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  1. #1
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    tan

    Hi,

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

    Thanks.
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  2. #2
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    Quote Originally Posted by lehder View Post
    Hi,

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

    Thanks.
    It is mainly hack work. A starting point is $\displaystyle \tan(5 \alpha) = \tan(4 \alpha + \alpha)$. If you need more help, please show all your work and say where you get stuck.
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  3. #3
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    Hello, lehder!

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


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

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


    Expand the right side:

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

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


    Equate real and imaginary components:

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


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

    . . . . . . . . . . . . . $\displaystyle =\;\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 $\displaystyle \cos^5\!x$

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


    . . $\displaystyle \boxed{\tan5x \;=\;\dfrac{5\tan x - 10\tan^3\!x + \tan^5\!x}{1 - 10\tan^2\!x + 5\tan^4\!x}} $
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