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

  1. #1
    hunisukl
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    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!
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  2. #2
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    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,
    $\displaystyle \sin x=-1/2$
    You can do this by using arcsine, but you should know what this angle is,
    $\displaystyle x=\frac{7\pi}{6}$
    and
    $\displaystyle x=\frac{11\pi}{6}$
    These are all the non-coterminal angles, thus, all solutions are,
    $\displaystyle x=\frac{7\pi}{6}+2\pi k$
    $\displaystyle x=\frac{11\pi}{6}+2\pi k$

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

    Thus, in your problem you get,
    $\displaystyle x=(-1)^k\sin^{-1}(-1/2)+\pi k$
    But $\displaystyle \sin (-1/2)=-\frac{\pi}{6}$
    Thus,
    $\displaystyle x=(-1)^k\frac{-\pi}{6}+\pi k=(-1)^{k+1}\frac{\pi}{6}+\pi k$
    ----------
    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 $\displaystyle x$. In fact, if were to write them out you will see you get the exact same thing.
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