1. ## 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.

2. Originally Posted by lehder
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.

3. 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 $\cos^5\!x$

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

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