## Trigonometry 7th Edition

$\cos\theta$ = - $\frac{3}{5}$
We know from first Pythagorean identity that- $\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$ As $\theta$ terminates in Q III, Therefore $\cos\theta$ will be negative- $\cos\theta$ = - $\sqrt (1-\sin^{2}\theta)$ substitute the given value of $\sin\theta$- $\cos\theta$ = - $\sqrt (1-(\frac{-4}{5})^{2})$ $\cos\theta$ = - $\sqrt (1-\frac{16}{25})$ $\cos\theta$ = - $\sqrt (\frac{25 - 16}{25})$ = $\sqrt (\frac{9}{25})$ $\cos\theta$ = - $\frac{3}{5}$