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?
 A: 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 $[\pi,2\pi]$.
A: 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.
A: $f$ cannot be an odd function. In fact, $f(-\theta)=1+\alpha sin(\theta)=\alpha sin(\theta)-1=-f(\theta)$ if and only if $1=-1$.
A: Consider the function $f(x) = 1 - \alpha \sin{(x)}$ in Cartesian coordinates.
Though $\alpha\sin{(x)}$ is an odd function, $1-\alpha\sin{(x)}$ is not, as shown by Mateus Figueiredo's answer.
Note also that shifting $\alpha\sin{(x)}$ to the left by $\frac{\pi}{2}$ makes it equal to $\cos{(x)}$, which is an even function.
Now let's consider the same function, but representing a polar coordinate
$$r=f(\theta)=1-\alpha\sin{(\theta)}$$
In Cartesian coordinates, the condition for symmetry about the y-axis was $f(x)=-f(-x)$.
In polar coordinates I am not quite sure what the general condition is, but here are two ways that a polar function $f(\theta)$ implies a graph that is symmetric relative to the y-axis:

*

*$f(\frac{\pi}{2}+\theta)$ is even, as noted by Arturo Magidin.


*$f$ is odd
Note the following

*

*Given a point in polar coordinates $(r,\theta)=(f(\theta),\theta)$ what can we say about $(-r,\theta)$?

$$(-r,\theta)=(r,\theta+\pi)=(f(\theta),\theta+\pi)\tag{1}$$
Proof of $1$
Let $\theta_1>0$ be a constant.
The lines $\theta = \frac{\pi}{2} + \theta_1$ and $\theta=\frac{\pi}{2}-\theta_1$ are symmetric relative to the y-axis. A given $r$ polar coordinate specifies the same $y$ coordinate on both lines. The corresponding pair of points, one on each line, with this $y$ coordinate are symmetric relative to the y axis.
$$y = r \sin{\left(\frac{\pi}{2} \pm \theta_1\right )}=r\sin{\left(\frac{\pi}{2} + \theta_1\right )}$$
$$x=r \cos{\left(\frac{\pi}{2} \pm \theta_1\right )}=\pm r \cos{\left(\frac{\pi}{2} + \theta_1\right )} $$
By assumption, $f(\frac{\pi}{2}+\theta)$ is even so
$$r=f(\frac{\pi}{2}+\theta)=f(\frac{\pi}{2}-\theta)$$
And the points
$$(f(\frac{\pi}{2}+\theta),\frac{\pi}{2}+\theta)=(r, \frac{\pi}{2}+\theta)$$
$$(f(\frac{\pi}{2}-\theta),\frac{\pi}{2}-\theta)=(r, \frac{\pi}{2}-\theta)$$
are symmetric because they are each on one of the lines $\theta=\frac{\pi}{2}\pm \theta_1$, and have the same $r$ (and also $y$) coordinate.
Proof of $f$ odd $\implies$ for every polar point $\left ( f(\frac{\pi}{2}+\theta), \frac{\pi}{2} +\theta \right )$ on the graph of $f$ there is another point on the graph of $f$, $\left ( f(\frac{\pi}{2}+\theta), \frac{\pi}{2} -\theta \right )$, symmetric to the first point relative to the $y$ axis.
Consider polar point
$$\left (f(-\frac{\pi}{2}-\theta), -\frac{\pi}{2} -\theta \right )$$
use $f$ odd
$$=\left (-f(\frac{\pi}{2}+\theta), -\frac{\pi}{2} -\theta \right )$$
use $(1)$
$$=\left (f(\frac{\pi}{2}+\theta), \frac{\pi}{2} -\theta \right )$$
$\implies$ $\left (f(\frac{\pi}{2}+\theta), \frac{\pi}{2} -\theta \right )$ is on the graph of $r=f(\theta)$.
But so is $\left (f(\frac{\pi}{2}+\theta), \frac{\pi}{2} +\theta \right )$
These two points are symmetric relative to the $y$ axis.
Now, considering the original question, the function $f(\theta)=1-\alpha\sin{(\theta)}$ is not odd, but condition $1$ above is satisfied:
$$f(\frac{\pi}{2} + \theta)=1-\alpha\sin{(\frac{\pi}{2}+\theta)}=1-\alpha\sin{(\frac{\pi}{2}-\theta)}=f(\frac{\pi}{2} - \theta)$$
Therefore, the graphs of $f(\theta)=1-\alpha\sin{(\theta)}$ are symmetric relative to the $y$ axis.
