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Thread: Identity manipulation but a tough one!

  1. #1
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    Identity manipulation but a tough one!

    Hi, I'm struggling with this question:

    Show that

    $\displaystyle 4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma) + sin(2\alpha + 2\beta + 2\gamma) = sin(2\alpha) + sin(2\beta) + sin(2\gamma)$

    I have covered identities but I get in a huge expansion that is so confusing and from past questions tend to have a neater way of doing things.

    Your help would be greatly appreciated.
    D
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  2. #2
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    Quote Originally Posted by dojo View Post
    Show that

    $\displaystyle 4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma) + sin(2\alpha + 2\beta + 2\gamma) = sin(2\alpha) + sin(2\beta) + sin(2\gamma)$

    I have covered identities but I get in a huge expansion that is so confusing and from past questions tend to have a neater way of doing things.
    I think that the tidiest way to do this is to use the addition formulas

    $\displaystyle (1)\;\sin x + \sin y = 2\sin\frac12(x+y)\cos\frac12(x-y),$
    $\displaystyle (2)\;\sin x - \sin y = 2\cos\frac12(x+y)\sin\frac12(x-y),$
    $\displaystyle (3)\;\cos x - \cos y = 2\sin\frac12(x+y)\sin\frac12(x-y).$

    Use those formulas to get

    $\displaystyle (4)\; \sin2(\alpha+\beta+\gamma) - \sin2\alpha = 2\cos(2\alpha+\beta+\gamma)\sin(\beta+\gamma)$ (from (2)),
    $\displaystyle (5)\; \sin2\beta + \sin2\gamma = 2\sin(\beta+\gamma)\cos(\beta - \gamma)$ (from (1)),

    and therefore

    $\displaystyle (6)\; \sin2(\alpha+\beta+\gamma) - \sin2\alpha - \sin2\beta - \sin2\gamma = 2\sin(\beta+\gamma)\bigl(\cos(2\alpha+\beta+\gamma ) - \cos(\beta - \gamma)\bigr)$ (subtracting (5) from (4)).

    Also,

    $\displaystyle (7)\; \cos(2\alpha+\beta+\gamma) - \cos(\beta - \gamma) = 2\sin(\alpha+\beta)\sin(\alpha+\gamma)$ (from (3)).

    Finally, substitute the result from (7) into (6) to get your formula.

    The interesting thing here is that although the formula is completely symmetrical in $\displaystyle \alpha$, $\displaystyle \beta$ and $\displaystyle \gamma$, it is necessary to break the symmetry in order to get an efficient proof (in this case, by treating the term $\displaystyle \sin2\alpha$ differently from its "partners" $\displaystyle \sin2\beta$ and $\displaystyle \sin2\gamma$).
    Last edited by Opalg; Nov 3rd 2010 at 02:54 AM.
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  3. #3
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    Thanks for running this through - its a great help and I feel a lot more confident with this type of question.

    However, when I rearrange and replace as you suggest I get:

    $\displaystyle 4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma) = - 4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma)$ which isn't true.

    Where am I going wrong??
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  4. #4
    MHF Contributor Unknown008's Avatar
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    $\displaystyle \cos x - \cos y = 2\sin\frac12(x+y)\sin\frac12(x-y)$

    This line should be:

    $\displaystyle \cos x - \cos y = -2\sin\frac12(x+y)\sin\frac12(x-y)$

    This makes:

    $\displaystyle \cos(2\alpha+\beta+\gamma) - \cos(\beta - \gamma) = -2\sin(\alpha+\beta)\sin(\alpha+\gamma)$

    Otherwise, it's a very interesting problem
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