anonymous
  • anonymous
does 1/2sin^2x= -1/4 cos(2x)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Expand out both sides to see if that's true
anonymous
  • anonymous
not sure if I know how to do that..
anonymous
  • anonymous
cos(2x)=cos(x+x). now tell me how to expand the right hand side?

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anonymous
  • anonymous
2sin(x)^2=-cos(2x) cos(2X)=1-2sin^2x so.. 2sin(x)^2=2sin(x)^2-1 ?
anonymous
  • anonymous
yeah, actually. you're done!
anonymous
  • anonymous
so not equal... :(
anonymous
  • anonymous
does
anonymous
  • anonymous
yeah, you arrived at a contradiction and it can't be true.
anonymous
  • anonymous
\[\int\limits_{?}^{?} cosxsinx dx\] = which one?
anonymous
  • anonymous
integral of cosx * sinx dx I got the 1/2 sin^2(x) but my answer key I have the other one
anonymous
  • anonymous
|dw:1433648150311:dw|
anonymous
  • anonymous
|dw:1433648346891:dw|
anonymous
  • anonymous
anyone? am I doing my U-sub wrong?
anonymous
  • anonymous
That's correct as far as I am concerned, the only thing missing is the Constant
anonymous
  • anonymous
so then 1/2 sin^2x should = -1/4 cos(2x) ??
anonymous
  • anonymous
In a question like this, aren't we proving that LHS = RHS, how did you resort to using integration?
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1433649465541:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
OK Thank you Concentrating!
anonymous
  • anonymous
You're welcome :)

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