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Thread: Proving Identity of Trigonometry Ratio

  1. #1
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    Proving Identity of Trigonometry Ratio

    Given that :
    $\displaystyle sinB/sinA=cos(A+B)$



    To prove :
    $\displaystyle tan(A+B)=2tanA$


    How to do this prove ? thank you very much
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  2. #2
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    Hello mandyn318

    Welcome to Math Help Forum!
    Quote Originally Posted by mandyn318 View Post
    Given that :
    $\displaystyle sinB/sinA=cos(A+B)$



    To prove :
    $\displaystyle tan(A+B)=2tanA$


    How to do this prove ? thank you very much
    This looks easy, but I think it's surprisingly tricky. I have a proof that assumes $\displaystyle \cos A > 0$ and $\displaystyle \cos B>0$. I'll leave it to you to check what happens if you take the negative values of the square roots.

    To make it easier to read, then, I'm going to let:
    $\displaystyle \sin A = a$ and $\displaystyle \sin B = b$
    So:
    $\displaystyle \cos A = \sqrt{1-a^2}$ and $\displaystyle \cos B = \sqrt{1-b^2}$, taking the positive square roots
    Therefore the original given equation:
    $\displaystyle \frac{\sin B}{\sin A} = \cos(A+B)$
    $\displaystyle =\cos A \cos B - \sin A \sin B$
    can be written as:
    $\displaystyle \frac ba=\sqrt{(1-a^2)(1-b^2)}-ab$

    $\displaystyle \Rightarrow b(1+a^2)=a\sqrt{(1-a^2)(1-b^2)}$
    ...(1)
    Now
    $\displaystyle \tan(A+B)=\sin(A+B)\cdot\frac{1}{\cos(A+B)}$
    $\displaystyle =\sin (A+B)\cdot\frac{\sin A}{\sin B}$, from the original given
    $\displaystyle =\frac{\sin A(\sin A \cos B + \cos A \sin B)}{\sin B}$

    $\displaystyle =\frac{a(a\sqrt{1-b^2}+b\sqrt{1-a^2})}{b}$

    $\displaystyle =\frac{a\Big(a\sqrt{(1-a^2)(1-b^2)}+b(1-a^2)\Big)}{b\sqrt{1-a^2}}$, multiplying top-and-bottom by $\displaystyle \sqrt{1-a^2}$

    $\displaystyle =\frac{a\Big(b(1+a^2)+b(1-a^2)\Big)}{b\sqrt{1-a^2}}$ from
    (1)

    $\displaystyle =\frac{2ab}{b\sqrt{1-a^2}}$

    $\displaystyle =2\tan A$
    Grandad
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  3. #3
    Super Member Quacky's Avatar
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    Wow! And my grandfather never knew what 'surd' meant!
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