Answer
$\dfrac{5}{13}$
Work Step by Step
We know that
$g(x)=\cos(x), f(x)=\sin(x)$
Thus we can write:
$g(f^{-1} (\dfrac{12}{13}))=\cos (\sin^{-1}(\dfrac{12}{13}))$
Suppose that $\theta=\sin^{-1} (\dfrac{12}{13})$
We know from SOH-CAH-TOA:
$\sin{\theta}=\dfrac{opposite}{hypotenuse}=\dfrac{12}{13}$
To find the missing adjacent side, we use the Pythagorean Theorem:
$adjacent^2+12^2=13^2$
$adjacent=\sqrt{(13)^2-(12)^2}=\sqrt {25}=5$
Therefore,
$\cos{\theta}=\dfrac{adjacent}{hypotenuse}=\dfrac{5}{13}$
