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Thread: Inverse Trigo Functions

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    Inverse Trigo Functions

    For the inverse cosine function, we have
    $\displaystyle \cos x=y$ implies $\displaystyle x=\cos^{-1}y$ and $\displaystyle 0 \leq x \leq \pi$

    Let's say if I want to find the inverse of $\displaystyle g(x)=\cos x$ where $\displaystyle -2\pi \leq x \leq -\pi$

    Is the following correct?

    $\displaystyle g(x)= \cos x = \cos(x+2\pi)$

    Let $\displaystyle y=g(x)$
    $\displaystyle y=cos(x+2\pi)$
    $\displaystyle x+2\pi=\cos^{-1}y$ since $\displaystyle 0 \leq x+2\pi \leq \pi$
    $\displaystyle x=\cos^{-1}y-2\pi$

    Hence, $\displaystyle g^{-1}(x)=\cos^{-1}x-2\pi$ where $\displaystyle -1 \leq x \leq 1$

    Is there any other way of doing this?

    What about finding the inverse of
    $\displaystyle h(x)=\cos x$ where $\displaystyle -\pi \leq x \leq 0$?

    Many thanks for those who can help
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    Quote Originally Posted by acc100jt View Post
    For the inverse cosine function, we have
    $\displaystyle \cos x=y$ implies $\displaystyle x=\cos^{-1}y$ and $\displaystyle 0 \leq x \leq \pi$

    Let's say if I want to find the inverse of $\displaystyle g(x)=\cos x$ where $\displaystyle -2\pi \leq x \leq -\pi$

    Is the following correct?

    $\displaystyle g(x)= \cos x = \cos(x+2\pi)$

    Let $\displaystyle y=g(x)$
    $\displaystyle y=cos(x+2\pi)$
    $\displaystyle x+2\pi=\cos^{-1}y$ since $\displaystyle 0 \leq x+2\pi \leq \pi$
    $\displaystyle x=\cos^{-1}y-2\pi$

    Hence, $\displaystyle g^{-1}(x)=\cos^{-1}x-2\pi$ where $\displaystyle -1 \leq x \leq 1$

    Is there any other way of doing this?

    What about finding the inverse of
    $\displaystyle h(x)=\cos x$ where $\displaystyle -\pi \leq x \leq 0$?

    Many thanks for those who can help
    what you're doing is fine. I just find the resulting angle from the defined range of $\displaystyle y = \arccos{x}$ and think about where it would be in another interval on the unit circle.

    for $\displaystyle -\pi \le x \le 0$

    $\displaystyle y = -\arccos{x} $
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