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Math Help - Inverse Trigo Functions

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    Inverse Trigo Functions

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

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

    Is the following correct?

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

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

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

    Is there any other way of doing this?

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

    Many thanks for those who can help
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  2. #2
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    Quote Originally Posted by acc100jt View Post
    For the inverse cosine function, we have
    \cos x=y implies x=\cos^{-1}y and 0 \leq x \leq \pi

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

    Is the following correct?

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

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

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

    Is there any other way of doing this?

    What about finding the inverse of
    h(x)=\cos x where -\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 y = \arccos{x} and think about where it would be in another interval on the unit circle.

    for -\pi \le x \le 0

    y = -\arccos{x}
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