Question

terminal point on the unit circle determined by t is the point (-1/3, (2sqrt2)/3), compute and simplify the following:

sin (pi - t)
i got (2sqrt2)/3

cos( pi + t)
-1/3

sec(-t)
3

cot (-pi - t)
sqrt2/4

tan (2pi - t)
2sqrt2

are my answers correct? and could someone show me how i'd graph either the cot or tan function?

Answer 1

tan = sine/cosine
cot = cosine/sine

Answer 2

Are you referring to graphing on a unit circle or plotting?

Answer 3

For these, you can only plot points

Answer 4

graphing on a unit circle like for ex if i have pi - t, i would show an arrow moving from (1, 0) counterclockwise to (-1, 0) for pi, and then going clockwise to wherever t is at. i know that tan = sin/cos and cot = cos/sin.

Answer 5

The only thing I've ever seen graphed on a unit circle are triangles and corresponding points on the circle. If you're going to do it, the (pi - t) part would be the angle. The unit circle has radius one.

Answer 6

well yeah the arrows are just to show how you got to that point, but what you said is true

Answer 7

i have to leave to my class, but once i get there if i have time i can get back on and continue. if you solve it all, thanks a lot : )

Answer 8

I believe (pi - t) would be in the second quadrant....but it could be in the first as well...depending on how long the length of t is

Answer 9

well, its a unit circle. it has got to be in the second quadrant, neither x or y of t are greater than 1
and we start from (-1, 0).

Answer 10

lol i'd appreciate it though if you could my answers. i have to go. ttyl and thanks

Answer 11

Yes, good job because x is negative.

Answer 12

It would look something like this:

Answer 13

and that angle from pi to t would be pi - t

Answer 14

Yes, that makes sense. Thx. Could you help me with either cot and tan?