Results 1 to 2 of 2

Math Help - Trig Question - Help Needed

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    8

    Trig Question - Help Needed

    If cos a/cos b + sin a/sin b=-1. Find the value of cos^3 b/cos a + sin^3 b/sin a.

    Thanks.
    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 dynamicsagar View Post
    If \cos a/\cos b + \sin a/\sin b=-1. Find the value of \cos^3 b/\cos a + \sin^3 b/\sin a.
    Stage 1. What does the given information \frac{\cos a}{\cos b} + \frac{\sin a}{\sin b}=-1 tell you? Multiply out the fractions and it becomes \cos a\sin b + \sin a\cos b = -\cos b\sin b. Using standard trig formulas for sines and cosines , you can write this as \sin(a+b) = -\frac12\sin(2b).

    Stage 2. Can you think of any examples where that condition is satisfied? One way to get \sin(a+b) = -\frac12\sin(2b) is if \sin(2b) = 1 and \sin(a+b) = -\frac12. That will occur for example if b = \pi/4 and a = -5\pi/12. Then \cos b = \sin b = 1/\sqrt2, and (see here) \cos a = \frac{\sqrt2}4(-1+\sqrt3),\ \,\sin a = \frac{\sqrt2}4(-1-\sqrt3). For that example, \dfrac{\cos^3 b}{\cos a} + \dfrac{\sin^3 b}{\sin a} = \dfrac1{2\sqrt2}\left(\dfrac4{\sqrt2(-1-\sqrt3)} + \dfrac4{\sqrt2(-1+\sqrt3)}\right), which simplifies (details omitted – do the calculation to check that it works) to 1.

    Stage 3. Make a wild guess. Could it always be true that \frac{\cos^3 b}{\cos a} + \frac{\sin^3 b}{\sin a} = 1 (given that \frac{\cos a}{\cos b} + \frac{\sin a}{\sin b}=-1)?

    Stage 4. See if you can prove that the guess in Stage 3 is correct. If it is, then multiplying out the fractions means that we want to show that \sin a\cos^3b + \cos a\sin^3b = \cos a\sin a. Let's see:

    \begin{aligned}\sin a\cos^3b + \cos a\sin^3b &= \sin a\cos b(1-\sin^2b) + \cos a\sin b(1-\cos^2b)\\ &= \sin a \cos b + \cos a\sin b - \sin b\cos b(\sin a\sin b + \cos a\cos b)\\ &= \sin(a+b) - \sin b\cos b\cos(a-b)\\ &= -\cos b\sin b + \sin(a+b)\cos(a-b)\quad\text{\footnotesize (using the given information)}\\ &= \ldots\quad\text{\footnotesize (can you complete the proof from there?)} \end{aligned}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help Needed with this Trig Equation!
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 6th 2011, 06:02 PM
  2. trig help needed
    Posted in the Trigonometry Forum
    Replies: 13
    Last Post: January 10th 2010, 10:22 PM
  3. Trig help needed
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 24th 2008, 07:12 PM
  4. Replies: 4
    Last Post: November 30th 2007, 09:52 AM
  5. Trig help needed
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 27th 2007, 10:56 AM

Search Tags


/mathhelpforum @mathhelpforum