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Step1 1996 q6

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Original post by Dog4444
Thanks, it seems to be true for K>1 though. :s-smilie:
So...?
Reply 21
Original post by DFranklin
So...?


Well, I just thought it needs some adjustment or something, and having done it again (IBP twice) I got:

[br]coskxcosx=(k2k21)((1/k)cosxsinkx(1/k2)sinxcoskx)[br][br]\int coskxcosx = (\frac{k^2}{k^2-1})((1/k)cosxsinkx-(1/k^2)sinxcoskx)[br]

What resolves this issue, and I hope anyone can verify that's right. :smile:

But I still try to find the other way to go about it, but spoilers didn't help me.
Your argument is valid for k not equal to 1. (That is, the method is OK, and I assume your answer is correct but I haven't checked the algebra).

k = 1 is a special case. Fortunately, it's not hard to find cosxcosxdx\int \cos x \cos x \,dx.
Reply 23
Anyone could help me on other ways to solve the last part?

I checked speleo's solution (http://www.thestudentroom.co.uk/showthread.php?t=358891&page=2&p=8027848#post8027848), but I can't really follow from place where turns sins into cosines and looks like he lost a "2" when he used double angle formula.
Reply 24
Original post by Dog4444
Anyone could help me on other ways to solve the last part?

I checked speleo's solution (http://www.thestudentroom.co.uk/showthread.php?t=358891&page=2&p=8027848#post8027848), but I can't really follow from place where turns sins into cosines and looks like he lost a "2" when he used double angle formula.


It comes from the identity

sinAsinB=cos(AB)cos(A+B)2\sin A \sin B = \dfrac{\cos (A-B) - \cos (A+B)}{2}

which is evident by considering where the ±\pm signs occur in the expansion of cos(A±B)\cos (A \pm B).

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