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Thread: Function

  1. #1
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    Function

    Is: $\displaystyle g:[-1,1]=>[0,\pi]$ defined by: $\displaystyle g(x)=arccos(x)$ The value of $\displaystyle g=-\frac{1}{2}$ is:

    a) $\displaystyle \frac{2\pi}{3}$
    b) $\displaystyle -\frac{2\pi}{3}$
    c) $\displaystyle \frac{\pi}{3}$
    d) $\displaystyle -\frac{\pi}{3}$
    e) $\displaystyle \frac{\pi}{6}$
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  2. #2
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    You're question, from what I can gather, asks: What is the value of the cosine function in radians at -1/2 if the function has a range of [-1, 1] and a domain of 0 to Pi.

    Take a look at the image below, Pi/2 occurs when y=0, and pi occurs when y= -1. Since we know that the point y=-1/2 lies between pi/2 and pi it cannot be pi/6 or pi/3. And radians are not negative since the domain is not between 0 and -pi. Therefore the answer has to be 2pi/3.

    If you're in doubt try cos of 120 or cos of 2pi/3 in your calculator, you'll get the answer -0.5 or -1/2.

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  3. #3
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    Quote Originally Posted by boy47 View Post
    Is: $\displaystyle g:[-1,1]=>[0,\pi]$ defined by: $\displaystyle g(x)=arccos(x)$ The value of $\displaystyle g=-\frac{1}{2}$ is:
    a) $\displaystyle \frac{2\pi}{3}$
    b) $\displaystyle -\frac{2\pi}{3}$
    c) $\displaystyle \frac{\pi}{3}$
    d) $\displaystyle -\frac{\pi}{3}$
    e) $\displaystyle \frac{\pi}{6}$
    Your question asks: $\displaystyle \cos (?) = - \frac{1}{2}$.
    The answer to that question is: $\displaystyle \frac{{2\pi }}{3}\,\& \, - \frac{{2\pi }}{3}$.
    BUT the domain of the arccosine function is $\displaystyle \left[ {0,\pi } \right]$.
    So what is the correct answer?
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