Right Angle Relationships

**~berserk** · February 12th 2011, 04:58 PM

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

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

**dwsmith** · February 12th 2011, 04:59 PM

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, $u>0$

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

and that's secant raised to the negative 1

type \sec^{-1}

**Soroban** · February 12th 2011, 05:38 PM

Hello, ~berserk!

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

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

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

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

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

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

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

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

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

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