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Math Help - Right Angle Relationships

  1. #1
    Member ~berserk's Avatar
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    Right Angle Relationships

    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})
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  2. #2
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    Quote Originally Posted by ~berserk View Post
    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}
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  3. #3
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    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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