1. ## Analytic trig question

2tanx/(1-(tan(x)^2) + 1/2(cosx^2)-1 = cosx+sinx/cosx-sinx

2. Originally Posted by bilbobaggins
2tanx/(1-(tan(x)^2) + 1/2(cosx^2)-1 = cosx+sinx/cosx-sinx
Will you please rewrite this using parenthesis so we can see what the actual problem is?

-Dan

3. Hello, bilbobaggins!

I think I've guessed what you meant . . .

We need a number of double-angle identities:

. . $\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta$

. . $\displaystyle \cos2\theta \:=\:\cos^2\!\theta - \sin^2\!\theta \:=\:2\cos^2\!\theta -1$

. . $\displaystyle \tan2\theta \:=\:\frac{2\tan\theta}{1 - \tan^2\!\theta}$

$\displaystyle \frac{2\tan x}{1-\tan^2\!x} + \frac{1}{2\cos^2\!x-1} \;=\; \frac{\cos x+\sin x}{\cos x-\sin x}$

The left side is: .$\displaystyle \tan2x + \frac{1}{\cos2x} \;=\;\frac{\sin2x}{\cos2x} + \frac{1}{\cos2x} \;=\;\frac{\sin2x + 1}{\cos2x} \;=\;\frac{2\sin x\cos x + 1}{\cos^2\!x -\sin^2\!x}$

Since $\displaystyle 1 \:=\:\sin^2\!x+\cos^2\!x$, the numerator becomes:

. . $\displaystyle 2\sin x\cos x + \sin^2\!x+\cos^2\!x \;= \;\cos^2\!x + 2\sin x\cos x + \sin^2\!x \;=\;(\cos x + \sin x)^2$

The fraction becomes: .$\displaystyle \frac{(\cos x + \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)} \;=\;\frac{\cos x + \sin x}{\cos x - \sin x}$