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Please help me prove the identity:


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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.




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


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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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