Results 1 to 2 of 2

Math Help - Please help with Trig equation sin(x) = -1/2

  1. #1
    hunisukl
    Guest

    Please help with Trig equation sin(x) = -1/2

    I can not figure out how to work this problem, can any one help???

    sin(x) = -1/2

    I would appreciate any help. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by hunisukl
    I can not figure out how to work this problem, can any one help???

    sin(x) = -1/2

    I would appreciate any help. Thanks!
    I think what is bothering you is that not that you can solve this equation for some x but to solve it for all x.

    You begin by finding one solution of,
    \sin x=-1/2
    You can do this by using arcsine, but you should know what this angle is,
    x=\frac{7\pi}{6}
    and
    x=\frac{11\pi}{6}
    These are all the non-coterminal angles, thus, all solutions are,
    x=\frac{7\pi}{6}+2\pi k
    x=\frac{11\pi}{6}+2\pi k

    The problem which many people have is finding what the angle has to be. Notice that if the problem was \sin x=1/2 then the angle is \frac{\pi}{6} you should have this memorized, but the problem asks for \sin x=-\frac{1}{2} thus, one such solution is, -\frac{\pi}{6}. Now find where this is located the on circle and in which quadrant. This is located in 4th quadrant producing angle of \frac{11\pi}{6}. How do you know what this angle is? Well, you now it is \frac{\pi}{6} backwards and the full circle has 2\pi thus if you subtract them properly you get \frac{11\pi}{6}. To find the other non-coterminal solution you now since it is negative it is in the 3rd quadrant. Thus, it is \frac{\pi}{6} below the \pi axis. Thus, a total of \frac{7\pi}{6} if you add them properly. Now since adding 2\pi or subtracting it does not change the sines values they are all solutions thus, this is what 2\pi is doing there.
    -------------------------
    There is a much easier way. Look at this:
    \sin x=a for -1\leq a\leq 1
    Then ALL the solutions for x are,
    x=(-1)^k\sin^{-1}(a)+\pi k
    Where \sin^{-1}(a) is the inverse sine function.

    Thus, in your problem you get,
    x=(-1)^k\sin^{-1}(-1/2)+\pi k
    But \sin (-1/2)=-\frac{\pi}{6}
    Thus,
    x=(-1)^k\frac{-\pi}{6}+\pi k=(-1)^{k+1}\frac{\pi}{6}+\pi k
    ----------
    You might be thinking then what is the difference between the first solution I gave and this one? The answer is there is no difference, the second one is a fancier way of expressing all the solutions for x. In fact, if were to write them out you will see you get the exact same thing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 08:00 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 08:07 AM
  3. Trig Equation with varied trig functions
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: April 12th 2010, 11:31 AM
  4. Replies: 1
    Last Post: July 24th 2009, 04:56 AM
  5. Replies: 1
    Last Post: July 24th 2009, 03:29 AM

Search Tags


/mathhelpforum @mathhelpforum