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anonymous

  • one year ago

does 1/2sin^2x= -1/4 cos(2x)

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  1. anonymous
    • one year ago
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    Expand out both sides to see if that's true

  2. anonymous
    • one year ago
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    not sure if I know how to do that..

  3. anonymous
    • one year ago
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    cos(2x)=cos(x+x). now tell me how to expand the right hand side?

  4. anonymous
    • one year ago
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    2sin(x)^2=-cos(2x) cos(2X)=1-2sin^2x so.. 2sin(x)^2=2sin(x)^2-1 ?

  5. anonymous
    • one year ago
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    yeah, actually. you're done!

  6. anonymous
    • one year ago
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    so not equal... :(

  7. anonymous
    • one year ago
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    does

  8. anonymous
    • one year ago
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    yeah, you arrived at a contradiction and it can't be true.

  9. anonymous
    • one year ago
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    \[\int\limits_{?}^{?} cosxsinx dx\] = which one?

  10. anonymous
    • one year ago
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    integral of cosx * sinx dx I got the 1/2 sin^2(x) but my answer key I have the other one

  11. anonymous
    • one year ago
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    |dw:1433648150311:dw|

  12. anonymous
    • one year ago
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    |dw:1433648346891:dw|

  13. anonymous
    • one year ago
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    anyone? am I doing my U-sub wrong?

  14. anonymous
    • one year ago
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    That's correct as far as I am concerned, the only thing missing is the Constant

  15. anonymous
    • one year ago
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    so then 1/2 sin^2x should = -1/4 cos(2x) ??

  16. anonymous
    • one year ago
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    In a question like this, aren't we proving that LHS = RHS, how did you resort to using integration?

  17. anonymous
    • one year ago
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    the question was the integration. What I got was the 1/2 sin^2 (x) and my prof answer key shows the -1/4 cos (2x) wondering who made the mistake of if they are the same

  18. anonymous
    • one year ago
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    |dw:1433649465541:dw|

  19. anonymous
    • one year ago
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    Well \(cos(2x) = 1-2sin^{2}x\) So \(-\frac{1}{4}cos(2x) = -\frac{1}{4}(1-2sin^{2}x) =-\frac{1}{4} + \frac{1}{2}sin^{2}x\) But keep in mind that we have that constant of integration, the +C. That plus C would absorb the -1/4 and leave us with simply \(\frac{1}{2}sin^{2}x + C\) Your answer is equivalent as far as Im concerned.

  20. anonymous
    • one year ago
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    OK Thank you Concentrating!

  21. anonymous
    • one year ago
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    You're welcome :)

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spraguer (Moderator)
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