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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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    Where does that first $\cos{(2x)}$ come from in your expansion?2011-06-23
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    2(2sinxcosx) I disribute the 2 to sin and cos.2011-06-23
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    You shouldn't distribute into or out of the argument of functions in general. Here you just have $2(2\sin(x)\cos(x))=(2\cdot 2)(\sin(x)\cos(x))=4\sin(x)\cos(x)$.2011-06-23
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    I dont know what an argument of function is.2011-06-23
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    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 $2\cos{x}$?2011-06-23
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    Is that not equal? My books without explanation expchanges 2 theta into 2costheta all the time.2011-06-23
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    The argument is like the input of a function, for $\sin(x)$, $x$ is the input/argument of the $\sin$ function.2011-06-23
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    Alright well I know that sin(2a)=2sinAcosA so that gives me sin2(2x) = 2sin2xCos2x which gives me 4sinxcos2x which is not an identity.2011-06-23
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    Now unwrap the $\sin(2x)$ in your last expression to get the identity.2011-06-23
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    4sinxcos2x I seem to be missing cosx right?2011-06-23
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    isn't it obvious?$\sin 2x = 2 \sin x \cos x$, use this identity twice and you will get what you need.2011-06-23
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    8sinxcos2x?????2011-06-23
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    (a) sin2(2x) = 2sin2xcos2x = 2[sin(2x)cos(2x)] (b) sin(2x) = **2[sin(x)cos(x)]**. Substituting (b) into the last expression on the right of (a) gives 2(**2(sin(x)cos(x))**cos(2x)) $= 2\cdot 2\left[sin(x)cos(x)cos(2x)\right] = 4sin(x)cos(x)cos(2x)$, as desired.2011-06-24
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    Where does B come from?2011-06-24
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    part (b) 's the same identitydo you mean? you mention in a comment above that you know that sin(2a) = 2 sin(a)cos(a). So in the expansion 2[ **sin(2x)** cos(2x), expanded the **sin(2x)** factor, getting sin(2x) = **2 sin(x)cos(x)** ...and than inserting that back into the expansion, replacing sin(2x) with its equivalent, getting 2[ **2sin(x)cos(x)** cos(2x)]. Go through Gerry's answer slowly, carefully.2011-06-24
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    Oh that makes more sense now, thanks.2011-06-24
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    @Adam: After a while, you also could be a mathemagician.2011-06-24

1 Answers 1

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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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    You said "We can use the double-angle on it to get" I know what that is, it is what I orginally did to get the equation but what did you actually do?2011-06-24
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    @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.2011-06-24
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    I get it now, its like inserting a formula into a formula.2011-06-24