Trigonometry Identity homework help Could you please help me prove this:
$${2\cos(\theta/2)-1-\cos\theta\over2\cos(\theta/2)+1+\cos\theta}={1-\cos(\theta/2)\over1+\cos(\theta/2)}$$
 A: we have the rules $\cos(\theta)=\cos^2(\theta)-\sin^2(\theta)$ and $1=\sin^2(\theta)+\cos^2(\theta)$
so $$
\frac{2\cos(\theta/2)-2\cos^2(\theta/2)}{2\cos(\theta/2+2\cos^2(\theta/2)}=\frac{2\cos(\theta/2)(1-\cos(\theta/2))}{2\cos(\theta/2)(1+\cos(\theta/2))}=\frac{1-\cos(\theta/2)}{1+\cos(\theta/2)}
$$
A: Using the double angle formula we have
$$\cos \theta = 2\cos^2\frac{\theta}{2} - 1$$
Substituting this in, we get
$$\begin{align}\frac{2\cos\frac{\theta}{2} - 1 - \cos\theta}{2\cos\frac{\theta}{2} + 1 + \cos\theta} &= \frac{2\cos\frac{\theta}{2} - 2\cos^2\frac{\theta}{2}}{2\cos\frac{\theta}{2} + 2\cos^2\frac{\theta}{2}}\\
&= \frac{1 - \cos\frac{\theta}{2}}{1 + \cos\frac{\theta}{2}}\end{align}$$
A: Let $\alpha = \theta/2$. Then the numerator of the expression on the left is
$$
2\cos\alpha - 1 - \cos2\alpha \\
= 2\cos\alpha - (\cos^2\alpha + \sin^2\alpha) - (\cos^2\alpha - \sin^2\alpha) \\
= 2\cos\alpha - 2\cos^2\alpha \\
= 2\cos\alpha(1 - \cos\alpha)
$$
Do the same sort of thing with the denominator. Cancel.
