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Math Help - Inverse Functions

  1. #1
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    Inverse Functions

    I'm not quite sure how to tackle this question. It asks;

    a) Find the value of x such that sin^{-1}x = cos^{-1}x.

    b) On the same diagram, sketch the graphs of y = sin^{-1}x, y = cos^{-1}x and y = sin^{-1}x + cos^{-1}x.

    With this second part, I'm able to sketch the graphs of y = sin^{-1}x and y = cos^{-1}x, but I'm not sure how to sketch the third...
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  2. #2
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    Quote Originally Posted by Flay View Post
    I'm not quite sure how to tackle this question. It asks;

    a) Find the value of x such that sin^{-1}x = cos^{-1}x.

    b) On the same diagram, sketch the graphs of y = sin^{-1}x, y = cos^{-1}x and y = sin^{-1}x + cos^{-1}x.

    With this second part, I'm able to sketch the graphs of y = sin^{-1}x and y = cos^{-1}x, but I'm not sure how to sketch the third...
    a) x = \frac{\pi}{4}:

    Let \cos^{-1} x = \alpha \Rightarrow x = \cos \alpha.

    Then \sin^{-1} x = \alpha \Rightarrow x = \sin \alpha \Rightarrow \cos \alpha = \sqrt{1 - x^2}.

    Therefore x = \sqrt{1 - x^2} \Rightarrow x^2 = 1 - x^2 \Rightarrow x = \pm \frac{1}{\sqrt{2}}.

    Discard extraneous solutions: x = \frac{\pi}{4}.

    This solution should be obvious to you in hindsight (if not foresight).


    b) Use addition of ordinates to sketch y = \sin^{-1}x + \cos^{-1}x. You will find that y = \frac{\pi}{2} and this result ought not surprise you in hindsight.
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