1. ## Right Angle Relationships

Hello, I am stuck on how to solve this problem:
Write each expression as an algebraic (nontrigonometric) function expression in u, $\displaystyle u>0$

$\displaystyle \sin(2\sec^{-1} \frac{u}{12})$

2. Originally Posted by ~berserk
Hello, I am stuck on how to solve this problem:
Write each expression as an algebraic (nontrigonometric) function expression in u, $\displaystyle u>0$

$\displaystyle \sin(2\sec^-1 \frac{u}{12})$

and that's secant raised to the negative 1
type \sec^{-1}

3. Hello, ~berserk!

Write the expression as an algebraic (nontrigonometric) expression in $\displaystyle u,\;u>0$

. . $\displaystyle \sin\left(2\sec^{\text{-}1}\!\dfrac{u}{12}\right)$

$\displaystyle \text{Let: }\,\theta \:=\:\sec^{\text{-}1}\!\dfrac{u}{12}$

$\displaystyle \text{Then: }\,\sec\theta \:=\:\dfrac{u}{12} \:=\:\dfrac{hyp}{adj}$

$\displaystyle \,\theta$ is in a right triangle with: .$\displaystyle adj = 12.\;hyp = u$

Pythagorus says: .$\displaystyle opp \,=\,\sqrt{u^2-144}$

. . $\displaystyle \text{Hence: }\;\begin{Bmatrix}\sin\theta &=& \dfrac{opp}{hyp} &=& \dfrac{\sqrt{u^2-144}}{u} \\ \\[-3mm] \cos\theta &=& \dfrac{adj}{hyp} &=& \dfrac{12}{u} \end{Bmatrix}$

$\displaystyle \text{Therefore: }\;\sin\left(2\sec^{\text{-}1}\!\dfrac{u}{12}\right) \;=\; \sin(2\theta) \;=\;2\sin\theta\cos\theta$

. . . . . . . . . $\displaystyle =\;2\left(\dfrac{\sqrt{u^2-144}}{u}\right)\left(\dfrac{12}{u}\right) \;=\;\dfrac{24\sqrt{u^2-144}}{u^2}$