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  1. #1
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    Help

    Help me, please.
    arcsin (sin(9Pi/5) = ?
    arctg (tg (7Pi/5) = ?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Netochka View Post
    Help me, please.
    arcsin (sin(9Pi/5) = ?
    arctg (tg (7Pi/5) = ?
    Hi

    If you can find k integer so that $\displaystyle -\frac{\pi}{2} \leq \theta + 2k \pi \leq \frac{\pi}{2}$ then $\displaystyle Arcsin(\sin(\theta)) = \theta + 2k \pi$

    If not then exist n integer so that $\displaystyle -\frac{\pi}{2} \leq -\theta + (2n+1) \pi \leq \frac{\pi}{2}$ and then $\displaystyle Arcsin(\sin(\theta)) = -\theta + (2n+1) \pi$

    For instance
    For $\displaystyle \theta = \frac{9\pi}{5}$
    $\displaystyle \theta -2 \pi = -\frac{\pi}{5}$ and $\displaystyle -\frac{\pi}{2} \leq -\frac{\pi}{5} \leq \frac{\pi}{2}$ then $\displaystyle Arcsin(\sin(\frac{9\pi}{5})) = -\frac{\pi}{5}$

    For $\displaystyle \theta = \frac{7\pi}{5}$
    $\displaystyle \theta -2 \pi = -\frac{3\pi}{5}$ and $\displaystyle -\frac{3\pi}{5} < -\frac{\pi}{2}$
    $\displaystyle -\theta + \pi = -\frac{2\pi}{5}$
    $\displaystyle -\frac{\pi}{2} \leq -\frac{2\pi}{5} \leq \frac{\pi}{2}$ then
    $\displaystyle Arcsin(\sin(\frac{7\pi}{5})) = -\frac{2\pi}{5}$


    For Arctan it is more simple
    $\displaystyle Arctan(\tan(\theta)) = \theta + k \pi$ so that $\displaystyle -\frac{\pi}{2} < \theta + k \pi < \frac{\pi}{2}$
    $\displaystyle Arctan(\tan(\frac{7\pi}{5})) = \frac{2\pi}{5}$
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  3. #3
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    Thank you very much
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