Identify the symmetries and sketch the curve $r=\sin (\theta/2)$

Question

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.

Answer 1

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.

Answer 2

Your computation is correct, but your conclusion is wrong. The equation has failed the symmetry test, but that does not mean that the graph is not symmetric with respect to x-axis. As a matter of fact if passing a symmetry test verifies that symmetry will be exhibited in a graph, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric. In others words, the graph of a polar equation can be symmetric with respect to one of these axes (or the pole) and not satisfy any of the test equations you wrote. This is because a graph can have many polar representations, so many tests are possible. There are many different ways of specifying a point in polar coordinates. You can realize this fact by thinking that every point $(r,\theta)$ could also be called $(r,\theta+2n \pi)$, with $n$ any integer. This fact can affect a symmetry test, even if I think it is possible to define alternative robust symmetry tests.

Answer 3

Plotting is the real check where all three these rules are verifiable.

In this (cardioid) curve it has same radius $r$ for $\pm \theta$ and that makes the polar curve symmetrical with respect to x-axis.

Answer 4

To check the symmetry about both axes and origin:
About X-axis
(r,θ): r = sin(θ/2) = ± { 1 – cos (θ)/ 2 }^1/2 (Half-angle property)
(r,-θ) : r = ± { 1 – cos (-θ) / 2 }^1/2
r = ± { 1 – cos (θ) / 2 }^1/2
Therefore, (r,θ) = (r,-θ)
The curve is symmetrical about X-axis
About Y-axis
(r,θ) : r = sin (θ/2)
= sin (θ/2)
= ± {1 – cos (θ)/ 2 }^1/2
(-r,-θ) : -r = ± {1 – cos (-θ)/2}^1/2
-r= ± {1 – cos (θ)/2 }^1/2
Therefore, (r,θ) ≠ (-r,-θ)
(r , π-θ ) : r = ± {1 – cos(π-θ)/2}^1/2

r = ± {1 + cos (θ) /2}^1/2

Therefore , (r,θ) ≠ (r , π-θ )
The curve isn’t symmetrical about Y-axis
About origin
(r,θ) ; r = ± { 1 - cos (θ) /2}^1/2
(r,π +θ) ; r = ±{1 – cos (π +θ)/2}^1/2
r = ± {1 – cos (θ)/2 }^1/2
Therefore, (r,θ) = (r, π +θ)
Thus the curve is symmetrical about origin.
hope it helps