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Math Help - Inverse trigonometric function

  1. #1
    Senior Member pankaj's Avatar
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    Inverse trigonometric function

    If A=\tan^{-1}(2\sqrt2-1) and B=3\sin^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left(\frac{3}{5}\right)

    Without using calculator prove that A>B
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by pankaj View Post
    If A=\tan^{-1}(2\sqrt2-1) and B=3\sin^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left(\frac{3}{5}\right)

    Without using calculator prove that A>B
    \tan(A) = 2\sqrt{2} - 1

    M = 3 \sin^{-1}\frac{1}{3}
    N = sin^{-1}\frac{3}{5}
    \sin \frac{M}{3} = \frac{1}{3}
    \sin N = \frac{3}{5}
    B=M + N
    \tan B = \tan (M + N)=\frac{tanM + tanN}{1 - tanM tanN}


    \tan N = \frac{\sin N}{\sqrt{1- \sin^2 N}}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}

    \tan M = \frac{\sin M}{\sqrt{1- \sin^2 M}}=?
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  3. #3
    MHF Contributor
    Grandad's Avatar
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    Inverse trig functions

    Hello pankaj
    Quote Originally Posted by pankaj View Post
    If A=\tan^{-1}(2\sqrt2-1) and B=3\sin^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left(\frac{3}{5}\right)

    Without using calculator prove that A>B
    First, let's note that A and B can have infinitely many values. Some will give A > B and some will give A < B. So I think we must assume that all inverse trig functions are assumed to be principle values.

    Secondly, on this assumption, I'm afraid that it just isn't true! Try it with your calculator, and you'll see.

    Could you please check that we have the correct question here?

    Grandad
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