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Math Help - Inverse trig domain

  1. #1
    Senior Member Stroodle's Avatar
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    Inverse trig domain

    Hi there,

    Just wanted to check if my answer to this question is correct. I need to find the implied domain of sin^{-1}\left [ 2cos(x+\frac{\pi}{2})\right ]

    I'm getting x\in \left [ -\frac{5\pi}{6} , -\frac{\pi}{6}\right ]

    Thanks.
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  2. #2
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    Quote Originally Posted by Stroodle View Post
    Hi there,

    Just wanted to check if my answer to this question is correct. I need to find the implied domain of sin^{-1}\left [ 2cos(x+\frac{\pi}{2})\right ]

    I'm getting x\in \left [ -\frac{5\pi}{6} , -\frac{\pi}{6}\right ]

    Thanks.
    We must have

    -1\leq 2cos\left(x+\frac{\pi}{2}\right)\leq 1

    -\frac{1}{2}\leq cos\left(x+\frac{\pi}{2}\right)\leq\frac{1}{2}

    Taking the inverse throughout gives:



    -\frac{\pi}{3}\leq x+\frac{\pi}{2}\leq \frac{\pi}{3}

    \frac{-5\pi}{6}\leq x \leq -\frac{\pi}{6}

    This gives me the same domain you obtained. So you are correct
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  3. #3
    Senior Member Stroodle's Avatar
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    Awesome. Thanks for the confirmation.
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  4. #4
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    Hello everyone

    Hold on a moment!

    Think about the graph of y = \cos \theta. When \cos\theta = \tfrac12,\; \theta = \tfrac{\pi}{3}, ...
    and when \cos\theta = -\tfrac12, \;\theta = \tfrac{2\pi}{3}, ...

    So:
    -\frac12\le\cos\left(x+\frac{\pi}{2}\right)\le\frac  12

    \Rightarrow \frac{\pi}{3}\le x+\frac{\pi}{2}\le\frac{2\pi}{3}
    I think you'd better have another look at it!

    Grandad
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  5. #5
    Senior Member Stroodle's Avatar
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    Oh yeah. Thanks for that! I always miss things with inverse trig domains
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