# Thread: Function

1. ## 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) $\displaystyle \frac{\pi}{6}$

2. 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.

3. Originally Posted by boy47
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?