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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.

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  • $\begingroup$ Please use \sin instead of sin in math mode. $\endgroup$
    – MrYouMath
    May 5, 2018 at 13:41
  • $\begingroup$ @MrYouMath I didn't know that! Thank you. Edited. :) $\endgroup$
    – Kode Ch
    May 5, 2018 at 13:52
  • $\begingroup$ so it won't be symmetrical about x-axis. You got it right. $\endgroup$
    – Vasili
    May 5, 2018 at 13:54
  • $\begingroup$ @Vasya But the correction says that it is symmetric on all three (if there are 2 symmetric then the third is there) $\endgroup$
    – Kode Ch
    May 5, 2018 at 13:56

4 Answers 4

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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".

enter image description here

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.

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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.

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  • $\begingroup$ Thank you for explaining it to me, but I do have to ask, our teacher told us that we find the symmetries directly from these three tests, and that would help us set an easier bound to work with (e.g. if it is symmetric on all then we work on $[0; \pi/2]$ and mirror everything else). So if I don't know from the test when it's symmetric, then I can't really draw it properly... $\endgroup$
    – Kode Ch
    May 8, 2018 at 12:14
  • $\begingroup$ If the test is successful you can conclude that the graph is symmetric, however if the test fails you cannot draw any conclusion. In this case you have to plot the graph to see if it is symmetric. $\endgroup$
    – Upax
    May 8, 2018 at 19:23
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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.

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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

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  • $\begingroup$ hi. You should format your answer with LaTeX. It is not hard, and it makes things much more readable. $\endgroup$
    – rschwieb
    Jan 19, 2021 at 22:03

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