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zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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?
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Just add the x's together?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Oh okay, so what's next?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
In this case a=x b=2x And we'll split it up the way I did with the definition above.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
\[\large \sin(x+2x)\quad =\quad \sin x \cdot \cos(2x)+\sin(2x)\cdot \cos x\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Try to compare it to the definition above with the sin(a+b), see if it makes sense or not.
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
It doesn't really seem like it makes sense
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Of course you will have to repeat the process again to split up sin(2x) and cos(2x), using a=x and b=x.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
\[\large \sin(a+b) \quad = \quad \sin(a)\cos(b)+\sin(b)\cos(a)\]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 \[\large \sin(a+b) \quad \text{where} \quad a=x \quad \text{and} \quad b=2x\] \[\sin(a)\cos(b)+\sin(b)\cos(a) \quad \rightarrow \quad \sin(x)\cos(2x)+\sin(2x)\cos(x)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
See how we replaced the A and B's? Too much for ya? D:
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Well I'm with you so far
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Yes I have them in my notes
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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:\[\large \cos(2x)=\cos^2(x)sin^2(x)\]
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Alrighty then
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
\[\sin(x)\cos(2x)+\sin(2x)\cos(x)\]\[\large \rightarrow \quad \sin(x)\left(\cos^2x\sin^2x\right)+\left(2\sin(x)\cos(x)\right)\cos(x)\]
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Ohh i see
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
So now just multiply them together and you should have your final answer?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

EasyMath Group TitleBest ResponseYou've already chosen the best response.0
Yes, I understand, thank you very much :)
 one year ago
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