Asked by Abbey
Test for symmetry with repsect to Q=pi/2, the polar axis, and the pole.
r=16cos 3Q
r=16cos(3pi/3 + 3h) = 16(cos3pi/2 3h- sin 3pi/2 sin h)
cos 3pi/2 = 0 sin 3pi/2 = -1
Not symmetrical
My teacher said that this was wrong!Why?
r=16cos 3Q
r=16cos(3pi/3 + 3h) = 16(cos3pi/2 3h- sin 3pi/2 sin h)
cos 3pi/2 = 0 sin 3pi/2 = -1
Not symmetrical
My teacher said that this was wrong!Why?