1. ## Double Angle Identities

I'm stuck on this question...

I've tried expressing tan^2 (x/2) as a function of tan x and tan x/2, then substituting for cos/sin but that doesn't seem to work (unless I've done the algebra wrong...) can anyone give me some pointers please?

Identities here:

2. ## Re: Double Angle Identities

$\tan^2(\frac x 2)=$

$\sec^2(\frac x 2)-1=$

$\dfrac 1 {\cos^2( \frac x 2)} - 1=$

$\dfrac 1 {\dfrac {1+\cos(x)}2}-1=$

$\dfrac 2 {1+\cos(x)} - 1=$

$\dfrac{1-\cos(x)}{1+\cos(x)}$

3. ## Re: Double Angle Identities

Originally Posted by romsek
$\tan^2(\frac x 2)=$

$\sec^2(\frac x 2)-1=$

$\dfrac 1 {\cos^2( \frac x 2)} - 1=$

$\dfrac 1 {\dfrac {1+\cos(x)}2}-1=$

$\dfrac 2 {1+\cos(x)} - 1=$

$\dfrac{1-\cos(x)}{1+\cos(x)}$
Thanks, as always!