SOLUTION: can someone help me with the following i'm kind of lost at what to do. thanks!! "Evaluate the following expressions:" a.) {{{cot(-pi/2)}}} b.) {{{sin^2(5pi/6)}}}...(note) that

Algebra ->  Trigonometry-basics -> SOLUTION: can someone help me with the following i'm kind of lost at what to do. thanks!! "Evaluate the following expressions:" a.) {{{cot(-pi/2)}}} b.) {{{sin^2(5pi/6)}}}...(note) that      Log On


   

Question 47978: can someone help me with the following i'm kind of lost at what to do. thanks!!
"Evaluate the following expressions:"
a.) cot%28-pi%2F2%29
b.) sin%5E2%285pi%2F6%29...(note) that is sin^2
c.) sec%5E2%2811pi%2F6%29...(note) that is sec^2
d.)sec%5E-1%28sec%28-30degrees%29%29 is finding out an angel first of all
i know that i have to use the unit circle but i dont know what to do.
any help would help thanks!!!
Answer by ChillyWiz282(3) About Me  (Show Source):
You can put this solution on YOUR website!
1. cot, cot%28-pi%2F2%29=%28cos%28-pi%2F2%29%2Fsin%28-pi%2F2%29%29, cos%28-pi%2F2%29=0, sin%28-pi%2F2%29=-1, then cot%28-pi%2F2%29=+0
2. sin%285pi%2F6%29 has a reference angle of pi%2F6} and in Quadrant 2. sin is negative in Quadrant 2. sin%28pi%2F6%29=1%2F2, so sin%285pi%2F6%29=-1%2F2.
Take the square of %28-1%2F2%29%5E2 gives us 1/4
3. Sec%2811pi%2F6%29 has a reference angle of pi%2F6 and in Quadrant 4. Sec is positive in Quadrant 4. sec%28pi%2F6%29=1%2Fcos%28pi%2F6%29 and cos%28pi%2F6%29=%28sqrt%283%29%29%2F2 then sec%28pi%2F6%29=2%2F%28sqrt%283%29%29
Squaring it gives %282%2F%28sqrt%283%29%29%29%5E2=4%2F3
4. sec%5E-1%28sec%28-30degrees%29%29=+30++or+-30+degree
sec%28-30degree%29=+2%2F%28sqrt%283%29%29
sec%5E-1%282%2F%28sqrt%283%29%29%29=+30++or+-30+degree
since in both Quadrant 1 and 4, sec(-30degree) is positive