Question

Find the values of the trigonometric functions in Exercise 6 to 10.
$\sin {765}^0$
$\cos ec\left(-{1410}^0\right)$
$\tan \frac{19\pi }{3}$
$\sin \left(-\frac{11\pi }{3}\right)$
$\cot \left(-\frac{15\pi }{4}\right)$

Answer 1

(a) $\sin {765}^o=\sin \left(2\right(360)+45)=\sin \left(45\right)$

$=\frac{1}{\sqrt{2}}$ $[\because \sin (360+\theta )=\sin \theta ]$.

(b) $\csc (-{1410}^o)=-\csc \left(1410\right)$

$=-\csc \left(\right(360)3+330)$

$=-\csc \left(330\right)$

$=-[-\csc (270+60\left)\right]$

$=\sec 60$

$=2$.

(c) $\tan \frac{19\pi }{3}=\tan [\frac{19}{3}\times 180]=\tan 1140$

$=\tan \left[\right(90\times 12)+60]$

$=\tan 60$ $[\because \tan (90+\theta )=\tan \theta ]$

$=\sqrt{3}$.

(d) $\sin \left(\frac{-11}{3}\pi \right)=-\sin \left(\frac{11}{3}\pi \right)$

$=-\sin (4\pi -\frac{1}{3}\pi )$

$\Rightarrow -\sin \left(\frac{-1\pi }{3}\right)=+\sin \frac{\pi }{3}$

$=\frac{\sqrt{3}}{2}$.

(e) $\cot \left(\frac{-15\pi }{4}\right)=-\cot \left(\frac{15\pi }{4}\right)=-\cot (4\pi -\frac{\pi }{4})$

$=-\cot \left(\frac{-\pi }{4}\right)$

$=\cot \left(\frac{\pi }{4}\right)$

$=1$.