## Calculus: Early Transcendentals (2nd Edition)

$\sin(\sec^{-1}\big(\frac{\sqrt{x^2+16}}{4}\big)) = \frac{x}{\sqrt{x^2+16}}$
We are trying to find $\sin(\sec^{-1}\big(\frac{\sqrt{x^2+16}}{4}\big))$. The $\sqrt{x^2+16}$ looks awfully like the hypotenuse of a right triangle with legs of lengths $x$ and $4$, which motivates us to draw a right triangle. Let $\theta$ be the angle between the side with length $4$ and the hypotenuse. This means that $\sec(\theta) = \frac{\sqrt{x^2+16}}{4}$ and solving for $\theta$, we get $\theta = \sec^{-1}\big(\frac{\sqrt{x^2+16}}{4}\big)$. This is in fact part of our original expression, so we can substitute this in: $\sin(\sec^{-1}\big(\frac{\sqrt{x^2+16}}{4}\big)) = \sin(\theta) = \frac{x}{\sqrt{x^2+16}}$.