Answer
$sin\theta=-\frac{\sqrt 5}{5}$,
$cos\theta=-\frac{2\sqrt 5}{5}$,
$tan\theta=\frac{1}{2}$,
$cot\theta=2$,
$csc\theta=-\sqrt 5$,
$sec\theta=-2\sqrt 5$
Work Step by Step
Step 1. With the line equation $4y-2x-1=0$, we can find its slope as $m=\frac{1}{2}$, the acute angle $\alpha$ formed by the line with the x-axis is then given by $tan\alpha=\frac{1}{2}$
Step 2. As the terminal side is parallel to the line, angle $\alpha$ will be the reference angle for $\theta$, considering $\theta$ is in Quadrant III for the signs, we can find the following function values:
$sin\theta=-\frac{1}{\sqrt {1^2+2^2}}=-\frac{\sqrt 5}{5}$,
$cos\theta=-\frac{2}{\sqrt {1^2+2^2}}=-\frac{2\sqrt 5}{5}$,
$tan\theta=tan\alpha=\frac{1}{2}$,
$cot\theta=\frac{1}{tan\theta}=2$,
$csc\theta=\frac{1}{sin\theta}=-\sqrt 5$,
$sec\theta=\frac{1}{cos\theta}=-2\sqrt 5$
