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Finally got to double angles. Anyways I need to show that these are identities.

$\sin(4x) = 4 \sin(x) \cos(x) \cos(2x)$

The book does some magic and gets $2(2\sin(x)\cos(x))\cos(2x)$

This makes no sense to me, if I expand that I get $4\sin(x)\cos(2x)\cos(2x)$ which is not equal.

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    @Adam: After a while, you also could be a mathemagician.2011-06-24

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Everything starts with $\sin(a+b)=\sin a\cos b+\cos a\sin b$ This is an identity, it holds for all $a$ and $b$. In particular, you're allowed to replace $b$ with $a$, so long as you do it consistently throughout, and you get $\sin2a=2\sin a\cos a$ Stop me if you didn't follow this. Now we can replace $a$ everywhere with $2x$ and get $\sin 4x=2\sin2x\cos2x$ Now there's a $\sin2x$ in that formula; we can use double-angle on it to get $\sin4x=2(2\sin x\cos x)\cos2x$ Now multiplication is associative, which means as long as all we're doing is multiplication, we don't need parentheses. On the right side, we're multiplying 5 things: $\sin4x=2\times2\times\sin x\times\cos x\times\cos2x$ Finally, $2\times2=4$, so $\sin4x=4\sin x\cos x\cos2x$

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    I get it now, its like inserting a formula into a formula.2011-06-24