Question 730425: Given sec t = -4 and t is in quadrant 3, find the other 5 trig functions of t.
At first since we are dealing with sec t I took the trig functions formulas which are: Sin = y/r csc = r/y, cos = x/r sec = r/x, and tan = y/x cot = x/y I took sec = r/x and figured there would be a 1 under the -4 to hold the place so r = -4, x = 1 and to find y I took x^2 + y^2 = r^2 and got the square root of -17. but square roots cant be negative so i'm not sure if i am right or not. Can you help me? Also, I wasn't aloud to use a calculator for this problem.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Given sec t = -4 and t is in quadrant 3, find the other 5 trig functions of t.
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let O=opposite side
let A=adjacent side
let H=Hypotenuse
..
sec t=-4=H/A
H=4, A=-1
O=√H^2-A^2)=-√(16-1)=-√15
..
sin t=O/H=-√15/4
cos t=A/H=-1/4
tan t=O/A=-√15/-1=√15
csc t=H/O=-4/√15
sec t=H/-1=-4
cot t=A/O=-1/-√15=1/√15
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