91Ó°ÊÓ

To test the polar equation for symmetry, we need to use the negative angle identity of cosine, which tells us that \( \cos(-\theta) = \cos(\theta) \). Let's substitute \( -\theta \) for \( \theta \) in the equation and check if the function is symmetric with respect to the origin. The function would be symmetric if after substitution, we can manipulate the equation to have the same form as the original one.When we do this substitution in the equation \( r = \frac{2}{1 - \cos \theta} \), it transforms into \( r = \frac{2}{1 - \cos(-\theta)} \).Since \( \cos(- \theta) = \cos(\theta) \), the equation remains the same. So, this indicates that the given polar equation is symmetric with respect to the origin.

To plot a polar function, we will convert polar coordinates to Cartesian ones using the relations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). In this case, substituting \( r = \frac{2}{1 - \cos \theta} \) into the Cartesian relations, we obtain: \( x = \frac{2 \cos \theta}{1 - \cos \theta} \) and \( y = \frac{2 \sin \theta}{1 - \cos \theta} \).Since the theta values range from 0 to \( 2\pi \), we can plot points (x,y) for these theta values and then connect these points to get the graph of the function. Remember, as the function is symmetric with respect to the origin, only need to graph for the interval \( 0 \leq \theta \leq \pi \) and the second half of the graph will be a reflection of the first half across the origin.