trigonometry identity I have some problem with proving this identity: 
$$2\left(1+\cos\alpha \right)-\sin^2\alpha=4\cos^4\frac{\alpha}{2}$$
I tried to start from the right side rewritting it to $(2\cos^2(2\frac{\alpha}{4}))^2$ but it's not working. 
 A: Replacing $\sin^2(x)=1-\cos^2(x)$ gives
$$
\begin{align}
2(1+\cos(\alpha))-\sin^2(\alpha)
&=1+2\cos(\alpha)+\cos^2(\alpha)\\[6pt]
&=(1+\cos(\alpha))^2\\
&=\left(2\cos^2\left(\frac\alpha2\right)\right)^2
\end{align}
$$
using the identity $\cos(\alpha)=2\cos^2\left(\frac\alpha2\right)-1$.
A: By considering the trigonommetric identity:
$$\cos^2{x} = \frac{1+\cos{2x}}{2},$$
you have:
$$2(1+\cos{\alpha}) = \sin^2 \alpha + 4 \left(\frac{1+\cos{\alpha}}{2}\right)^2 = \sin^2\alpha + 1 + 2 \cos\alpha + \cos^2\alpha, $$
which becomes an identity if you perform further simplification. 
Cheers! 
A: The most straightforward way is to use the Weierstrass substitution. Note that $\tan^2(\alpha/2)=\frac{1}{\cos^2(\alpha/2)} -1,$ that should help you with the RHS.
A: We have the double angle identities $sin(x)=2sin(\frac{x}{2})cos(\frac{x}{2})$ and $cos(x)=2cos^2(\frac{x}{2})-1$
Substituting into the left hand side of your expression, we have 
$2(1+2cos^2(\frac{\alpha}{2})-1) - (2sin(\frac{x}{2})cos(\frac{x}{2}))^2$
$=4cos^2(\frac{\alpha}{2})-4sin^2(\frac{\alpha}{2})cos^2(\frac{\alpha}{2})$
$=4cos^2(\frac{\alpha}{2})[1-sin^2(\frac{\alpha}{2})]$
$=4cos^2(\frac{\alpha}{2})[cos^2(\frac{\alpha}{2})]$
$=4cos^4(\frac{\alpha}{2})$
A: Let there is a function $f(x) = \cos x + 1$, then $$f^2(x) = \cos^2 x + 2\cos x + 1 = 2(1 + \cos x) - \sin^2 x.$$ While $$f(x) = 2\cos^2\frac{x}{2} = \sqrt{4\cos^4\frac{x}{2}}.$$ Therefore $$ 4\cos^4\frac{x}{2} = 2(1 +\cos x) - \sin^2x.$$
A: begin by writing sin a = 2sin a/2 cos a/2 in left side ,easy by factoring you will have the answer
