Results 1 to 5 of 5

Math Help - [SOLVED] Solve cos54cos@ + sin54sin@ = sin2@

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    Talking [SOLVED] Solve cos54cos@ + sin54sin@ = sin2@

    Find the general solution of
    cos54cos@ + sin54sin@ = sin2@
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    Quote Originally Posted by differentiate View Post
    Find the general solution of
    cos54cos@ + sin54sin@ = sin2@
    see below
    Last edited by bigwave; January 15th 2010 at 10:33 AM. Reason: revised option
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    Quote Originally Posted by differentiate View Post
    Find the general solution of
    cos54cos@ + sin54sin@ = sin2@
    On your LHS employ the identity \cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b) to simplify.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello differentiate
    Quote Originally Posted by differentiate View Post
    Find the general solution of
    cos54cos@ + sin54sin@ = sin2@
    My best guess is that you want the general solution of
    \cos54^o\cos\alpha+\sin54^o\sin\alpha = \sin2\alpha
    in which case, as pickslides has suggested:
    \cos(54^o-\alpha) =\sin2\alpha
    =\cos(90^o-2\alpha)
    Now the general solution of
    \cos A = \cos B
    is (in degrees):
    A = 360n^o \pm B, n =0, \pm1, \pm2, ...
    So here we get:
    54^o-\alpha = 360n^o\pm (90^o-2\alpha)
    Taking the + sign:
    54^o-\alpha = 360n^o+ (90^o-2\alpha)
    =360n^o+90-2\alpha
    \Rightarrow \alpha =360n^o+36^o
    Taking the - sign:
    54^o-\alpha = 360n^o- (90^o-2\alpha)
    =360n^o-90+2\alpha
    \Rightarrow -3\alpha =360n^o-144^o

    \Rightarrow \alpha = 120n^o+48^o (noting that this is now a 'new' n)
    So there's the general solution: \alpha = 360n^o+36^o or 120n^o+48^o, n = 0,\pm1,\pm2, ...

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    you are the best grandad!!!!!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: July 16th 2011, 03:31 PM
  2. Finding θ in r=1/32vēsin2θ?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 8th 2011, 04:14 PM
  3. How to find sinθ given sin2θ?
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: July 13th 2010, 09:55 PM
  4. [SOLVED] Solve for a?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 18th 2009, 01:28 PM
  5. [SOLVED] X=1 Solve
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 23rd 2009, 04:58 PM

Search Tags


/mathhelpforum @mathhelpforum