## Precalculus (6th Edition) Blitzer

$\theta=\sin^{-1} (\dfrac{opposite}{hypotenuse})$
The trigonometric ratios are as follows: $\sin \theta =\dfrac{opposite}{hypotenuse}$ Suppose $\theta =\sin^{-1} [\dfrac{opposite}{hypotenuse} ]$ Now, $\sin \theta =\sin [\sin^{-1} (\dfrac{opposite}{hypotenuse})]$ Thus, we have $\theta=\sin^{-1} (\dfrac{opposite}{hypotenuse})$