anonymous
  • anonymous
Rewrite with only sin(x) and cos(x). sin(3x)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
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?
anonymous
  • anonymous
Just add the x's together?
zepdrix
  • zepdrix
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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anonymous
  • anonymous
Oh okay, so what's next?
zepdrix
  • zepdrix
In this case a=x b=2x And we'll split it up the way I did with the definition above.
zepdrix
  • zepdrix
\[\large \sin(x+2x)\quad =\quad \sin x \cdot \cos(2x)+\sin(2x)\cdot \cos x\]
zepdrix
  • zepdrix
Try to compare it to the definition above with the sin(a+b), see if it makes sense or not.
anonymous
  • anonymous
It doesn't really seem like it makes sense
anonymous
  • anonymous
Of course you will have to repeat the process again to split up sin(2x) and cos(2x), using a=x and b=x.
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
\[\large \sin(a+b) \quad = \quad \sin(a)\cos(b)+\sin(b)\cos(a)\]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 \[\large \sin(a+b) \quad \text{where} \quad a=x \quad \text{and} \quad b=2x\] \[\sin(a)\cos(b)+\sin(b)\cos(a) \quad \rightarrow \quad \sin(x)\cos(2x)+\sin(2x)\cos(x)\]
zepdrix
  • zepdrix
See how we replaced the A and B's? Too much for ya? D:
anonymous
  • anonymous
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
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
Well I'm with you so far
zepdrix
  • zepdrix
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
anonymous
  • anonymous
Yes I have them in my notes
zepdrix
  • zepdrix
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:\[\large \cos(2x)=\cos^2(x)-sin^2(x)\]
anonymous
  • anonymous
Alrighty then
zepdrix
  • zepdrix
\[\sin(x)\cos(2x)+\sin(2x)\cos(x)\]\[\large \rightarrow \quad \sin(x)\left(\cos^2x-\sin^2x\right)+\left(2\sin(x)\cos(x)\right)\cos(x)\]
anonymous
  • anonymous
Ohh i see
anonymous
  • anonymous
So now just multiply them together and you should have your final answer?
zepdrix
  • zepdrix
yes.
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
Yes, I understand, thank you very much :)

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