# Trig identities

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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Where does that first $\cos{(2x)}$ come from in your expansion? –  t.b. Jun 23 '11 at 23:24
2(2sinxcosx) I disribute the 2 to sin and cos. –  Adam Jun 23 '11 at 23:29
You shouldn't distribute into or out of the argument of functions in general. Here you just have $2(2\sin(x)\cos(x))=(2\cdot 2)(\sin(x)\cos(x))=4\sin(x)\cos(x)$. –  yunone Jun 23 '11 at 23:31
I dont know what an argument of function is. –  Adam Jun 23 '11 at 23:31
But no distribution here, $a(bc) = abc$ not $(ab)(ac)$. Also I'm wondering how you got $\cos{(\mathbf{2}x)}$ there. Did you mean $2\cos{x}$? –  t.b. Jun 23 '11 at 23:32

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 $$\sin4x=4\sin x\cos x\cos2x$$
@Adam, do you mean, what did I do at that step? I saw the $\sin2x$, and replaced it with $2\sin x\cos x$, which is OK, since the double-angle formula says $\sin2x=2\sin x\cos x$. That's the only difference between the 3rd displayed equation and the 4th. –  Gerry Myerson Jun 24 '11 at 0:55