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Math Help - inverse trig

  1. #1
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    inverse trig

    show that  arcsinx + arccosx = \frac{\pi}{2}
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  2. #2
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    Quote Originally Posted by Tweety View Post
    show that  arcsinx + arccosx = \frac{\pi}{2}
    I think I can use calculus to solve your problem, since you wrote some threads in calculus sub-forum, which means that you know the derivatives.

    For your problem:
    In other words, you want to prove that:
    sin^{-1}(x) = \frac{\pi}{2} - cos^{-1}(x)
    Let f(x)=sin^{-1}(x), and g(x)=\frac{\pi}{2} - cos^{-1}(x)
    If you proved that f^{'}(x)=g^{'}(x)
    then, f(x)=g(x)+C.
    By substituting x = anynumber \in [-1,1] in the last equation you will get the value of C.

    Try it.
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  3. #3
    MHF Contributor
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    Hi Tweety,

    Draw a right-angled triangle.
    Label one acute angle A and the other B.
    Label the side opposite A as x and the hypotenuse 1.

    Inverse Sine and Inverse Cosine give you angle from the ratios of

    \frac{opposite}{hypotenuse} and \frac{adjacent}{hypotenuse}

    For this triangle

    SinA=x and CosB=x

    Sin^{-1}x=A and Cos^{-1}x=B

    Since the sum of the acute angles in a right-angled triangle is 90^o=\frac{\pi}{2}

    your statement is proven.
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