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EasyMath

  • 3 years ago

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

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  1. zepdrix
    • 3 years ago
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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?

  2. EasyMath
    • 3 years ago
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    Just add the x's together?

  3. zepdrix
    • 3 years ago
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    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.

  4. EasyMath
    • 3 years ago
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    Oh okay, so what's next?

  5. zepdrix
    • 3 years ago
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    In this case a=x b=2x And we'll split it up the way I did with the definition above.

  6. zepdrix
    • 3 years ago
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    \[\large \sin(x+2x)\quad =\quad \sin x \cdot \cos(2x)+\sin(2x)\cdot \cos x\]

  7. zepdrix
    • 3 years ago
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    Try to compare it to the definition above with the sin(a+b), see if it makes sense or not.

  8. EasyMath
    • 3 years ago
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    It doesn't really seem like it makes sense

  9. henpen
    • 3 years ago
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    Of course you will have to repeat the process again to split up sin(2x) and cos(2x), using a=x and b=x.

  10. zepdrix
    • 3 years ago
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    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

  11. zepdrix
    • 3 years ago
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    \[\large \sin(a+b) \quad = \quad \sin(a)\cos(b)+\sin(b)\cos(a)\]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 \[\large \sin(a+b) \quad \text{where} \quad a=x \quad \text{and} \quad b=2x\] \[\sin(a)\cos(b)+\sin(b)\cos(a) \quad \rightarrow \quad \sin(x)\cos(2x)+\sin(2x)\cos(x)\]

  12. zepdrix
    • 3 years ago
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    See how we replaced the A and B's? Too much for ya? D:

  13. EasyMath
    • 3 years ago
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    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

  14. zepdrix
    • 3 years ago
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    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.

  15. EasyMath
    • 3 years ago
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    Well I'm with you so far

  16. zepdrix
    • 3 years ago
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    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

  17. EasyMath
    • 3 years ago
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    Yes I have them in my notes

  18. zepdrix
    • 3 years ago
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    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:\[\large \cos(2x)=\cos^2(x)-sin^2(x)\]

  19. EasyMath
    • 3 years ago
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    Alrighty then

  20. zepdrix
    • 3 years ago
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    \[\sin(x)\cos(2x)+\sin(2x)\cos(x)\]\[\large \rightarrow \quad \sin(x)\left(\cos^2x-\sin^2x\right)+\left(2\sin(x)\cos(x)\right)\cos(x)\]

  21. EasyMath
    • 3 years ago
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    Ohh i see

  22. EasyMath
    • 3 years ago
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    So now just multiply them together and you should have your final answer?

  23. zepdrix
    • 3 years ago
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    yes.

  24. zepdrix
    • 3 years ago
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    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.

  25. EasyMath
    • 3 years ago
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    Yes, I understand, thank you very much :)

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