Question 889604: the values for x(3pi/2) and y(3pi/2) for x(t)=cos(t),y(t)=sin(t) are
A)(0,-1)
B)(.5, sqrt(3)/2)
C)(-.5,0)
D)(-.5,-sqrt(3)/2)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe your solution is selection A which is (0,-1)
3pi/2 radians * 180 / pi = 3*180/2 degrees = 270 degrees.
cos(3pi/2) = cos(270) = 0
sin(3pi/2) = sin(270) = -1
the reference angle for 270 is equal to 270 - 180 if you assume quadrant 3, or 360 - 270 if you assume quadrant 4.
since 270 is on the border between quadrant 3 and quadrant 4, it can be assumed to be in either one, both of which get you the same answer.
so the reference angle is 270 - 180 = 90, or the reference angle is 360 - 270 = 90.
the reference angle is 90 degrees.
the cosine of 90 degrees is 0
the cosine of 90 degrees is 1
that would be in quadrant 1.
in quadrant 3, the cosine is negative which doesn't matter since the cosine is 0.
in quadrant 4, the cosine is positive which doesn't matter since the cosine is 0.
in quadrant 3, the sine is negative.
in quadrant 4, the sine is negative.
either way, you have:
cos(270) = 0
sin(270) = -1
that's selection A.
you are dealing with functional notation.
x(t) = cos(t)
y(t) = sin(t)
x(3pi/2) becomes cos(3pi/2)
y(2pi/2) becomes sin(3pi/2)
in both cases, you replace t with 3pi/2.
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