If Sin (value) returns 0, how do I know whether the angle was 0, 180 or 360?

Question

Obviously this cannot happen in a right rectangle, but otherwise - as Sin(0) or 180 or 360 equals 0, I guess there is no way to find out what the original angle was?

Answer 1

In general if $\sin x = M$ and $-180 < x \le 180$ degrees there are two possible answers to what $x$ is.

There is one $k$ where $0 \le k \le 90$ or $-90 \le k < 0$. That one is called $k=\arcsin M$.

Then there is another one $m$ where $x \ge 90$ or $k \le -90$. These are related in that $m = 180 - \arcsin M> 90$ if $0\le \arcsin M < 90$; or $m = -180 - \arcsin M$ if $-90\le \arcsin M < 0$.

(The only time these equal each other and there is only one answer is when $\pm 1$ and $\pm 180 -\arcsin M = \arcsin M = \pm 90$.)

However, it's easier to consider that as angles revolve every $360$ degrees and trig functions repeat with a period of $360$ degrees we think of angles $x$ and $x + 360$ and $x - 360$ and $x +360k$ for any integer $k$ are considered equivalent. (So $0^\circ$ and $360^\circ$ are considered to be the "same" thing. As are $... \equiv - 450^{\circ} \equiv-90^\circ \equiv 270^{\circ}\equiv 630^{\circ} \equiv ....$)

And if we have $\sin x = M$ we say there are two solutions $\arcsin M$ and $180 - \arcsin M$ where $\arcsin M$ is the value in the first and fourth quadrant and $180 - \arcsin M$ is the value in the second or third. And whether we refer to angles in the third and fourth quadrant as greater than $180$ or as negative is just a matter of convenience.

Also, once you start do any serious trig you should just stop doing angles and start doing radians.

$\pi \approx 180^\circ$; $\frac \pi 2 \approx 90^\circ$; $2\pi \approx 360^\circ$ etc.

And $x \equiv x + 2k\pi; k \in \mathbb Z$ and $-\frac \pi 2 \equiv \frac {3\pi} 2$.

So If $\sin x = 0$ then either $x = \arcsin 0 = 0$ or $x = \pi - \arcsin 0 = \pi - 0 =\pi$. So $x = 0$ or $ \pi$.

Answer 2

The general solution of the equation is as follows. Note that $\pi \text{ rad}=180^{\circ}$. $$\sin \theta=0\implies \theta=n\pi, n\in \mathbb{Z}=\{..., -\pi, 0,\pi,2\pi,...\}\equiv \theta=n\cdot180^{\circ}, n\in\mathbb{Z}$$

Answer 3

This is the inverse problem. The inverse sin function $\arcsin(x)$ is not one-to-one https://en.wikipedia.org/wiki/Injective_function as you pointed out. In general this can type of situation can make it hard to solve many problems in both math and physics. Usually only approximate methods can be used, and in this case because there is no extra information (e.g., $x$ was an angle between 90 and 200) there is no way to tell what value of $x$ generated the zero.

Answer 4

Caution:

The equation

$$\sin(x)=0$$

has infinitely many solutions, each an integer multiple of a half turn.

But

$$\arcsin(0)=0$$

and nothing else, because the arc sine is defined to be a function.