1. ## Trig identities

1-{(sin^2)/(1+cos x)}=cos x
How?

2. Is this what you mean?

Prove that,

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

If so, then it is not as hard as you think. Everything is already changed to $\displaystyle sine$ and $\displaystyle cosine$, so half of the work is done.

First, multiply both sides by $\displaystyle 1 + cos x$:

$\displaystyle 1 + cos x - sin^2 x = cos x + cos^2 x$

Then subtract cos x from both sides:

$\displaystyle 1 - sin^2 x = cos^2 x$

Add $\displaystyle sin^2 x$to both sides and we now have that:

$\displaystyle cos^2 x + sin^2 x = 1$

This is a trigonometric identity, and therefore you have proved that:

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

3. Hello, dolphinlover527!

Prove: .$\displaystyle 1-\frac{\sin^2\!x}{1+\cos x} \:=\:\cos x$

We have: .$\displaystyle 1 - \frac{\sin^2\!x}{1+\cos x} \;\;=\;\;1 - \frac{1-\cos^2\!x}{1+\cos x} \;\;=\;\;1 - \frac{(1-\cos x)(1+\cos x)}{1+\cos x}$

Reduce: .$\displaystyle 1 - (1 - \cos x) \;\;=\;\;1 - 1 + \cos x \;\;=\;\;\cos x$