<s>[INST] What are the roots of $\sin(ax) + \sin((a + 2)x)$? I was playing around with $\sin(5x) + \sin(7x)$, wondering where the roots of the function are. I graphed it on wolframalpha and from the list of solutions I guessed that the solutions to $\sin(5x) + \sin(7x) = 0$ are exactly the roots of $\sin(6x)$.
After playing around with similar values, I have to wonder, is it true that solving 
$$\sin(a\cdot x) + \sin((a + 2)\cdot x) = 0$$
is equivalent to
$$\sin((a + 1)\cdot x) = 0$$

This problem seems to want induction.
It is true that the roots of $\sin(x) + \sin(3x)$ are the same roots as $\sin(2x)$ (again checked via wolframalpha). So that takes care of the base case.
However, I have no idea how to get from the $\sin(a\cdot x) + \sin((a+2)\cdot x)$ to the $\sin((a+1)\cdot x)$ for any $a$. It's just been guesswork.

Is it true that $\sin(a\cdot x) + \sin((a + 2)\cdot x)$ has the same roots as $\sin((a + 1)\cdot x)$?
 [/INST] We transform that sum into a product. An easy way to deduce it is: We begin with $\sin X + \sin Y$. Let $X = c + d$ and $Y = c - d$. This gives us $c = \frac{X + Y}{2}$ and $d = \frac{X - Y}{2}$. This way: $$\begin{align} \sin X + \sin Y &= \sin(c+d)+ \sin(c-d) \\ &= \sin c \cos d + \sin c \cos d + \sin c \cos d - \sin d \cos c \\ &= 2 \sin c \cos d \\ &= 2 \sin\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)\end{align}$$
Making $X = ax$ and $Y = (a+2)x$ gives: $$\sin(ax) + \sin((a+2)x) = 2\sin((a+1)x))\cos(-x) = 2\sin((a+1)x)~\cos(x)$$
since $\cos$ is an even function.
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