## Precalculus (10th Edition)

$\frac{12}{13}$.
$g(f(\frac{12}{13}))=\sin(\cos^{-1}{\frac{5}{13}})$. Let $\theta=\cos^{-1}{\frac{5}{13}}$, hence we know that because of the definition of the cosine function: $\cos{\theta}=\frac{5}{13}=\frac{\text{adjacent}}{\text{hypotenuse}}$ The Pythagorean Theorem says that for a right triangle (if $z$ is the hypotenuse and $x,y$ are the other sides): $x^2+y^2=z^2$. Hence here: $\text{opposite}=\sqrt{13^2-5^2}=\sqrt{144}=12.$ $\sin{\theta}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{12}{13}$.