ahhhhh!!! MATH!! please help!!!!?

Question

i'm trying to evaluate the trigonometric function of the quadrant angle sec(pie). i really just dont understand this whole concept of using reference angles and such. can anyone please show me how to do this problem and explain how it is done. please help as i have a quiz on this stuff tomorrow and i really just dont understand. i would be extremely grateful :). thanx!

actually that was the wrong problem sorry :( but thank you all so much. here is the actual problem. cot pi/2

Answer 1

sec pi
sec(pi-0)
-sec0
=1/cos0
=-1

Answer 2

First you need to understand radians. If you have a unit circle (radius = 1), then the diameter will be 2pi (not pie). You always start at 0 degrees (this is like the 3 o' clock position on a clock). You measure angles going counterclockwise. 90 degrees (straight up) is equivalent to pi/2. 180 degrees (9 o' clock) is pi. 270 degrees (6 o'clock) is 3/2 pi. Finally 360 degrees (back to 3 o'clock) is 2 pi.

0 to pi/2 is the first quadrant.
pi/2 to pi is the second quadrant
pi to 3/2 pi is the third quadrant
3/2pi to 2 pi is the fourth quadrant.

Now, secant is the same as 1/cos(angle). For any angle, sin(angle) is the distance up or down (from the axis) and cos(angle) is the distance left or right (from the axis).

If you go to the edge of the circle, and go around 180 degrees (pi) you are at the 9 o'clock position. sin(pi) is 0 (right on the axis). cos(pi) is -1 (1 unit left from the axis).

And 1/cos(pi) is 1/-1 = -1

So sec(pi) is -1.