1-{(sin^2)/(1+cos x)}=cos x
How?
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:
$\displaystyle 1 - \frac{sin^2 x}{1 + cos x} = cos x$
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: .$\displaystyle 1 - (1 - \cos x) \;\;=\;\;1 - 1 + \cos x \;\;=\;\;\cos x$