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EasyMath
 3 years ago
Rewrite with only sin(x) and cos(x).
sin(3x)
EasyMath
 3 years ago
Rewrite with only sin(x) and cos(x). sin(3x)

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zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7Oh 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?

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0Just add the x's together?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7No, 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.

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0Oh okay, so what's next?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7In this case a=x b=2x And we'll split it up the way I did with the definition above.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7\[\large \sin(x+2x)\quad =\quad \sin x \cdot \cos(2x)+\sin(2x)\cdot \cos x\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7Try to compare it to the definition above with the sin(a+b), see if it makes sense or not.

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0It doesn't really seem like it makes sense

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7Easy, 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
 3 years ago
Best ResponseYou've already chosen the best response.7\[\large \sin(a+b) \quad = \quad \sin(a)\cos(b)+\sin(b)\cos(a)\]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
 3 years ago
Best ResponseYou've already chosen the best response.7See how we replaced the A and B's? Too much for ya? D:

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0No 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
 3 years ago
Best ResponseYou've already chosen the best response.7Yes, 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.

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0Well I'm with you so far

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7Hmm 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

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0Yes I have them in my notes

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7Ok 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:\[\large \cos(2x)=\cos^2(x)sin^2(x)\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7\[\sin(x)\cos(2x)+\sin(2x)\cos(x)\]\[\large \rightarrow \quad \sin(x)\left(\cos^2x\sin^2x\right)+\left(2\sin(x)\cos(x)\right)\cos(x)\]

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0So now just multiply them together and you should have your final answer?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.7the 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.

EasyMath
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, I understand, thank you very much :)
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