Results 1 to 4 of 4

Math Help - Proving an identity

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Proving an identity

    Hi all, this is my first post here, i need some help proving this identity
    (sinAsin2A +sin3Asin6A)/(sinAcis2A+sin3Acos6A) =tan5A
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Use these formulas:

    \sin a\sin b=\frac{1}{2}[\cos(a-b)-\cos(a+b)]

    \sin a\cos b=\frac{1}{2}[\sin(a+b)+\sin(a-b)]

    And the right hand member must be -\tan 5A
    Last edited by red_dog; July 1st 2009 at 10:11 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    527
    You'll have to use the product-to-sum formulas:

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

    So,
    \frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}
    \begin{aligned}<br />
&= \frac{\frac{\cos(A - 2A) - \cos(A + 2A)}{2} + \frac{\cos(3A - 6A) - \cos(3A + 6A)}{2}}{\frac{\sin(A + 2A) + \sin(A - 2A)}{2} + \frac{\sin(3A + 6A) + \sin(3A - 6A)}{2}} \\<br />
&= \frac{\cos(-A) - \cos 3A + \cos (-3A) - \cos 9A}{\sin 3A + \sin(-A) + \sin 9A + \sin(-3A)} \\<br />
&= \frac{\cos A - \cos 3A + \cos 3A - \cos 9A}{\sin 3A - \sin A + \sin 9A - \sin 3A} \\<br />
&= \frac{\cos A - \cos 9A}{\sin 9A - \sin A}<br />
\end{aligned}

    Then, use the sum-to-product formulas:
    \sin x \pm \sin y = 2\sin \left(\frac{x \pm y}{2}\right)\cos\left(\frac{x \mp y}{2}\right) AND
    \cos x - \cos y = -2\sin \left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right)

    \frac{\cos A - \cos 9A}{\sin 9A - \sin A}
    \begin{aligned}<br />
&= \frac{-2\sin \left(\frac{A + 9A}{2}\right)\sin\left(\frac{A - 9A}{2}\right)}{2\sin \left(\frac{A - 9A}{2}\right)\cos\left(\frac{A + 9A}{2}\right)}<br />
 \\<br />
&= \frac{-2\sin 5A \sin (-4A)}{2\sin (-4A) \cos 5A} \\<br />
&= \frac{-\sin 5A}{\cos 5A} \\<br />
&= -\tan 5A<br />
\end{aligned}

    Hmm, I'm getting -tan 5A...


    01


    EDIT: posted too late. red dog also got -tan 5A.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    In my book it DOES say that tan5A so i guess it would be a printing error because there is at least one other problem where the "-" sign is missing.
    but thanks anyway!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving an identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 17th 2011, 12:23 PM
  2. Proving Another Identity
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: June 6th 2011, 10:34 AM
  3. Help proving an identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 22nd 2010, 11:57 AM
  4. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 01:30 PM
  5. proving an identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 16th 2008, 10:03 AM

Search Tags


/mathhelpforum @mathhelpforum