Pre-Calc double and half angle formulas

[sarah25]

The problem is tan(4x)
i have simplified it to (1-tan^2(x))/(1-tan^2(x))^2 - 4tan^2(x)

Where do i go from here?

 

[soroban]

Hello, sarah25!

\(\displaystyle \text{The problem is: }\,\tan 4x \)

Yes, but what is the question?

Are you supposed to write it terms of \(\displaystyle \tan x\,?\)

Then you need this identity: .\(\displaystyle \tan 2A \:=\:\dfrac{2\tan A}{1-\tan^2\!A}\)


We have: .\(\displaystyle \tan4x \:=\:\dfrac{2\tan2x}{1-\tan^22x}\)


Apply the identity again:

. . \(\displaystyle \tan 4x \;=\;\dfrac{2\left(\dfrac{2\tan x}{1-\tan^2\!x}\right)}{1 - \left(\dfrac{2\tan x}{1-\tan^2\!x}\right)^2}\)


. . . . . . . .\(\displaystyle =\;\dfrac{\dfrac{4\tan x}{1-\tan^2x}}{\dfrac{(1-\tan^2x)^2 - (2\tan x)^2}{(1-\tan^2\!x)^2}} \)


. . . . . . . .\(\displaystyle =\;\dfrac{4\tan x}{1-\tan^2\!x} \cdot \dfrac{(1-\tan^2\!x)^2}{(1-\tan^2\!x)^2 - (2\tan x)^2} \)


. . . . . . . .\(\displaystyle =\;\dfrac{4\tan x(1 - \tan^2x)}{1-2\tan^2\!x + \tan^4\!x - 4\tan^2\!x}\)


. . . . . . . .\(\displaystyle =\;\dfrac{4\tan x - 4\tan^3\!x}{1 - 6\tan^2\!x + \tan^4\!x}\)