Results 1 to 4 of 4

Math Help - sin4x=cos3x

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    77

    sin4x=cos3x

    How do I find all the solutions to sin4x=cos3x, [0, 2pi[ ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, weasley74!

    Find solutions to: . \sin4x\:=\:\cos3x\;\;\text{on }[0,\,2\pi]
    Code:
                        *
                      * *
                    * 3x*
               c  *     *
                *       * a
              *         *
            * 4x        *
          * * * * * * * *
                  b

    We see that: . \sin4x \:=\:\frac{a}{c} \:=\:\cos3x

    . . and that 4x and 3x are complementary.


    In general: . 4x + 3x \:=\:\frac{\pi}{2} + 2\pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{14} + \frac{2\pi}{7}n


    Therefore: . x \;=\;\frac{\pi}{14},\;\frac{5\pi}{14},\;\frac{9\pi  }{14},\;\frac{13\pi}{14},\;\frac{17\pi}{14},\;\fra  c{21\pi}{14},\;\frac{25\pi}{14}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Soroban View Post
    Hello, weasley74!

    Code:
                        *
                      * *
                    * 3x*
               c  *     *
                *       * a
              *         *
            * 4x        *
          * * * * * * * *
                  b

    We see that: . \sin4x \:=\:\frac{a}{c} \:=\:\cos3x

    . . and that 4x and 3x are complementary.


    In general: . 4x + 3x \:=\:\frac{\pi}{2} + 2\pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{14} + \frac{2\pi}{7}n


    Therefore: . x \;=\;\frac{\pi}{14},\;\frac{5\pi}{14},\;\frac{9\pi  }{14},\;\frac{13\pi}{14},\;\frac{17\pi}{14},\;\fra  c{21\pi}{14},\;\frac{25\pi}{14}

    There's one more, I'm sorry to say ......

    x = \frac{\pi}{2} + 2 n \pi.

    So, within the given domain, x = \frac{\pi}{2}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    There's one more, I'm sorry to say ......

    x = \frac{\pi}{2} + 2 n \pi.

    So, within the given domain, x = \frac{\pi}{2}.
    Another way of doing the question (which is how I picked up the missing solution):

    Since \sin A = \cos\left( \frac{\pi}{2} - A \right) (supplementary angle formula), the equation can be re-written as: \, \cos\left( \frac{\pi}{2} - 4x \right) = \cos(3x).

    Case 1: \, \frac{\pi}{2} - 4x = 3x + 2m \pi \Rightarrow 7x = \frac{\pi}{2} + 2n \pi (where n = -m) and you have the solutions given by Soroban. Note that these solutions include \frac{21 \pi}{14} = \frac{3 \pi}{2}.


    Case 2: Since \, \cos (-\theta) = \cos(\theta) it can also be said that

    - \left( \frac{\pi}{2} - 4x \right) = 3x + 2n \pi \Rightarrow \Rightarrow -\frac{\pi}{2} + 4x = 3x + 2n \pi  \Rightarrow x  = \frac{\pi}{2} + 2n \pi.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help working out a Taylor series of cos3x
    Posted in the Differential Geometry Forum
    Replies: 17
    Last Post: April 4th 2011, 10:38 AM
  2. Prove Cos3x=4(cos^3)x-3cosx
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: November 15th 2010, 11:57 AM
  3. integrate sinx/sin4x
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 13th 2010, 08:00 AM
  4. Trig. Identities (cos3x)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 19th 2010, 01:49 AM
  5. Trig Proof/Verification : cot 2x = (1+cos4x)/sin4x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 10th 2008, 08:25 PM

Search Tags


/mathhelpforum @mathhelpforum