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    Hi, I'm having trouble with my Reduction Formula homework.

    Consider In = (Integral between pi/2 and 0 of) sin2n(theta)/sin (theta) , where n is a non negative integer.

    Using sinA - sinB = 2cos[(A+B)/2]sin[(A-B)/2], derive the reduction formula

    In - I(n-1) = [2(-1)^(n-1)] / (2n -1)





    I realise it may be hard to understand as I can't use latex but any help would be much appreciated!
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    I think the hint is in the I_n - I_{n-1}. If you write this out then you get \displaystyle \int_0^{\frac{\pi}{2}} \left( \dfrac{ \sin 2n\theta - \sin 2(n-1)\theta }{\sin \theta} \right) \, d\theta, and then you can apply your identity to what's in the brackets. You'll notice that nice things cancel and you get something you can integrate.
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    Thanks for the help

    I had a feeling it was something simple. But then I wondered whether using what they had given was sort of cheating and I had to derive that some other way.
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    (Original post by mknox669)
    I had a feeling it was something simple. But then I wondered whether using what they had given was sort of cheating and I had to derive that some other way.
    Usually when a hint is given it's because it's the best way to go about it and other ways of tackling it will make you hit brick walls -- use all the hints you're given :p: It's definitely not cheating.
 
 
 
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