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Math Help - How do I simplify this

  1. #1
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    How do I simplify this

    arcsin(sin(x+\pi)) for x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]

    I can see that arcsin(sin(x+\pi)) = x+\pi

    How does x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right] apply to my answer?

    (I have an answer x-2\pi and didn't understand the logic)

    Thanks.
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  2. #2
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    Re: How do I simplify this

    Quote Originally Posted by terrorsquid View Post
    arcsin(sin(x+\pi)) for x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]
    I can see that arcsin(sin(x+\pi)) = x+\pi
    How does x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right] apply to my answer?
    (I have an answer x-2\pi and didn't understand the logic
    First, take note that x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]
    is the same domain as x\in \left[\frac{-\pi}{2},\frac{\pi}{2}\right].
    That happens to be the range of \arcsin(x).
    Observe that \frac{3\pi}{2}-2\pi=\frac{-\pi}{2}
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  3. #3
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    Re: How do I simplify this

    I can see it's domain, but I am still confused as to how I can derive an answer from that information. There is a disconnect between x+\pi and the domain for me. How do I combine them to form an answer?
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  4. #4
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    Re: How do I simplify this

    Quote Originally Posted by terrorsquid View Post
    I can see it's domain, but I am still confused as to how I can derive an answer from that information. There is a disconnect between x+\pi and the domain for me. How do I combine them to form an answer?
    To transform x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right] to x\in \left[\frac{-\pi}{2},\frac{\pi}{2}\right] we use x-2\pi.

    I don't know any other way to explain it.
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  5. #5
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    Re: How do I simplify this

    Hmm ok.What was the point of the original equation and finding arcsin(sin(x+ \pi)) =  x+ \pi then if x-2\pi is found just by looking at the domain?
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  6. #6
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    Re: How do I simplify this

    Quote Originally Posted by terrorsquid View Post
    Hmm ok.What was the point of the original equation and finding arcsin(sin(x+ \pi)) =  x+ \pi then if x-2\pi is found just by looking at the domain?
    Frankly, I do not understand the wording of the original question.
    I was trying to give you a reason for the proposed answer.

    I can see that \arcsin(\sin(x+\pi))=\arcsin(-\sin(x)) that is a simplification.
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