Finding the value of trigonometric functions

This is probably one of the easiest concepts but I do not get it, so I am going to give the two problems that are giving me the most trouble on my very long worksheet I have to do, maybe you guys can help me understand how to do it all.

1.) $\tan(-3π/2)$;

2.) $\cos(17π/3)$

Steps and explanations would be very helpful, thanks and please no judging on my lack of knowledge in math! I said this in the comments below but thought I would put it up here too. My teacher has us writing it out like this, for example, tan(-3π/2)= -y/x=____. We have to write the problem, show that we know the definition (hence the y/x part) and then look on our trig wheel and fill in the x and y coordinates. I don't know why I am confused, I guess I am just having trouble locating it on the trig wheel, but if you know how to work it this way can anyone show me step by step?

• Do you have memorized some very basic values of sin and cos? Say, for $0$, $\pi/6$, $\pi/4$, $\pi/3$, and $\pi/2$? Usually you can reduce most questions you will be asked to those values using some simple trig identities.
– Seth
Commented Apr 15, 2014 at 22:54
• In particular for these problems use the definition of tan and periodicity of cos and sin
– Seth
Commented Apr 15, 2014 at 22:56
• My teacher handed me the trig wheel today so I am in the process of memorizing it all. Commented Apr 15, 2014 at 22:56
• The important thing, beyond @Seth’s suggestion, is to remember that in view of the periodicity of $\tan$ and $cos$, you can add or subtract multiples of $\pi$ from the argument of $\tan$ without changing the value, similarly, you can add or subtract multiples of $2\pi$ from the argument of $\cos$. Commented Apr 15, 2014 at 22:56

Remember that sin and cos are $2\pi$ periodic. Also remember that cos is even and sin is odd.
$\tan(-3\pi/2)=\frac{\sin(-3\pi/2)}{\cos(-3\pi/2)}=\frac{\sin(\pi/2)}{\cos(\pi/2)}$
$\cos(17\pi/3)=\cos(-\pi/3)=\cos(\pi/3)$