# Test for symmetry with respect to theta = pi/2, the polar axis, and the pole? r=16cos3thetar=6sin(theta) please help!!

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### 1 Answer

r=16cos3theta

r=6sintheta

1)To test for symmetry with respect to theta= pi/2:

replace theta with pi-theta,

r=16cos3theta

r= 16cos3(pi-theta)= 16cos(3pi-3theta) = -16cos3theta

thus we get r= -16cos3theta,we do not get the same curve.

thus r= 16cos3theta is not symmetric about the line theta= pi/2

similarly let us check for symmetry for the curve r= 6sintheta:

replace theta with pi-theta,

r= 6sin(pi-theta)= 6sintheta, thus we get r= 6sintheta,

therefore the curve is symmetric about the line theta= pi/2

2) To test for symmetry about polar axis:

replace theta with -theta

r= 16cos3(pi-(-theta)) = 16cos(3pi+3theta) =16. (-1)^3cos3theta = -16cos3theta

thus we get r= -16cos3theta, we do not get the same curve,

therefore r= 16cos3theta is not symmetric about the polar axis

replace theta with -theta in r= 6sintheta

we get r= 6sin(-theta) = -6sintheta

r= -6sintheta, we do not get the same curve

therefore r=6sintheta is not symmetric about the polar axis

3) To test for symmetry about the pole:

replace r with -r in both the curves

we get -r= 16cos3theta,

-r = 6sintheta,

we do not get the same curves in both the cases hence they are not symmetric about the pole.