Results 1 to 3 of 3

Math Help - Hence what? rsin(x+alpha) and solving a trig equation.

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    21

    Hence what? rsin(x+alpha) and solving a trig equation.

    Hiya, either I'm missing something or my book is being weird:
    Q) Express 4\sin x + 3\cos x in the form r\sin(x+\alpha).
    Hence find all the values of x in the range 0\le x \le 360^\circ for which \cos 3x=\cos 2x
    Q END)

    I can get the r\sin(x+\alpha) form of 4\sin x + 3\cos x:
    5\sin(x+36.9^\circ), but I don't see how that helps me with the second part of the question.

    I think I can do the second part of the question anyway. By using \cos 3x= \cos(2x + x) and then using substitutions to change all 2x into single x and to get rid of all the sinx that appear, I get:
    4\cos^3 x -2\cos^2 x -3\cos x + 1 = 0
    Then subsituting y=\cos x and solving for y gives (y-1)(4y^2 +2y-1) \Rightarrow y=1, y=\frac{-1\pm\sqrt{5}}{4}
    \Rightarrow x=0^\circ,360^\circ,72^\circ,288^\circ,144^\circ,2  16^\circ

    But the book gives as an answer: 119.6^\circ and 346.7^\circ
    I made the graph and I think it backs up my answer, the brown line has roots where the blue and green lines cross. So why on earth do they say 'hence' as if the first part of the question should help solve the second part? Is the question just fubared or am I missing something(s)?
    Thanks.
    Attached Thumbnails Attached Thumbnails Hence what? rsin(x+alpha) and solving a trig equation.-graph1.1.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello tleave2000
    Quote Originally Posted by tleave2000 View Post
    Hiya, either I'm missing something or my book is being weird:
    Q) Express 4\sin x + 3\cos x in the form r\sin(x+\alpha).
    Hence find all the values of x in the range 0\le x \le 360^\circ for which \cos 3x=\cos 2x
    Q END)

    I can get the r\sin(x+\alpha) form of 4\sin x + 3\cos x:
    5\sin(x+36.9^\circ), but I don't see how that helps me with the second part of the question.
    I must say, I don't see that either. But the easiest way to solve \cos3x =\cos2x is as follows:

    The general solution of \cos A = \cos B is

    A = 2n\pi \pm B, n = 0, \pm1, \pm2, ...

    (Check out the graph of y = \cos x if you want to confirm this.)

    So \cos3x =\cos2x \Rightarrow 3x = 2n\pi \pm 2x

    Taking the plus sign on the RHS gives x = 2n\pi, and taking the minus sign gives 5x = 2n\pi; i.e. x = \tfrac25n\pi

    So in the range 0\le x \le 2\pi, the solutions are:

    x = 0, \tfrac25\pi, \tfrac45\pi, \tfrac65\pi, \tfrac85\pi, 2\pi

    Or, in degrees, the solutions you found: 0^o,72^o,144^o,216^o,288^o,360^o

    The book's answers are clearly wrong: they just don't give \cos3x = \cos2x. So, I'm on your side here! I suspect someone has got some wires crossed somewhere.

    Grandad
    Last edited by Grandad; May 20th 2009 at 12:54 AM. Reason: Correct typos
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    21
    Thanks alot. Interesting to use a general solution that way, hadn't seen that before.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  2. Replies: 2
    Last Post: February 9th 2011, 06:12 AM
  3. Solving trig equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 20th 2009, 07:54 AM
  4. Replies: 3
    Last Post: May 15th 2009, 07:50 AM
  5. Replies: 4
    Last Post: April 6th 2009, 12:43 PM

Search Tags


/mathhelpforum @mathhelpforum