Adding sin(x + a) + sin(x + b)

Question

I'm trying to prove $$\forall a,b \in \mathbb{R} \exists c,d,e\in \mathbb{R}: f(x):=\sin(x+a) + \sin(x+b) = c \sin(x+d) + e$$

I attempted using $\sin(s) = \frac{e^{is} - e^{-is}}{2i}$ and ended up with an equality similar to $e^{ix} - \alpha e^{-ix}$, where usually $\alpha \neq 1$, which indicates that this way is impossible.

I also looked at the series expansion of sine, but had no idea where to even start.

This is part of a larger problem, and likely not required in the solution, but I am curious how this can be solved.

Answer 1

Do you get access to calculus to do this?

Your function $f$ is given by $$f(x)=\sin(x+a)+\sin(x+b)$$

And you propose that it is equal to $g$ given by $$g(x)=c\sin(x+d)+e$$

First of all, the midline of $f$ is clearly(?) at $y=0$, so if this will work, $e$ must be $0$.

Now note that both functions satisfy $y''=-y$. So if they have the same value at $x=0$ and derivative at $x=0$, then they are indeed the same function. To make this happen requires $$\begin{align} \sin(a)+\sin(b)=c\sin(d)\\ \cos(a)+\cos(b)=c\cos(d) \end{align}$$

Which is a system of equations from which you could eliminate $c$ to solve for $d$ in terms of $a$ and $b$. And then you could solve for $c$ too.

This results in $$\sin(x+a)+\sin(x+b)=\frac{\sin(a)+\sin(b)}{\sin\left(\arctan\left(\frac{\sin(a)+\sin(b)}{\cos(a)+\cos(b)}\right)\right)}\sin\left(x+\arctan\left(\frac{\sin(a)+\sin(b)}{\cos(a)+\cos(b)}\right)\right)$$

(And the trig(inverse trig) parts could be simplified using $\sin(\arctan(z))=\frac{z}{\sqrt{1+z^2}}$.)

Answer 2

Don't use complex exponentials, just expand everything using the addition formula for sine and group terms alike.

Answer 3

Another approach to prove this to yourself would be to choose special points like $x = \pi + a$ and $x = -b$, and so on, and solve for $c,d,e$. After having solved for your constants, you can then go and verify your answer with as many points as you want (or, as said above, by using the addtion formulae).

I will hint that it will be easiest to use your initial equation and its first derivative.