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Math Help - Solving inverse trig equations

  1. #1
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    Solving inverse trig equations

    Hey, I have two problems here that I really don't know how to start. Hopefully someone can help me out!
    Question
    Prove that cos(sin^{-1}x) = \sqrt{1-x^2}

    Solution
    I don't know where to start with this problem. I know that sin^{-1}x = y -> siny = x . I don't know how to apply this to this equation. A similar problem in the examples of my text is: "simplify:  cos(tan^{-1}x) .
    To simplify this the text first let y =  tan^{-1}x . Then it jumps to  sec^2y = 1 + tan^2y = 1 + x^2 . I don't understand this step in their example and this may prove to be the problem I am having.

    I hope I am making sense here! I have another problem, but I figure once I get help with this one I can force my way though the second. Please help!
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  2. #2
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    Quote Originally Posted by Kakariki View Post
    Hey, I have two problems here that I really don't know how to start. Hopefully someone can help me out!
    Question
    Prove that cos(sin^{-1}x) = \sqrt{1-x^2}

    Solution
    I don't know where to start with this problem. I know that sin^{-1}x = y -> siny = x . I don't know how to apply this to this equation. A similar problem in the examples of my text is: "simplify:  cos(tan^{-1}x) .
    To simplify this the text first let y =  tan^{-1}x . Then it jumps to  sec^2y = 1 + tan^2y = 1 + x^2 . I don't understand this step in their example and this may prove to be the problem I am having.

    I hope I am making sense here! I have another problem, but I figure once I get help with this one I can force my way though the second. Please help!
    There have been several questions quite similar to this one posted in recent days. Have you reviewed the threads in this subforum?
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  3. #3
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    Quote Originally Posted by Kakariki View Post
    Hey, I have two problems here that I really don't know how to start. Hopefully someone can help me out!
    Question
    Prove that cos(sin^{-1}x) = \sqrt{1-x^2}

    Solution
    I don't know where to start with this problem. I know that sin^{-1}x = y -> siny = x . I don't know how to apply this to this equation. A similar problem in the examples of my text is: "simplify:  cos(tan^{-1}x) .
    To simplify this the text first let y =  tan^{-1}x . Then it jumps to  sec^2y = 1 + tan^2y = 1 + x^2 . I don't understand this step in their example and this may prove to be the problem I am having.

    I hope I am making sense here! I have another problem, but I figure once I get help with this one I can force my way though the second. Please help!
    Here's another way to view it.........

    You've gotten to

    Sin^{-1}x=y\Rightarrow\ x=Siny

    therefore y is an angle (make it acute in a right-angled triangle) and

    Siny=\displaystyle\frac{opposite}{hypotenuse}

    hence this ratio is

    x=\displaystyle\frac{kx}{k}

    so the opposite is "kx" and the hypotenuse is "k"

    Using Pythagoras' theorem.... (adjacent)^2+(kx)^2=k^2\Rightarrow\ adj=\sqrt{k^2-(kx)^2}=k\sqrt{1-x^2}

    You can work this through to calculate

    \displaystyle\ Cosy=\frac{adjacent}{hypotenuse}
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  4. #4
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    For the similar example, they've let y = \tan^{-1}x which means x = \tan{y}.


    \sec^2{y} = 1 + \tan^2{y} is a well-known variation of the Pythagorean Identity. So

    \sec^2{y} = 1 + x^2 because x = \tan{y}

    \sec{y} = \sqrt{1 + x^2}

    \frac{1}{\cos{y}} = \sqrt{1 + x^2}

    \cos{y} = \frac{1}{\sqrt{1 + x^2}}

    \cos{(\tan^{-1}{x})} = \frac{1}{\sqrt{1 + x^2}} bceause y = \tan^{-1}{x}.
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