# Prove the trigonometric identity

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

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Oct 14 '12 at 13:25

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.

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

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

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

$\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)=\frac{1}{2}\cdot 2\sin^2\alpha$

$\sin^2\alpha=\sin^2\alpha$

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This is badly formatted: One says in effect that if a certain equality holds, then $\sin^2\alpha=\sin^2\alpha$, and concludes that that equality holds. One should be "$=$" between, for example, $\cos^2\alpha-\cos^4\alpha+\sin^4\alpha$ and the thing on the line after it, $\cos^2\alpha(1-\cos^2\alpha+\sin^4\alpha$, and so on. – Michael Hardy Sep 26 '12 at 13:09

$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\cos^2\alpha+(\sin^4\alpha-\cos^4\alpha)=$$ $$=\cos^2\alpha+(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)=\cos^2\alpha+\sin^2\alpha-\cos^2\alpha=$$ $$=\sin^2\alpha=1/2-1/2\cos2\alpha$$ Over!

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Yes ,youare right! – Riemann Sep 26 '12 at 11:46
This is a better answer than the "accepted" one. – Michael Hardy Sep 26 '12 at 13:09

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

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