Results 1 to 3 of 3

Math Help - tan 70 = tan 20 + 2 tan 40 +4 tan 10

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    152

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

    prove




    tan 70 = tan 20 + 2 tan 40 +4 tan 10
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,245
    Thanks
    1
    We have the identity tan x -cotan x=-2cotan 2x.
    Then we have
    tan20-cotan20=-2cotan40
    2tan40-2cotan40=-4cotan80
    4tan10=4cotan80
    Adding the three equalities it results
    tan20+2tan40+4tan10=cotan20=tan70
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,663
    Thanks
    603
    Hello, perash!

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


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


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

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

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

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

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

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

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

    . . = \;\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:

    \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}

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

    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum