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?

It looks like there was a mistake in your calculations. Let's go step by step to determine the symmetry of the given equation.

The equation is r = 16cos(3Q).

1. Symmetry with respect to Q = pi/2:
To test the symmetry with respect to Q = pi/2, we need to replace Q with pi/2 - Q in the equation.

r = 16cos(3(pi/2 - Q))
= 16cos((3pi/2) - 3Q)
= 16cos(3pi/2)cos(-3Q) + 16sin(3pi/2)sin(-3Q)
= 0cos(-3Q) + (-16)sin(-3Q)
= -16sin(-3Q)
= -16sin(3Q)

Since the resulting equation is not equal to the original equation r = 16cos(3Q), it is not symmetric with respect to Q = pi/2.

2. Symmetry with respect to the polar axis (Q = pi or Q = -pi):
To test the symmetry with respect to the polar axis, we need to replace Q with -Q in the equation.

r = 16cos(3(-Q))
= 16cos(-3Q)

Since the resulting equation is not equal to the original equation r = 16cos(3Q), it is not symmetric with respect to the polar axis.

3. Symmetry with respect to the pole (Q = 0 or Q = 2pi):
To test the symmetry with respect to the pole, we need to replace Q with pi - Q in the equation.

r = 16cos(3(pi - Q))
= 16cos(3pi - 3Q)
= 16cos(3pi)cos(-3Q) + 16sin(3pi)sin(-3Q)
= 16cos(-3Q)
= 16cos(3Q)

Since the resulting equation is equal to the original equation r = 16cos(3Q), it is symmetric with respect to the pole.

Therefore, the correct answer is that the equation is symmetric with respect to the pole, but not symmetric with respect to Q = pi/2 or the polar axis.