So the exact value of Cos(-pi/3) is 1/2.

[riri]

Hello!
Simple as it sounds, I would greatly appreciate help on finding the exact value of Cos(\frac{-\pi}{3}
If I need to find a negative value of Cos, does it become the inverse?
So, since if I draw on unit circle, I get (\frac{\sqrt{3}}{2}, \frac{-1}{2}), would the value of Cos(-pi/3) = the y-value of -1/2?
Thank you!

 

[kaliprasad]

Hello!
Simple as it sounds, I would greatly appreciate help on finding the exact value of Cos(\frac{-\pi}{3}
If I need to find a negative value of Cos, does it become the inverse?
So, since if I draw on unit circle, I get (\frac{\sqrt{3}}{2}, \frac{-1}{2}), would the value of Cos(-pi/3) = the y-value of -1/2?
Thank you!

No. you should get $(\frac{1}{2}, \frac{-\sqrt{3}}{2})$ and you should take the x value as $\cos = \frac{x}{hypotenuse }$ and value of cos is x as hypotenuse is 1

 

[MarkFL]

Hello!
Simple as it sounds, I would greatly appreciate help on finding the exact value of Cos(\frac{-\pi}{3}
If I need to find a negative value of Cos, does it become the inverse?
So, since if I draw on unit circle, I get (\frac{\sqrt{3}}{2}, \frac{-1}{2}), would the value of Cos(-pi/3) = the y-value of -1/2?
Thank you!

You could also use the fact that cosine is an even function, that is $\cos(-\theta)=\cos(\theta)$ to write:

\(\displaystyle \cos\left(-\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\)