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Math Help - Exact Values

  1. #1
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    Exact Values

    Hey i need help finding exact values for 2 problems

    1) sin[2sin^-1(1/2)]

    2) sec[tan^-1(1/2)]
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  2. #2
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    Quote Originally Posted by DjamaalRobinson View Post
    Hey i need help finding exact values for 2 problems

    1) sin[2sin^-1(1/2)]
    I do this one, \sin^{-1} \tfrac{1}{2} = \tfrac{\pi}{6}.
    Thus, \sin [2\cdot \tfrac{\pi}{6}] = \sin \tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2}.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by DjamaalRobinson View Post
    Hey i need help finding exact values for 2 problems

    1) sin[2sin^-1(1/2)]

    2) sec[tan^-1(1/2)]
    1) \sin\left[2\sin^{-1}\left(\tfrac{1}{2}\right)\right]

    Depending on the restriction on \vartheta, we see that \vartheta=\sin^{-1}\left(\tfrac{1}{2}\right)=\frac{\pi}{6}~or~\frac  {5\pi}{6}

    Thus, we can either have \sin\left(2\cdot\frac{\pi}{6}\right)=\sin\left(\fr  ac{\pi}{3}\right)=\frac{\sqrt{3}}{2}, or  \sin\left(2\cdot\frac{5\pi}{6}\right)=\sin\left(\f  rac{5\pi}{3}\right)=-\frac{\sqrt{3}}{2}

    2) \sec\left[\tan^{-1}\left(\tfrac{1}{2}\right)\right]

    Construct a triangle to represent the angle \vartheta=\tan^{-1}\left(\tfrac{1}{2}\right)

    This implies that we have an opposite of length 1, and an adjacent of length 2. Thus, the length of the hypoteuse can be found by applyin the pythagorean forumla:

    h^2=1+4\implies h=\sqrt{5}

    Thus, \sec\vartheta=\color{red}\boxed{\frac{\sqrt{5}}{2}  }

    I hope this makes sense!

    --Chris
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Depending on the restriction on \vartheta
    The arcsine function is defined with range [-\pi/2,\pi/2]
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