Answer
$\cos\theta=-\frac{1}{\sqrt {17}}$ and $\sin\theta =-\frac{4}{\sqrt {17}}$
Work Step by Step
$\sec^{2}\theta=1+\tan^{2}\theta=1+16=17$
$\cos^{2}\theta=\frac{1}{\sec^{2}\theta}=\frac{1}{17}$
$\cos \theta=±\frac{1}{\sqrt {17}}$
Since $\theta$ lies in third quadrant, $\cos \theta$ is negative. Therefore
$\cos\theta=-\frac{1}{\sqrt {17}}$
$\sin^{2}\theta= 1-\cos^{2}\theta=1-\frac{1}{17}=\frac{16}{17}$
$\sin \theta=±\frac{4}{\sqrt {17}}$
Since $\theta$ lies in third quadrant, $\sin \theta$ is negative. Therefore
$\sin\theta =-\frac{4}{\sqrt {17}}$
