Find the values of $x$ and $y$ that satisfies $\sin(x+y)=\sin x+\sin y$.

Question

I know that in general the following equality does not hold: $\sin(x+y)=\sin x + \sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:

$\sin(x+y)=\sin x\cos y + \sin y \cos x = \sin x + \sin y$. Then from this equation, I got

$\sin x(1-\cos y)+\sin y(1-\cos x)=0$. There are two possibilities:

Case 1: $\sin x(1-\cos y)=0$ and $\sin y(1-\cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2\pi k$ for some integer $k$.

Case 2: $\sin x(1-\cos y)=n$ and $\sin y(1-\cos x)=-n$ where $n$ is a nonzero real number. Then we have

$\sin x(1-\cos y)=\sin y(\cos x-1)$ Since $n\neq0$ then we can divide both sides of this equation by $\sin x \sin y)$ to obtain $\csc y - \cot y = \cot x - \csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...

Answer 1

We know that

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

and by sum to product formula we have

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

therefore

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

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

then we need to consider two cases

$$2\sin\left(\frac{x+y}{2}\right)=0 $$

or otherwise we can cancel out $2\sin\left(\frac{x+y}{2}\right)$ and obtain

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

Answer 2

use that $$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ and $$\sin(x+y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$

Answer 3

Try $$x=y\pm2k\pi$$ and $$x=-y\pm 2k\pi$$

See if these are the only solutions.