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- anonymous

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

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- anonymous

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

- chestercat

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- anonymous

Expand out both sides to see if that's true

- anonymous

not sure if I know how to do that..

- anonymous

cos(2x)=cos(x+x). now tell me how to expand the right hand side?

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- anonymous

2sin(x)^2=-cos(2x)
cos(2X)=1-2sin^2x so..
2sin(x)^2=2sin(x)^2-1 ?

- anonymous

yeah, actually. you're done!

- anonymous

so not equal... :(

- anonymous

does

- anonymous

yeah, you arrived at a contradiction and it can't be true.

- anonymous

\[\int\limits_{?}^{?} cosxsinx dx\] = which one?

- anonymous

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

- anonymous

|dw:1433648150311:dw|

- anonymous

|dw:1433648346891:dw|

- anonymous

anyone? am I doing my U-sub wrong?

- anonymous

That's correct as far as I am concerned, the only thing missing is the Constant

- anonymous

so then 1/2 sin^2x should = -1/4 cos(2x) ??

- anonymous

In a question like this, aren't we proving that LHS = RHS, how did you resort to using integration?

- 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

|dw:1433649465541:dw|

- 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

OK Thank you Concentrating!

- anonymous

You're welcome :)

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