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