Precalculus Someone please help
posted by Josh .
Evaluate all six trigonometric functions at t, where the given point lies on the terminal side of an angle of t radians in standard position. (40/41, -9/41)
draw the triangle, but first, scale it (that wont change the angle). (40,-9) works. That makes the hypotenuse sqrt(40^2+9^2)-=41
so you have a triangle, in the fourth quadrant sides 40, -9, with a hypotenuse of 41
You take it from here.
You know you are in quadrant IV
by CAST, the cosine, (and the secant), are positive in that quadrant. All other trig ratios are negative.
You also know you are dealing with a 9-40-41 right-angled triangle, since 9^2 + 40^2 = 41^2.
so x=40, y = -9 , and r = 41
Now you can state the other 5 trig ratios.
e.g. tan t = -9/41 (because tanØ = y/x, etc)
e.g. tan t = -9/40