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Math Help - more trigonometry

  1. #1
    Senior Member slevvio's Avatar
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    more trigonometry

     sin^2(\frac{\pi}{8}+\frac{A}{2})-sin^2(\frac{\pi}{8}-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA

    I was having trouble proving this and was wondering if anyone could help me out... thanks in advance.
    Last edited by slevvio; November 10th 2007 at 01:52 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slevvio View Post
     sin^2(\frac{\pi}{2}+\frac{A}{2})-sin^2(\frac{\pi}{2}-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA

    I was having trouble proving this and was wondering if anyone could help me out... thanks in advance.
    Not true:
    sin \left ( \frac{\pi}{2} + \frac{A}{2} \right ) = sin \left ( \frac{\pi}{2} \right ) cos \left ( \frac{A}{2} \right ) + cos \left ( \frac{\pi}{2} \right ) sin \left ( \frac{A}{2} \right ) = cos \left ( \frac{A}{2} \right )

    and
    sin \left ( \frac{\pi}{2} - \frac{A}{2} \right ) = sin \left ( \frac{\pi}{2} \right ) cos \left ( \frac{A}{2} \right ) - cos \left ( \frac{\pi}{2} \right ) sin \left ( \frac{A}{2} \right ) = cos \left ( \frac{A}{2} \right )

    So
    sin^2 \left ( \frac{\pi}{2} + \frac{A}{2} \right ) - sin^2 \left ( \frac{\pi}{2} + \frac{A}{2} \right ) = <br />
cos^2 \left ( \frac{A}{2} \right ) - cos^2 \left ( \frac{A}{2} \right ) = 0

    Unless you are to solve for A? In that case then we have
    0 = \frac{sin(A)}{\sqrt{2}}
    means that
    A = n \pi
    where n is an integer.

    -Dan
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  3. #3
    Senior Member slevvio's Avatar
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    Thanks a lot, no I only needed to prove the statement I didnt realise that \cos(\frac{\pi}{2})=0 hence difficulties... Maybe i shouldn't try maths problems at 10:30 at night lol...
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by slevvio View Post
    Thanks a lot, no I only needed to prove the statement I didnt realise that \cos(\frac{\pi}{2})=0 hence difficulties... Maybe i shouldn't try maths problems at 10:30 at night lol...
    Anytime you're taking a trigonometric function of a constant with a fraction of pi in it, it is a good idea to check your unit circle. There are a few angles which are extremely common, pi/2 being one of them, you will see these angles all the time, because they come out cleanly (well, i guess if you consider square root of 3, divided by 2 clean lol). Anyway, many equations use these angles, and much of the time it is necessary to be able to convert them.
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  5. #5
    Senior Member slevvio's Avatar
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    Woops! In my exhaustion I copied down the fraction wrong into the post so I will edit it now and hopefully someone will prove that which is actually true!
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by slevvio View Post
     sin^2(\frac{\pi}{8}+\frac{A}{2})-sin^2(\frac{\pi}{8}-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA

    I was having trouble proving this and was wondering if anyone could help me out... thanks in advance.
     sin^2\frac{1}{2} \left( {\frac{\pi}{4}+A} \right) -sin^2\frac{1}{2} \left( {\frac{\pi}{4}-A} \right) = \frac{1 - cos (\frac{\pi}{4}+A) - (1 - cos (\frac{\pi}{4}-A))}{2}
     = \frac{cos (\frac{\pi}{4}-A) - cos (\frac{\pi}{4}+A)}{2}

     = \frac{cos \frac{\pi}{4}cosA + sin \frac{\pi}{4} sin A - cos \frac{\pi}{4}cosA + sin \frac{\pi}{4} sin A}{2}

     = sin \frac{\pi}{4} sin A = \frac{1}{\sqrt 2}sin A
    Last edited by kalagota; November 10th 2007 at 03:45 AM.
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  7. #7
    Senior Member slevvio's Avatar
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    Thanks for the help! I was wondering if you could tell me which rule you used in the first line of working i.e. sin^2\frac{1}{2}(\frac{\pi}{4}+A)-sin^2\frac{1}{2}(\frac{\pi}{4}-A)=\frac{1-cos(\frac{\pi}{4}+A)-(1-cos(\frac{\pi}{4}-A)}{2}?

    I eventually got it to work using the addition formula:

     cos\theta - cos\phi = -2sin(\frac{\theta - \phi}{2})sin(\frac{\theta + \phi}{2}) which gives the same result as yours but after about 2 pages of working - do you know a quicker way? Or did you just omit your working? Thanks again.
    Last edited by slevvio; November 10th 2007 at 03:17 AM.
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by slevvio View Post
    Thanks for the help! I was wondering if you could tell me which rule you used in the first line of working i.e. sin^2\frac{1}{2}(\frac{\pi}{4}+A)-sin^2\frac{1}{2}(\frac{\pi}{4}-A)=\frac{1-cos(\frac{\pi}{4}+A)-(1-cos(\frac{\pi}{4}-A)}{2}?

    I eventually got it to work using the addition formula:

     cos\theta - cos\phi = -2sin(\frac{\theta - \phi}{2})sin(\frac{\theta + \phi}{2}) which gives the same result as yours but after about 2 pages of working - do you know a quicker way? Or did you just omit your working? Thanks again.
    ahh okay..

    note that sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - cos \theta}{2}}

    so, sin^2 \frac{\theta}{2} = \frac{1 - cos \theta}{2}

    okay already?
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