91Ó°ÊÓ

We test for symmetry about (a) x-axis, (b) y-axis, and (c) Origin (Polar Axis). - (a) For symmetry in x-axis, replace \( \theta \) by \(- \theta \) and see if we obtain the same equation. Here it's not true because \( \sin(- \theta) \neq \sin(\theta) \) and \( \cos(-\theta) = \cos(\theta) \). So it's not symmetric about the x-axis (Polar Axis).- (b) For symmetry in y-axis, replace \( \theta \) by \( \pi - \theta \). We get \( r= \sin(\pi - \theta) + \cos(\pi - \theta) \), which simplifies to \( r = \sin(\theta) - \cos(\theta) \). Hence, it's not the same as the given equation, and it's not symmetric about the y-axis.- (c) For symmetry about origin, replace \( r \) by \( -r \) and \( \theta \) by \( \pi + \theta \). The equation becomes \( -r = \sin(\pi + \theta) + \cos(\pi + \theta) \), which simplifies to \( -r = -\sin(\theta) - \cos(\theta) \), again not equivalent to the initial equation, so it's not symmetric about the origin.

Once we have confirmed the lack of symmetry, we can proceed to draw the graph. To do this, we need to replace \( \theta \) with various angles and calculate the resultant \( r \). It's easier to start at \( \theta = 0 \) and work up to \( \theta = 2\pi \) in increments of \( \frac{\pi}{4} \). This will give a clear picture of the graph and the amount of detail needed.