I've been at this for a while and I can't think clearly so I'm definitely doing something wrong.
The question:
Identify the symmetries of the curves in Exercises 1–12. Then sketch
the curves.$r = \sin (\frac{\theta}{2})$
In the book it states that first to find symmetry we have to check the following:
Symmetry about x-axis
$$( r ; – \theta )$$
$$( -r ; \pi-\theta )$$
Symmetry about y-axis
$$( -r ; -\theta )$$
$$( r ; \pi-\theta )$$
Symmetry about origin
$$( -r ; \theta )$$
$$( r ; \pi+\theta )$$
So I checked for the first one and this is what I got:
For $-\theta$:
$r = \sin (\frac{-\theta}{2})$
$-r = \sin (\frac{\theta}{2})$
Which turns out to satisfy the y-axis.
Now here's what I don't understand. I want to check for $\pi-\theta$, So I replace that where $\theta$ would be.
$r = \sin (\frac{\pi-\theta}{2})$
$r = \sin (\frac{\pi}{2} – \frac{\theta}{2})$
Following the formula for sin(A-B), then:
$r = \sin (\frac{\pi}{2})\cos(\frac{\theta}{2}) – \sin (\frac{\theta}{2})\cos(\frac{\pi}{2})$
$\cos(\frac{\pi}{2}) = 0$ and $\sin (\frac{\pi}{2}) = 1$ then:
$r = \cos(\frac{\theta}{2})$
Which obviously doesn't make sense. So I checked online and found that it should actually be $r = \sin (\pi – \frac{\theta}{2})$ and then that it would satisfy the symmetry about x-axis. But I checked it (I'm sure by now I've done something major wrong):
$r = \sin (\pi – \frac{\theta}{2})$
$r = \sin (\pi)\cos(\frac{\theta}{2}) – \sin(\frac{\theta}{2})\cos(\pi)$
$\cos(\pi) = -1$ and $\sin (\pi) = 0$ then:
$r = \sin (\frac{\theta}{2})$
Which satisfies y-axis and not x-axis.
So my question is this, why is it that wherever I look it should be
$r = \sin (\pi – \frac{\theta}{2})$ and not $r = \sin (\frac{\pi-\theta}{2})$ and what did I do wrong in the proofs and how can I make sure not to make the same mistakes again?
Thank you.
Best Answer
An alternative approach:
Let us first focus on the equation
$$\rho=\sin\theta$$ or, in Cartesian coordinates,
$$\rho^2=x^2+y^2=\rho\sin\theta=y.$$
This is the circle $$x^2+\left(y-\frac12\right)=1$$ with its center at $\left(0,\dfrac12\right)$ and radius $\dfrac12$.
Now if you consider
$$\rho=\sin\frac\theta2$$ this curve is a transformed version of the circle such that the polar angle "rotates twice as fast" and the circle is "unrolled".
The symmetry of the circle wrt. the axis $y$ becomes a symmetry wrt. $x$. As $\theta\in[0,\pi]$ and $\theta\in[\pi,2\pi]$ describes twice the circle, $\dfrac\theta2$ just needs to describe a full turn.