Results 1 to 3 of 3

Math Help - tan

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    62

    tan

    Hi,

    I tried to express tan(5 \alpha) in terms of tan( \alpha) but ican't do it, can you help me please??

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by lehder View Post
    Hi,

    I tried to express tan(5 \alpha) in terms of tan( \alpha) but ican't do it, can you help me please??

    Thanks.
    It is mainly hack work. A starting point is \tan(5 \alpha) = \tan(4 \alpha + \alpha). If you need more help, please show all your work and say where you get stuck.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,866
    Thanks
    745
    Hello, lehder!

    If we are allowed to use DeMoivre's Theorem,
    . . there is a clever solution.


    \text{Express }\tan 5x\text{ in terms of }\tan x

    We are told that: . \cos5x + i\sin5x \;=\;(\cos x + i\sin x)^5


    Expand the right side:

    . . \cos5x + i\sin 5x \;=\; \cos^5\!x + 5i\cos^4\!x\sin x - 10\cos^3\!x\sin^2\!x
    . . . . . . . . . . . . . . . . . . - 10i\cos^2\!x\sin^3\!x + 5\cos x \sin^4\!x + i\sin^5\!x

    . . . . . . . . . . . . . . =\;(\cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x)
    . . . . . . . . . . . . . . . . . . + i(5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x)


    Equate real and imaginary components:

    . . \begin{array}{ccc}\cos5x &=& \cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x \\ \sin5x &=& 5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x \end{array}


    We have: . \tan5x \;=\;\dfrac{\sin5x}{\cos5x}

    . . . . . . . . . . . . . =\;\dfrac{5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x}{\cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x}


    Divide numerator and denominator by \cos^5\!x

    . . \displaystyle \tan5x \;=\;\frac{\;\dfrac{5\cos^4\!x\sin x}{\cos^5\!x} - \dfrac{10\cos^2\!x\sin^3\!x}{\cos^5\!x} + \dfrac{\sin^5\!x}{\cos^5\!x}\;}<br />
{\dfrac{\cos^5\!x}{\cos^5\!x} - \dfrac{10\cos^3\!x\sin^2\!x}{\cos^5\!x} + \dfrac{5\cos x\sin^4\!x}{\cos^5\!x}}


    . . \boxed{\tan5x \;=\;\dfrac{5\tan x - 10\tan^3\!x + \tan^5\!x}{1 - 10\tan^2\!x + 5\tan^4\!x}}
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum