Point P has a y-coordinate of 4/5 in quadrant II.
Can you help me find:
Sin t
cos t
tan t
sec t?
I thought you had to use x^2 + 4/5^2 = 1 to find x but I am not sure
+130511 If we have a unit circle......x = -sqrt (1 - (4/5)^2 ) = -sqrt ( 25/25 - 16/25) = - sqrt (9/25) = -3/5
[Remember, x is negative in the 2nd quadrant ]
So
sin(t ) = y/r = (4/5) / 1 = 4/5
cos (t) = x / r = (-3/5)/ 1 = -3/5
tan(t) = y/x = [4/5] / [-3/5] = [4/5] * [ -5/3] = -4/3
sec (t) = reciprocal of cos (t) = -5/3
