Results 1 to 2 of 2

Math Help - Trigo surds question

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Singapore
    Posts
    4

    Trigo surds question

    (a) Prove that sin 3A = 3 sin A - 4 sin3 A.
    (b) If A = 36o, show that sin 3A = sin 2A.
    (c) Deduce that cos 36o = (1 + 51/2)/ 4.

    I've managed to solve (a) & (b), but I'm stuck at (c). I've figured out (c) has something to do with the fact that cos 36o = cos (63 - 27)o and that the angles 63o, 27o and 90o form a right angle, which matches the hypotenuse (51/2), adjacent (1) and opposite (2) somehow. I tried substitution of (a) and (b) into the working for (c) too, and I get the left hand side as 0.5(3 - 4 sin2 A) and the right hand side as 0.5(sin A - 1)/ cos A. Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Trigo surds question

    Using parts a) and b), we may write:

    3\sin A-4\sin^3A=2\sin A\cos A

    Since \sin A\ne0 we may divide through by this to get:

    3-4\sin^2A=2\cos A

    Now, using a Pythagorean identity, we have:

    3-4(1-\cos^2A)=2\cos A

    Now, arrange this as a quadratic in \cos A and take the positive root.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surds question 1
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 20th 2010, 09:30 AM
  2. Surds question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 18th 2010, 01:29 PM
  3. Help with question involving surds
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 14th 2010, 08:46 AM
  4. Simple surds question I can't work out!!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 4th 2010, 03:22 AM
  5. surds fraction question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 30th 2009, 07:14 AM

Search Tags


/mathhelpforum @mathhelpforum