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?
To understand why your answer may be incorrect, let's go through the process of testing for symmetry with respect to the given conditions.
In this case, we need to test for symmetry with respect to Q = pi/2, the polar axis, and the pole.
To test for symmetry with respect to Q = pi/2, we need to substitute (-Q) in place of Q in the equation and check if the resulting equation remains unchanged. So, let's replace Q with (-pi/2) in the equation r = 16cos(3Q):
r = 16cos(3(-pi/2))
r = 16cos(-3pi/2)
Now, we can simplify the above expression using the identity cos(-x) = cos(x):
r = 16cos(3pi/2)
Now, we compare this with the original equation r = 16cos(3Q). Since r = 16cos(3pi/2) does not match the original equation r = 16cos(3Q), we can conclude that the curve is not symmetric with respect to Q = pi/2.
To test for symmetry with respect to the polar axis, we need to substitute (pi - Q) in place of Q in the equation and check if the resulting equation remains unchanged. Let's replace Q with (pi - Q) in the equation r = 16cos(3Q):
r = 16cos(3(pi - Q))
r = 16cos(3pi - 3Q)
Using the cosine difference formula (cos(a - b) = cos(a)cos(b) + sin(a)sin(b)), we can simplify this expression:
r = 16(cos(3pi)cos(3Q) + sin(3pi)sin(3Q))
r = 16((-1)cos(3Q) + 0sin(3Q))
r = -16cos(3Q)
Now, we compare this with the original equation r = 16cos(3Q). Again, we can see that r = -16cos(3Q) does not match the original equation r = 16cos(3Q), indicating that the curve is not symmetric with respect to the polar axis.
Finally, to test for symmetry with respect to the pole (Q = 0), we need to replace (-Q) in place of Q in the equation and check if the resulting equation remains unchanged. Replacing Q with (-Q) in the equation r = 16cos(3Q):
r = 16cos(3(-Q))
r = 16cos(-3Q)
Using the cosine difference formula, we can simplify this expression:
r = 16(cos(0)cos(3Q) + sin(0)sin(3Q))
r = 16(1cos(3Q) + 0sin(3Q))
r = 16cos(3Q)
Now, we compare this with the original equation r = 16cos(3Q). We can see that r = 16cos(3Q) matches the original equation, indicating that the curve is symmetric with respect to the pole (Q = 0).
In summary, the curve is symmetric with respect to the pole (Q = 0), but it is not symmetric with respect to Q = pi/2 or the polar axis. If your teacher said your answer was wrong, it is possible that there was a mistake in the calculations or some misunderstanding of the concept of symmetry in polar coordinates. It may be helpful to double-check the calculations and clarify any doubts with your teacher.
I apologize for any confusion. The equation you mentioned, r = 16cos(3Q), describes a polar curve. To test symmetry, we need to check for symmetry with respect to the polar axis, the pole, and the given angle Q = pi/2.
1. Symmetry with respect to the polar axis: Substitute Q = -Q in the equation and see if the resulting equation remains the same. In this case, substituting Q = -Q gives r = 16cos(-3Q). Simplifying this equation, we get r = 16cos(3Q). Since the equation remains the same, the curve is symmetric with respect to the polar axis.
2. Symmetry with respect to the pole: Substitute Q = pi - Q in the equation and see if the resulting equation remains the same. In this case, substituting Q = pi - Q gives r = 16cos(3(pi - Q)). Simplifying this equation, we get r = 16cos(3pi - 3Q). Now, using the periodicity of cosine, we can rewrite this as r = 16cos(-3Q). Since the equation remains the same, the curve is symmetric with respect to the pole.
3. Symmetry with respect to Q = pi/2: To check for symmetry with respect to Q = pi/2, we need to substitute Q = pi/2 - Q in the equation. Substituting Q = pi/2 - Q gives r = 16cos(3(pi/2 - Q)). Simplifying this equation, we get r = 16cos(3pi/2 - 3Q). Now, using the periodicity of cosine, we can rewrite this as r = 16cos(3pi/2 + 3Q). Since the equation does not remain the same, the curve is not symmetric with respect to Q = pi/2.
Therefore, the correct conclusion is that the curve given by r = 16cos(3Q) is symmetric with respect to the polar axis and the pole, but not symmetric with respect to Q = pi/2.