}{3}$
x lies in the second quadrant. Therefore, cot x is negative.

Question:3 Find the values of other five trigonometric functions $\small \cot x = \frac{3}{4}$ , x lies in third quadrant.

Answer:

$\cot x= \frac {3}{4}$

$\tan x = \frac{1}{\cot x}= \frac{1}{\frac {3}{4}} = \frac {4}{3}$
$1 + \tan ^ {2}x = \sec ^{2}x\\ 1+\frac{4^2}{3^2} = \sec ^{2}x\\ \\ 1 + \frac {16}{9} = \sec ^{2}x\\ \frac {25}{9} = \sec ^{2}x\\ \sec x = \sqrt {\frac {25}{9}} = \pm \frac {5}{3}$
x lies in x lies in third quadrant. therefore sec x is negative
$\sec x = -\frac{5}{3}$

$\cos x = \frac {1}{\sec x} = \frac {1}{-\frac{5}{3}} = -\frac{3}{5}$
$\sin ^{2 }x+ \cos ^{2}x = 1\\ \sin ^{2 }x = 1 - \cos ^{2}x\\ \sin ^{2 }x = 1 -\left ( -\frac{3}{5} \right )^{2}\\ \sin ^{2 }x = 1 - \frac {9}{25}\\ \sin ^{2 }x = \frac{16}{25}\\ \sin x = \sqrt {\frac{16}{25}} = \pm \frac{4}{5}$
x lies in the third quadrant. Therefore, sin x is negative
$\sin x = -\frac {4}{5}$
$cosec x = \frac {1}{\csc} = \frac {1}{-\frac{4}{5}} = - \frac{5}{4}$

Question:4 Find the values of other five trigonometric functions $\small \sec x = \frac{13}{5}$ , x lies in fourth quadrant.

Answer:
$\sec x = \frac {13}{5}$
$\cos x = \frac {1}{\sec x} = \frac{1}{\frac {13}{5}} = \frac {5}{13}$
$\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \frac {5}{13}\\ \sin^{2}x = 1 - \frac {25}{169} = \frac {144}{169}\\ \sin x = \sqrt { \frac {144}{169}} = \pm \frac {12}{13}$
lies in the fourth quadrant. Therefore sin x is negative
$\sin x =- \frac {12}{13}$
$\csc x = \frac {1}{\sin x} = \frac {1}{-\frac {12}{13}} = -\frac {13}{12}$
$\tan x = \frac {\sin x}{\cos x} = \frac {-\frac{12}{13}}{\frac{5}{13}} = -\frac {12}{5}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{12}{5}} = -\frac{5}{12}$

Question:5 Find the values of the other five trigonometric functions $\small \tan x = -\frac{5}{12}$ , x lies in second quadrant.

Answer:
$\tan x = -\frac {5}{12}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{5}{12}} = -\frac {12}{5}$
$1 + \tan^{2}x = \sec^{2}x\\ 1 + \left ( -\frac{5}{12} \right )^{2} = \sec^{2}x\\ 1 + \frac {25}{144} = \sec^{2}x\\ \\ \frac {169}{144} = \sec^{2}x\\ \sec x = \sqrt {\frac {169}{144}} = \pm \frac {13}{12}$
x lies in the second quadrant. Therefore the value of sec x is negative
$\sec x = - \frac {13}{12}$
$\cos x = \frac{1}{\sec x}= \frac{1}{-\frac{13}{12}} = -\frac {12}{13}$
$\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \left ( -\frac{12}{13} \right )^{2}\\ \sin^{2}x = 1 - \frac{144}{169}\\ \sin^{2}x = \frac {25}{169}\\ \sin x = \sqrt {\frac{25}{169}} = \pm \frac{5}{13}$
x lies in the second quadrant. Therefore, the value of sin x is positive
$\sin x = \frac {5}{13}$
$\csc = \frac {1}{\sin x} = \frac {1}{\frac {5}{13}} = \frac {13}{5}$

Question:6 Find the values of the trigonometric functions $\small \sin 765^\circ$

Answer:
We know that values of sin x repeat after an interval of $2\pi\ or\ 360\ degree$

$\sin765^\circ = \sin (2\times360^\circ + 45^\circ ) = \sin45^\circ\\ sin45^\circ = \frac {1}{\sqrt{2}}$

Question:7 Find the values of the trigonometric functions $\small cosec \ (-1410^\circ)$

Answer:

We know that value of cosec x repeats after an interval of $2\pi \ or \ 360^\circ$
$cosec (-1410^\circ) = cosec (-1410^\circ + 360^\circ\times4)\\ cosec\ 30^\circ = 2$

or

$cosec(-1410^\circ)= - cosec(1410^\circ)\\= -cosec(4 \times 360^\circ - 30^\circ)= - cosec(-30^\circ) = 2$

Question:8 Find the values of the trigonometric functions $\small \tan \frac{19\pi }{3}$

Answer:

We know that tan x repeats after an interval of $\pi$ or 180 degrees
$\tan (\frac{19\pi}{3}) = \tan (6\pi+\frac{\pi}{3})= \tan \frac{\pi}{3} =\tan 60^\circ = \sqrt{3}$

Question:9 Find the values of the trigonometric functions $\sin\left ( -\frac{11\pi}{3} \right )$

Answer:

We know that sin x repeats after an interval of $2\pi or 360^\circ$
$\sin \left ( -\frac{11\pi}{3} \right ) = \sin \left ( -4\pi +\frac{\pi}{3} \right ) = \sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}$

Question:10 Find the values of the trigonometric functions $\small \cot \left ( -\frac{15\pi }{4} \right )$

Answer:

We know that cot x repeats after an interval of $\pi or 180^\circ$
$\cot \left ( -\frac{15\pi}{4} \right ) = \cot \left ( -4\pi +\frac {\pi}{4} \right ) = \cot \left ( \frac{\pi}{4} \right ) = 1$