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

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

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

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



Everything starts with $$sin(a+b)=sin acos b+cos asin b$$ 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 $$sin2a=2sin acos a$$ Stop me if you didn"t follow this. Now we can replace $a$ everywhere with $2x$ and get $$sin 4x=2sin2xcos2x$$ Now there"s a $sin2x$ in that formula; we can use double-angle on it to get $$sin4x=2(2sin xcos x)cos2x$$ 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: $$sin4x=2 imes2 imessin x imescos x imescos2x$$ Finally, $2 imes2=4$, so $$sin4x=4sin xcos xcos2x$$


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