Prove the trigonometric identity Please help me prove the identity: 

$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$$

 A: $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\cos^2\alpha+(\sin^4\alpha-\cos^4\alpha)=$$
$$=\cos^2\alpha+(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)=\cos^2\alpha+\sin^2\alpha-\cos^2\alpha=$$
$$=\sin^2\alpha=1/2-1/2\cos2\alpha$$ Over!
A: Use the identities, $\sin^2\alpha+\cos^2\alpha=1$ and $\cos2\alpha=1-2\sin^2\alpha$
Since, $\cos^2\alpha-\cos^4\alpha=\cos^2\alpha(1-\cos^2\alpha)=\cos^2\alpha\cdot\sin^2\alpha$
So, $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\cos^2\alpha\cdot\sin^2\alpha+\sin^4\alpha$$ $$=\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)$$ $$=\sin^2\alpha=\frac{1-\cos2\alpha}{2}$$
A: Implement the formula:
1) $1-\cos^2\alpha=\sin^2\alpha$
2) $\cos2\alpha=\cos^2\alpha-\sin\alpha$
3) $1=\sin^2\alpha+\cos^2\alpha$ 
Now turn the proof given identity.
$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$
$\cos^2\alpha(1-\cos^2\alpha)+\sin^4\alpha=\frac{1}{2}(1-\cos2\alpha)$
$\cos^2\alpha\sin^2\alpha+\sin^4\alpha=\frac{1}{2}(\sin^2\alpha+\cos^2\alpha-\cos^2\alpha+\sin^2\alpha)$
$\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)=\frac{1}{2}\cdot 2\sin^2\alpha$
$\sin^2\alpha=\sin^2\alpha$
