Find the values of the following trigonometric ratios:
Question:

Find the values of the following trigonometric ratios:

(i) $\sin \frac{5 \pi}{3}$

 

(ii) $\sin 17 \pi$

(iv) $\cos \left(-\frac{25 \pi}{4}\right)$

 

(v) $\tan \frac{7 \pi}{4}$

(vi) $\sin \frac{17 \pi}{6}$

 

(vii) $\cos \frac{19 \pi}{6}$

(viii) $\sin \left(-\frac{11 \pi}{6}\right)$

 

(ix) $\operatorname{cosec}\left(-\frac{20 \pi}{3}\right)$

(x) $\tan \left(-\frac{13 \pi}{4}\right)$

 

(xi) $\cos \frac{19 \pi}{4}$

(xii) $\sin \frac{41 \pi}{4}$

 

(xiii) $\cos \frac{39 \pi}{4}$

(xiv) $\sin \frac{151 \pi}{6}$

Solution:

(i) We have :

$\frac{5 \pi}{3}=\left(\frac{5}{3} \times 180\right)^{\circ}=300^{\circ}=\left(90^{\circ} \times 3+30^{\circ}\right)$

$300^{\circ}$ lies in the fourth quadrant in which the sine function is negative. Also, 3 is an odd integer.

$\therefore \sin \left(\frac{5 \pi}{3}\right)=\sin \left(300^{\circ}\right)=\sin \left(90^{\circ} \times 3+30^{\circ}\right)=-\cos 30^{\circ}=-\frac{\sqrt{3}}{2}$

(ii) We have :

$\sin 17 \pi=\sin 3060^{\circ}$

$3060^{\circ}=90^{\circ} \times 34+0^{\circ}$

Clearly $3060^{\circ}$ is in the negative direction of the $x$-axis, i.e. on the boundary line of the

II and III quadrant $s$.

Also, 34 is an even integer.

$\therefore \sin \left(3060^{\circ}\right)=\sin \left(90^{\circ} \times 34+0^{\circ}\right)=-\sin 0^{\circ}=0$

(iii) We have :

$\frac{11 \pi}{6}=\left(\frac{11}{6} \times 180\right)^{\circ}=330^{\circ}=\left(90^{\circ} \times 3+60^{\circ}\right)$

$330^{\circ}$ lies in the fourth quadrant in which the tangent function is negative.

Also, 3 is an odd integer.

$\therefore \tan \left(\frac{11 \pi}{6}\right)=\tan \left(330^{\circ}\right)=\tan \left(90^{\circ} \times 3+60^{\circ}\right)=-\cot 60^{\circ}=-\frac{1}{\sqrt{3}}$

(iv) We have :

$\cos \left(-\frac{25 \pi}{4}\right)=\cos \left(1125^{\circ}\right)$

$\cos \left(-1125^{\circ}\right)=\cos \left(1125^{\circ}\right)=\cos \left(90^{\circ} \times 12+45^{\circ}\right)$

$1125^{\circ}$ lies in the first quadrant in which the cosine function is positive.

Also, 12 is an even integer.

$\therefore \cos \left(-1125^{\circ}\right)=\cos \left(1125^{\circ}\right)=\cos \left(90^{\circ} \times 12+45^{\circ}\right)=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

(v) We have :

$\tan \frac{7 \pi}{4}=\tan 315^{\circ}$

$315^{\circ}=\left(90^{\circ} \times 3+45^{\circ}\right)$

$315^{\circ}$ lies in the fourth quadrant in which the tangent function is negative.

Also, 3 is an odd integer.

$\therefore \tan \left(315^{\circ}\right)=\tan \left(90^{\circ} \times 3+45^{\circ}\right)=-\cot 45^{\circ}=-1$

(vi) We have :

$\sin \frac{17 \pi}{6}=\sin 510^{\circ}$

$510^{\circ}=\left(90^{\circ} \times 5+60\right)^{\circ}$

$510^{\circ}$ lies in the second quadrant in which the sine function is positive.

Also, 5 is an odd integer.

$\therefore \sin \left(510^{\circ}\right)=\sin \left(90^{\circ} \times 5+60^{\circ}\right)=\cos \left(60^{\circ}\right)=\frac{1}{2}$

(vii) We have :

$\cos \frac{19 \pi}{6}=\cos 570^{\circ}$

$570^{\circ}=\left(90^{\circ} \times 6+30^{\circ}\right)$

$570^{\circ}$ lies in the third quadrant in which the cosine function is negative.

Also, 6 is an odd integer.

$\therefore \cos \left(570^{\circ}\right)=\cos \left(90^{\circ} \times 6+30^{\circ}\right)=-\cos \left(30^{\circ}\right)=-\frac{\sqrt{3}}{2}$

(viii) We have :

$\sin \left(-\frac{11 \pi}{6}\right)=\sin \left(-330^{\circ}\right)$

$\sin \left(-330^{\circ}\right)=-\sin \left(330^{\circ}\right)=-\sin \left(90^{\circ} \times 3+60^{\circ}\right)$

$330^{\circ}$ lies in the fourth quadrant in which the sine function is negative.

Also, 3 is an odd integer.

$\therefore \sin \left(-330^{\circ}\right)=-\sin \left(330^{\circ}\right)=-\sin \left(90^{\circ} \times 3+60^{\circ}\right)=-\left(-\cos 60^{\circ}\right)=-\left(-\frac{1}{2}\right)=\frac{1}{2}$

(ix) We have:

$\operatorname{cosec}\left(-\frac{20 \pi}{3}\right)=\operatorname{cosec}\left(-1200^{\circ}\right)$

$\operatorname{cosec}\left(-1200^{\circ}\right)=-\operatorname{cosec}\left(1200^{\circ}\right)=-\operatorname{cosec}\left(90^{\circ} \times 13+30^{\circ}\right)$

$1200^{\circ}$ lies in the second quadrant in which the cosec function is positive.

Also, 13 is an odd integer.

$\therefore \operatorname{cosec}\left(-1200^{\circ}\right)=-\operatorname{cosec}\left(1200^{\circ}\right)=-\operatorname{cosec}\left(90^{\circ} \times 13+30^{\circ}\right)=-\sec 30^{\circ}=-\frac{2}{\sqrt{3}}$

(x) We have:

$\tan \left(-\frac{13 \pi}{4}\right)=\tan \left(-585^{\circ}\right)$

$\tan \left(-585^{\circ}\right)=-\tan \left(585^{\circ}\right)=-\tan \left(90^{\circ} \times 6+45^{\circ}\right)$

$585^{\circ}$ lies in the third quadrant in which tangent function is positive.

Also, 6 is an even integer.

$\therefore \tan \left(-585^{\circ}\right)=-\tan \left(585^{\circ}\right)=-\tan \left(90^{\circ} \times 6+45^{\circ}\right)=-\tan 45^{\circ}=-1$

(xi) We have:

$\cos \frac{19 \pi}{4}=\cos 855^{\circ}$

$855^{\circ}=90^{\circ} \times 9+45^{\circ}$

$855^{\circ}$ lies in the second quadrant in which the cosine function is negative.

Also, 9 is an odd integer.

$\therefore \cos \left(855^{\circ}\right)=\cos \left(90^{\circ} \times 9+45^{\circ}\right)=-\sin \left(45^{\circ}\right)=-\frac{1}{\sqrt{2}}$

(xii) We have :

$\sin \frac{41 \pi}{4}=\sin 1845^{\circ}$

$1845^{\circ}=90^{\circ} \times 20+45$

$1845^{\circ}$ lies in the first quadrant in which the sine function is positive.

Also, 20 is an even integer.

$\therefore \sin \left(1845^{\circ}\right)=\sin \left(90^{\circ} \times 20+45^{\circ}\right)=\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

(xii) We have:

$\cos \frac{39 \pi}{4}=\cos 1755^{\circ}$

$1755^{\circ}=90^{\circ} \times 19+45^{\circ}$

$1755^{\circ}$ lies in the fourth quadrant in which the cosine function is positive.

Also, 19 is an odd integer.

$\therefore \cos \left(1755^{\circ}\right)=\cos \left(90^{\circ} \times 19+45^{\circ}\right)=\sin \left(45^{\circ}\right)=\frac{1}{\sqrt{2}}$

(xiv)

$\sin \frac{151 \pi}{6}=\sin 4530^{\circ}$

$4530^{\circ}=90^{\circ} \times 50+30^{\circ}$

$4530^{\circ}$ lies in the third quadrant in which the sine function is negative.

Also, 50 is an even integer.

$\therefore \sin \left(4530^{\circ}\right)=\sin \left(90^{\circ} \times 50+30^{\circ}\right)=-\sin \left(30^{\circ}\right)=-\frac{1}{2}$

 

 

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