Is $f(\theta) = 1 - \alpha \sin{\theta}$ an odd function?

Consider equations in polar coordinates of the form

$$r = f(\theta) = 1 - \alpha \sin{\theta}$$

When I plot a few of these polar functions, I always get a graph that is symmetric relative to the $y$ axis, indicating that $f(\theta)$ is in fact odd.

Is $f(\theta)$ an odd function?

$$f(-\theta)=1-\alpha\sin{(-\theta)}$$

Since $\sin{(-\theta)}=-\sin{(\theta)}$

$$f(-\theta)=1+\alpha\sin{(\theta)}$$

$$-f(-\theta)=-1-\alpha\sin{(\theta)}=-(1+\alpha\sin{(\theta)}) \neq 1-\alpha\sin{(\theta)}=f(\theta)$$

Is $f(\theta)$ odd, in which case how do I show this? Or is there some edge case where it's not odd?

No, the function $f$ is not odd. But, yes, the image of the curve $\theta\mapsto c(\theta)=f(\theta)(\cos\theta,\sin\theta)$ is symmetric with respect to the $y$-axis. That's so because$$c(\pi-\theta)=f(\theta)(-\cos\theta,\sin\theta),$$which is the reflection of $c(\theta)$ on the $y$-axis. So, when $\theta$ goes from $0$ to $\pi$, for every point $(x,y)$ from its range, $(-x,y)$ also belongs to its range. And the same thing occurs on $[\pi,2\pi]$.

Odd functions: $f(-x)=-f(x)$

Let's try this odd function statement with yours: Assuming function $f$ is odd, $$f(-x)=-f(x)\\ f(-\theta)=-[1+\alpha \sin(\theta)]\\ [1+\alpha \sin(-\theta)]=-[1+\alpha \sin(\theta)]\\ 1-\alpha \sin(\theta) =-1-\alpha \sin(\theta) $$

Adding $\alpha \sin(\theta)$ to either side, we get: $1 =-1$. Now we know that $1 \neq -1$, therefore $f(\theta)=1+\alpha \sin(\theta)$ is not an odd function.