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Please help me prove the identity:

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

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3 Answers

up vote 1 down vote accepted

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