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Thread: tan 70 = tan 20 + 2 tan 40 +4 tan 10

  1. #1
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    tan 70 = tan 20 + 2 tan 40 +4 tan 10

    prove




    tan 70 = tan 20 + 2 tan 40 +4 tan 10
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have the identity $\displaystyle tan x -cotan x=-2cotan 2x$.
    Then we have
    $\displaystyle tan20-cotan20=-2cotan40$
    $\displaystyle 2tan40-2cotan40=-4cotan80$
    $\displaystyle 4tan10=4cotan80$
    Adding the three equalities it results
    $\displaystyle tan20+2tan40+4tan10=cotan20=tan70$
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  3. #3
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    Hello, perash!

    red_dog had a brilliant solution!
    . . Did you follow it?


    Prove: .$\displaystyle \tan70 \;= \;\tan20 + 2\tan40 + 4\tan10$
    We have the identity: .$\displaystyle \tan\theta - \cot\theta \:=\:-2\cot2\theta$ .**


    So we have: .$\displaystyle \begin{array}{ccc}\tan20 - \cot20 & = & -2\cot40 \\ 2(\tan40 - \cot40) & = & -4\cot80 \\ 4\tan10 & = & 4\cot 80 \end{array}$

    Add the three equations:
    . . $\displaystyle \tan20 -\cot20 + 2\tan40 - 2\cot40 + 4\tan10 \;=\;-2\cot40 - 4\cot80 + 4\cot80$

    which simplifies to:
    . . $\displaystyle \tan20 - \cot20 + 2\tan40 + 4\tan10 \:=\:0$

    . . . . . . . $\displaystyle \tan20 + 2\tan40 + 4\tan10 \;=\;\cot20$

    . . . . . . . $\displaystyle \tan20 + 2\tan40 + 4\tan10 \;=\;\tan70$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    We have: .$\displaystyle \tan\theta - \cot\theta \;=\;\tan\theta - \frac{1}{\tan\theta} \;=\;\frac{\tan^2\theta - 1}{\tan\theta} \;=\;\frac{-(1 - \tan^2\theta)}{\tan\theta}$

    . . $\displaystyle = \;\frac{-2(1-\tan^2\theta)}{2\tan\theta} \;=\;\frac{-2}{\frac{2\tan\theta}{1-\tan^2\theta}} \;=\;\frac{-2}{\tan2\theta} \;=\;-2\cot2\theta$

    . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Alternate proof:

    $\displaystyle \tan\theta - \cot\theta \;=\;\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\sin\theta} \;= \;\frac{\sin^2\theta - \cos^2\theta}{\sin\theta\cos\theta} \;=\;\frac{-(\cos^2\theta - \sin^2\theta)}{\sin\theta\cos\theta}$

    . . $\displaystyle = \;\frac{-2(\cos^2\theta - \sin^2\theta)}{2\sin\theta\cos\theta} \;=\;\frac{-2\cos2\theta}{\sin2\theta} \;=\;-2\cot2\theta$

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