Is there a simpler/better proof of this simple trigonometric property?

Question

The sine function has the following nice property : for any $x,y$, we have $\sin(x)+\sin(y)=\sin(x+y)$ iff at least one of $x,y,x+y$ is $0$ modulo $2\pi$.

I sketch below my current proof of it, which I find somewhat unsatisfying. Does anyone know a better proof ?

Let $\xi=\sin(x)$ and $\eta=\sin(y)$. Then $\sin(x+y)=\xi\cos(y)+\eta\cos(x)$, so

$$ \begin{array}{lcl} \sin^2(x+y)&=&\xi^2(1-\eta^2)+\eta^2(1-\xi^2)+2\xi\eta\cos(x)\cos(y) \\ &=& \xi^2+\eta^2-2\eta^2\xi^2+2\xi\eta\cos(x)\cos(y) \end{array} \tag{1}$$

Whence

$$ \big(\sin^2(x+y)-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2= 4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \tag{2} $$

If $\sin(x)+\sin(y)=\sin(x+y)$, it follows then that $A(\xi,\eta)=0$ where

$$ \begin{array}{lcl} A(\xi,\eta)&=&\big((\xi+\eta)^2-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2- 4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\ &=& \big(2\xi\eta-2\eta^2\xi^2\big)^2- 4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\ &=& \big(2\xi\eta\big(1-\eta\xi\big)\big)^2- 4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\ &=& 4(\xi\eta)^2 \Bigg(\big(1-\eta\xi\big)^2-(1-\xi^2)(1-\eta^2) \Bigg) \\ &=& 4(\xi\eta)^2 \Bigg(\big(1-2\eta\xi+\eta^2\xi^2\big)-\big(1-\xi^2-\eta^2+\eta^2\xi^2\big) \Bigg) \\ &=& 4(\xi\eta)^2 \big(\xi-\eta\big)^2 \\ \end{array} \tag{3}$$

So one of $\xi,\eta,\xi-\eta$ must be zero. A little more case analysis then shows that in the end one of $x,y,x+y$ must be zero modulo $2\pi$ as wished.

Answer 1

We have $\DeclareMathOperator{\Ima}{Im}$

$$\begin{align} \sin (x+y) - \sin x - \sin y &= \Ima \left(e^{i(x+y)} - e^{ix} - e^{iy}\right)\\ &= \Ima \left(e^{i(x+y)} - e^{ix} - e^{iy} + 1\right)\\ &= \Ima \left(\left(e^{ix}-1\right)\left(e^{iy}-1\right)\right)\\ &= \Ima \left(e^{i(x+y)/2}\left(e^{ix/2} - e^{-ix/2}\right)\left(e^{iy/2} - e^{-iy/2}\right)\right)\\ &= \Ima \left(e^{i(x+y)/2} \left(2i\sin \tfrac{x}{2}\right)\left(2i\sin\tfrac{y}{2}\right) \right)\\ &= - 4 \sin \frac{x+y}{2}\sin \frac{x}{2} \sin \frac{y}{2}. \end{align}$$

That product is $0$ if and only if at least one of the factors is $0$.

Answer 2

Here is my version. Let $u=e^{ix}$, $v=e^{iy}$. By assumption $z=uv-u-v\in\mathbb{R}$. This is equivalent to $$ uv-u-v=\frac{1}{uv}-\frac{1}{u}-\frac{1}{v}, $$ and again,this is equivalent to $$ (1-u)(1-v)=\left(1-\frac{1}{u}\right)\left(1-\frac{1}{v}\right)= \frac{(1-u)(1-v)}{uv} $$ or $$ (1-u)(1-v)\left(1-\frac{1}{uv}\right)=0 $$ so at least one of $u$, $v$, $uv$ is equal to one.

Answer 3

HINT:

For a proof in the real numbers use the fact that for a function with period $a$, $f(x+a\cdot n) = f(x)$ for $n \in \mathbb{Z}$. It's not easy but it avoids complex numbers.

Answer 4

Well here is (yet another) solution, this using only real variables.

By the sum formula we have

$$\sin x(1-\cos y)+\sin y(1-\cos x)=0$$

multiplying by $(1+\cos x)(1+\cos y)$ we get

$$\sin x \sin y [\sin y(1+\cos x)+\sin x(1+\cos y)]=0$$ so either $\sin x=0$ or $\sin y=0$ or $$\sin y(1+\cos x)+\sin x(1+\cos y))=0$$ which is equivalent to $$\sin x +\sin y +\sin(x+y) =0$$

so $$\sin(x+y) =0$$

Then it is easy to derive the result.

Answer 5

Here is a different derivation of Daniel Fischer's beautiful formula, We have the well known, $$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$$

And $$\sin (x+y)=2\sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x+y}{2}\right)$$

Now $$\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)= 2\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$ so $$\sin x+\sin y-\sin (x+y)=4\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$