Answer
The proof is below.
Work Step by Step
Because cosine and sine are cofunctions, cotangent and tangent are cofunctions, and cosecant and secant are cofunctions, the coefficient for the x-term is negative. For example, $cos(x) = sin(\frac{\pi}{2}-x)$ where the coefficient for the x-term is $-1$. By applying the chain rule, $g'$ is the negative of the cofunction of $f'$.
For example:
Let $g(x) = cos(x)$ and $f(x) = sin(x)$
We apply the chain rule:
$g'(x) = \frac{d}{dx} cos(x) = \frac{d}{dx} sin(\frac{\pi}{2}-x) = cos(\frac{\pi}{2}-x)*(\frac{d}{dx} [\frac{\pi}{2} -x]) = -cos(\frac{\pi}{2}-x) = -sin(x)$
$f'(x) = cos(x)$
The cofunction of $f'(x)$ is therefore $sin(x)$.
Thus, $g(x) = -sin(x) = - \text{Cofunction of $f(x)$}$
