# Thread: Help on trig identity.

1. ## Help on trig identity.

Establish each identity.
$\displaystyle cos(2x)/1 + sin(2x) = cotx - 1/ cotx + 1$

Now in the solutions manual, this is what they have in the middle:
$\displaystyle (cosx- sinx)(cosx+sinx)/cos^2x + sin^2x + 2sinxcosx$ =
$\displaystyle (cosx- sinx)(cosx+sinx)/(cosx+sinx)(cosx+sinx)$

How did they get rid of $\displaystyle 2sinxcosx$?
Thanks.

2. Originally Posted by ktcrew
Establish each identity.
$\displaystyle cos(2x)/1 + sin(2x) = cotx - 1/ cotx + 1$

Now in the solutions manual, this is what they have in the middle:
$\displaystyle (cosx- sinx)(cosx+sinx)/cos^2x + sin^2x + 2sinxcosx$ =
$\displaystyle (cosx- sinx)(cosx+sinx)/(cosx+sinx)(cosx+sinx)$

How did they get rid of $\displaystyle 2sinxcosx$?
Thanks.
they factorized. remember, $\displaystyle a^2 + 2ab + b^2 = (a + b)(a + b) = (a + b)^2$

if you take for instance, $\displaystyle a = \sin x$ and $\displaystyle b = \cos x$, then you get $\displaystyle \sin^2 x + 2 \sin x \cos x + \cos^2 x = (\sin x + \cos x)(\sin x + \cos x)$

3. Ok thanks. Man, I have trouble seeing the forest through the trees.