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Math Help - So confused..... domain and range of composite trig functions

  1. #1
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    So confused..... domain and range of composite trig functions

    How can I identify the domain and range of problems such as sin(cos^-1(2x)) without using a calculator???

    Other problems would be sin(sin^-1(x-1/2)) or cos^-1(2sin(x)).

    Please help asap because I have a math test tomorrow and this is the only concept I do not know.

    Thanks.
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  2. #2
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    Hello HubridNoxx

    Welcome to Math Help Forum!
    Quote Originally Posted by HubridNoxx View Post
    How can I identify the domain and range of problems such as sin(cos^-1(2x)) without using a calculator???

    Other problems would be sin(sin^-1(x-1/2)) or cos^-1(2sin(x)).

    Please help asap because I have a math test tomorrow and this is the only concept I do not know.

    Thanks.
    I assume that by \sin^{-1}(x), you mean \arcsin(x), etc, in which case my advice is to get rid of the inverse functions as soon as you can. They're nasty things!

    First, though, you'll find this page useful.

    Then, in the first one, let:
    y = \arccos(2x)
    So the domain is -1\le 2x \le 1, or -\tfrac12 \le x \le \tfrac12. And, assuming we're just taking the principle values, the range of values of y is 0\le y \le\pi.

    Now if we re-write the above equation, we get:
     2x = \cos y

    \Rightarrow \sin(\arccos(2x)) =  \sin y=\sqrt{1-\cos^2y} = \sqrt{1-4x^2}
    noting that we only need the positive square root, because for 0\le y \le \pi,\; \sin y \ge 0.

    Therefore range of values of \sin(\arccos(2x))\;( =\sqrt{1-4x^2}) for -\tfrac12le \le \tfrac12 is [0, 1].

    See if you can work out the others in the same way.

    Grandad
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  3. #3
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    Thanks, this should help me on my test.
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