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anonymous

  • 5 years ago

How do I convert: y = 6 sqrt(2) sin(x)-6 sqrt(2) cos(x) into the form y = A sin(Bx-C) that has the same graph as the above? Wolfram|Alpha tells me I'm supposed to get: y = -12sin(pi/4 - x) But I want to know how I get there.

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  1. anonymous
    • 5 years ago
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    im waiting for some one to respond to see how they figure this out!

  2. anonymous
    • 5 years ago
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    Me too..

  3. anonymous
    • 5 years ago
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    Original Equation for better readability: \[y = 6 \sqrt(2) \sin(x) - 6 \sqrt(2) \cos(x)\] And the equation in the \[y=A \sin(Bx-C)\] form: \[y = -12\sin(\pi/4 - x)\] ....How do I get from original to final equation posted?

  4. anonymous
    • 5 years ago
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    You want to put a sum of sine and cosine into one sine. You can consider the following:\[r \sin(x + \alpha) = r \cos \alpha \sin x + r \sin a \cos x\]

  5. anonymous
    • 5 years ago
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    For the two expressions to be identical, you have to equate the coefficients on either side and solve for r and alpha.

  6. anonymous
    • 5 years ago
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    Does this make sense?

  7. anonymous
    • 5 years ago
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    So\[r \cos \alpha = 6 \sqrt{2}, r \sin \alpha = -6 \sqrt{2}\]

  8. anonymous
    • 5 years ago
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    Divide sine by cosine to give\[\frac{r \sin \alpha}{r \sin \alpha}=\tan \alpha = -1 \rightarrow \alpha = \tan^{-1}(-1) = -\frac{\pi}{4}\]

  9. anonymous
    • 5 years ago
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    Substitute this into either one of the sine or cosine expressions to get r:\[r \cos \alpha = r \cos (-\pi/4)=r \frac{1}{\sqrt{2}}=6 \sqrt{2} \rightarrow r = 12\]

  10. anonymous
    • 5 years ago
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    what is happenin here

  11. anonymous
    • 5 years ago
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    \[6\sqrt{2}\sin x -6\sqrt{2} \cos x = 12\sin(x - \frac{\pi}{4})\]

  12. anonymous
    • 5 years ago
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    oh sweet, I think I can pick it up from here. Thanks so much!

  13. anonymous
    • 5 years ago
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    No worries.

  14. anonymous
    • 5 years ago
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    dam your goo at it

  15. anonymous
    • 5 years ago
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    Your Wolframalpha answer is the same - \[12\sin(x - \pi/4) = 12 \sin (-(\pi/4 -x))=-12\sin (\pi/4-x)\]

  16. anonymous
    • 5 years ago
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    thanks

  17. anonymous
    • 5 years ago
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    oh, btw, on the step where you're dividing sin by cos to get tan, why did you divide 2 sins? Is that just a typo or am I missing something?

  18. anonymous
    • 5 years ago
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    yeah, typo...

  19. anonymous
    • 5 years ago
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    no prob, thanks a ton man, it makes sense now.

  20. anonymous
    • 5 years ago
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    you're welcome

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