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Thread: 2 hard trig problems

  1. #1
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    2 hard trig problems

    I have tried to solve these through every way i know but i get stuck halfway.

    $\displaystyle tan^{2}10^{\circ}+tan^{2}50^{\circ}+tan^{2}70^{\ci rc}=9$

    $\displaystyle tan^{2}36^{\circ}\times tan^{2}72^{\circ}=5$
    Last edited by ldx2; Dec 15th 2009 at 11:57 AM.
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  2. #2
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    Hello ldx2
    Quote Originally Posted by ldx2 View Post
    I have tried to solve these through every way i know but i get stuck halfway.

    $\displaystyle tan^{2}10^{\circ}+tan^{2}50^{\circ}+tan^{2}70^{\ci rc}=9$

    $\displaystyle tan^{2}36^{\circ}\times tan^{2}72^{\circ}=5$
    I haven't got a solution for part 1 yet, but here's an answer to part 2, based on the fact that $\displaystyle \sin18^o = \frac{\sqrt5-1}{4}$ (which we proved in an earlier post).

    First note that
    $\displaystyle \sin18^o = \frac{\sqrt5-1}{4}$
    $\displaystyle \Rightarrow \sin^218^o = \frac{6-2\sqrt5}{16}=\frac{3-\sqrt5}{8}$
    Then, noting that $\displaystyle \tan72^o = \cot18^o = \frac{1}{\tan18^o}$:
    $\displaystyle \tan36^o\tan72^o = \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}$
    $\displaystyle =\frac{2}{1-\dfrac{\sin^218^o}{\cos^218^o}}$

    $\displaystyle =\frac{2(1-\sin^218^o)}{1-2\sin^218^o}$

    $\displaystyle = \frac{2\left(1-\dfrac{3-\sqrt5}{8}\right)}{1-\dfrac{3-\sqrt5}{4}}$

    $\displaystyle =$ ... etc

    $\displaystyle = \sqrt5$ (Can you supply the missing lines?)
    Hence $\displaystyle \tan^236^o\tan^272^o=5$

    Grandad
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  3. #3
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    I understood all but this part :
    $\displaystyle \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}$

    Can you explain what identity you used ??
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  4. #4
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    Hello ldx2
    Quote Originally Posted by ldx2 View Post
    I understood all but this part :
    $\displaystyle \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}$

    Can you explain what identity you used ??
    $\displaystyle \tan2A = \frac{2\tan A}{1-\tan^2A}$. So $\displaystyle \tan36^o =$ ?

    Grandad
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  5. #5
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    Yeah i get it now, thank you very much for your help. Any luck with the first problem ?
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