S sarah25 New member Joined Nov 27, 2011 Messages 1 Nov 27, 2011 #1 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?
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?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 27, 2011 #2 Hello, sarah25! \(\displaystyle \text{The problem is: }\,\tan 4x \) Click to expand... 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: . . \(\displaystyle \tan 4x \;=\;\dfrac{2\left(\dfrac{2\tan x}{1-\tan^2\!x}\right)}{1 - \left(\dfrac{2\tan x}{1-\tan^2\!x}\right)^2}\) . . . . . . . .\(\displaystyle =\;\dfrac{\dfrac{4\tan x}{1-\tan^2x}}{\dfrac{(1-\tan^2x)^2 - (2\tan x)^2}{(1-\tan^2\!x)^2}} \) . . . . . . . .\(\displaystyle =\;\dfrac{4\tan x}{1-\tan^2\!x} \cdot \dfrac{(1-\tan^2\!x)^2}{(1-\tan^2\!x)^2 - (2\tan x)^2} \) . . . . . . . .\(\displaystyle =\;\dfrac{4\tan x(1 - \tan^2x)}{1-2\tan^2\!x + \tan^4\!x - 4\tan^2\!x}\) . . . . . . . .\(\displaystyle =\;\dfrac{4\tan x - 4\tan^3\!x}{1 - 6\tan^2\!x + \tan^4\!x}\)
Hello, sarah25! \(\displaystyle \text{The problem is: }\,\tan 4x \) Click to expand... 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: . . \(\displaystyle \tan 4x \;=\;\dfrac{2\left(\dfrac{2\tan x}{1-\tan^2\!x}\right)}{1 - \left(\dfrac{2\tan x}{1-\tan^2\!x}\right)^2}\) . . . . . . . .\(\displaystyle =\;\dfrac{\dfrac{4\tan x}{1-\tan^2x}}{\dfrac{(1-\tan^2x)^2 - (2\tan x)^2}{(1-\tan^2\!x)^2}} \) . . . . . . . .\(\displaystyle =\;\dfrac{4\tan x}{1-\tan^2\!x} \cdot \dfrac{(1-\tan^2\!x)^2}{(1-\tan^2\!x)^2 - (2\tan x)^2} \) . . . . . . . .\(\displaystyle =\;\dfrac{4\tan x(1 - \tan^2x)}{1-2\tan^2\!x + \tan^4\!x - 4\tan^2\!x}\) . . . . . . . .\(\displaystyle =\;\dfrac{4\tan x - 4\tan^3\!x}{1 - 6\tan^2\!x + \tan^4\!x}\)
