# Another trig problem

## Homework Statement

Evaluate the following exactly.

cos 300°
sin(-120°)

Unit circle

## The Attempt at a Solution

I seem to have a problem with the remembering if it is going to be the position on the unit circle such as in these cases I thought it was
cos 300°= 5∏/3 but it is actually I think 1/2 so I got the problem wrong
then you have
sin(-120°) = 4∏/3 but this is wrong and I think it is -√3 / 2 so yeah then I got this one right
csc120°= 2√3 / 3
I just never know when I am supposed to use which one for the exact value. Could someone help out please?

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Mark44
Mentor

## Homework Statement

Evaluate the following exactly.

cos 300°
sin(-120°)

Unit circle

## The Attempt at a Solution

I seem to have a problem with the remembering if it is going to be the position on the unit circle such as in these cases I thought it was
cos 300°= 5∏/3 but it is actually I think 1/2 so I got the problem wrong
You're really garbling up a bunch of different stuff. cos 300° ≠ 5∏/3, but 300° = 5∏/3. The reference angle for 5∏/3 is -∏/3, and this angle has the same cosine as ∏/3.

Writing cos 300°= 5∏/3 is incorrect, but cos(300°)= cos(5∏/3) would be correct. Be mindful that the cosine of something is not the same as the something.
then you have
sin(-120°) = 4∏/3 but this is wrong and I think it is -√3 / 2 so yeah then I got this one right
csc120°= 2√3 / 3
I just never know when I am supposed to use which one for the exact value. Could someone help out please?

So then why is arcsin(-1/2) = -Pi/6 Is the inverse just going back to the reference angle then?

HallsofIvy
Yes, because the trig functions are periodic, they are not "one to one" and so do not have true "inverses". In order to talk about inverse for the trig functions, we have to restrict their domains. The standard convention for "sine" is to restrict to between $-\pi/2$ (-90 degrees) and $\pi/2$ (90 degrees). If x is positive, $sin^{-1}(x)$ is between 0 and $\pi/2$. If x is negative, $sin^{-1}(x)$ is between $-\pi/2$ and 0.