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Rewrite with only sin(x) and cos(x). sin(3x)

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Oh boy this one is going to be a pain in the butt :) We'll want to apply the Sum Formula for Sine. \[\large \sin(a+b)=\sin a \cdot \cos b+\sin b \cdot \cos a\]So we can rewrite our problem like this,\[\large \sin(3x) \qquad \rightarrow \qquad \sin(x+2x)\]See what we're going to do with it?
Just add the x's together?
No, we're doing the opposite actually.. we STARTED with the X's added together, and we're separating them, writing them as addition. So we can apply the Addition Formula for Sine.

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Other answers:

Oh okay, so what's next?
In this case a=x b=2x And we'll split it up the way I did with the definition above.
\[\large \sin(x+2x)\quad =\quad \sin x \cdot \cos(2x)+\sin(2x)\cdot \cos x\]
Try to compare it to the definition above with the sin(a+b), see if it makes sense or not.
It doesn't really seem like it makes sense
Of course you will have to repeat the process again to split up sin(2x) and cos(2x), using a=x and b=x.
Easy, The thing with the a+b at the very top. That is an identity, you just have to take that for granted. Just believe it!! :D The question is, does our sin(x+2x) match the way we split up the (a+b). Maybe I can give a little more detail.. Hmm
\[\large \sin(a+b) \quad = \quad \sin(a)\cos(b)+\sin(b)\cos(a)\]Try not to get confused by that part ok? You just have to believe that it's true XD In our problem, we've rewritten\[\large \sin(3x) \quad \text{as} \quad \sin(x+2x)\]We can think of this as \[\large \sin(a+b) \quad \text{where} \quad a=x \quad \text{and} \quad b=2x\] \[\sin(a)\cos(b)+\sin(b)\cos(a) \quad \rightarrow \quad \sin(x)\cos(2x)+\sin(2x)\cos(x)\]
See how we replaced the A and B's? Too much for ya? D:
No i understand that part completely :p its just it seems like there's something else we have to do because these are my answer choices: 1) 2 sin x cos2x + cos x 2) 2 sin x cos2x + sin3x 3) sin x cos2x - sin3x + cos3x 4) 2 cos2x sin x + sin x - 2 sin3x
Yes, we're only half done..... I'm trying to lead you along... I don't wanna go too far into this problem if you're confused.
Well I'm with you so far
Hmm so we have sin(2x) and cos(2x) terms now. We COULD do the same thing we did before, rewriting it as sin(x+x) and cos(x+x) and applying the sum formulas again. But that would be very tedious. From here, we'd rather apply the Double Angle Formulas. Do you remember those? :D
Yes I have them in my notes
Ok cool :) So we'll make use of these identities. Sine Double Angle Formula:\[\large \sin(2x)=2\sin(x)\cos(x)\] Cosine Double Angle Formula:\[\large \cos(2x)=\cos^2(x)-sin^2(x)\]
Alrighty then
\[\sin(x)\cos(2x)+\sin(2x)\cos(x)\]\[\large \rightarrow \quad \sin(x)\left(\cos^2x-\sin^2x\right)+\left(2\sin(x)\cos(x)\right)\cos(x)\]
Ohh i see
So now just multiply them together and you should have your final answer?
the sine in the first term needs to be distributed to each term in the brackets. and thennnnn on the right.. i think you end up with a cos^2 since you have 2 cosines and what not.
Yes, I understand, thank you very much :)

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