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Math Help - Reduction formulae.

  1. #1
    Super Member Showcase_22's Avatar
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    Reduction formulae.



    These two questions are what i'm stuck on. I managed to get a fair way through the first question:



    It clearly isn't finished and i'm not really sure what the next step would be.

    Please help!
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    For the first one.

    If you're stuck, imagine what I_{n-1} is :
    I_{n-1}=\int \frac{\sin(2(n-1) \theta)}{\sin(\theta)} ~d \theta

    You can see that what you have to find is also an expression for I_n-I_{n-1} :

    I_n-I_{n-1}=\int \frac{\sin(2n \theta)}{\sin(\theta)} - \frac{\sin(2(n-1) \theta)}{\sin(\theta)} ~d \theta

    =\int \frac{\color{red}\sin(2n \theta)-\sin(2(n-1)\theta)}{\sin(\theta)} ~d \theta

    Now, there's a useful formula :

    \sin(x)-\sin(y)=2 \cos(\tfrac{x+y}{2}) \sin (\tfrac{x-y}{2})

    So {\color{red}\sin(2n \theta)-\sin(2(n-1)\theta)}=2 \cos \left(\frac{2(n+n-1) \theta}{2}\right) \sin \left(\frac{2(n-(n-1)) \theta}{2}\right)
    =2 \cos ((2n-1) \theta) \sin(\theta)

    I'm sure you can finish
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  3. #3
    Super Member Showcase_22's Avatar
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    Thanks as ever Moo!

    I would post the remainder of my method but i've done it on my whiteboard and it won't fit on my scanner.

    Whenever I do these, I always try and get the I_n-1 from integrating by parts. This time it didn't work!

    Anyone have any thoughts from the second one?

    Don't worry! I managed to use the previous method to do the second question. Thanks again moo! =D
    Last edited by Showcase_22; September 9th 2008 at 12:31 PM. Reason: Applied method to the second question.
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  4. #4
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    Thanks as ever Moo!

    I would post the remainder of my method but i've done it on my whiteboard and it won't fit on my scanner.

    Whenever I do these, I always try and get the I_n-1 from integrating by parts. This time it didn't work!

    Anyone have any thoughts from the second one?
    Do it the same way, find a formula for I_n-I_{n-2}

    Most of the time, you'll have to use integration by parts. But it's quite often straightforward. So if you can't find anything interesting or that blocks you, you have the method of being inspired by the answer :
    I_n=a \cdot I_{n-1}+\dots for example, so find a formula of I_n-a \cdot I_{n-1} and prove it equals \dots
    Last edited by Moo; September 9th 2008 at 12:35 PM.
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  5. #5
    Super Member Showcase_22's Avatar
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    Thanks Moo!

    I did it the same way as that method yesterday (I wrote it in my last post).

    The trigonometric identity was particularly interesting. Cheers!
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