Results 1 to 4 of 4

Math Help - Identity manipulation but a tough one!

  1. #1
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92

    Identity manipulation but a tough one!

    Hi, I'm struggling with this question:

    Show that

     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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by dojo View Post
    Show that

     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

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

    Use those formulas to get

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

    and therefore

    (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,

    (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 \alpha, \beta and \gamma, it is necessary to break the symmetry in order to get an efficient proof (in this case, by treating the term \sin2\alpha differently from its "partners" \sin2\beta and \sin2\gamma).
    Last edited by Opalg; November 3rd 2010 at 02:54 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    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:

    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??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    \cos x - \cos y = 2\sin\frac12(x+y)\sin\frac12(x-y)

    This line should be:

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

    This makes:

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

    Otherwise, it's a very interesting problem
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Set manipulation
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 25th 2011, 11:10 PM
  2. Set Manipulation
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 30th 2009, 03:17 PM
  3. Matrix manipulation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 17th 2009, 03:34 AM
  4. complex manipulation?????
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 3rd 2008, 01:10 PM
  5. tough tough gauss law
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: April 15th 2008, 04:28 AM

Search Tags


/mathhelpforum @mathhelpforum