We are given that $\cos x+\cos y=1$ and $\sin x+\sin y=1$. Show that $x+y=(2n+\frac{1}{2})\pi$, when $n$ is an integer. Also show that $x$ or $y$ must be a multiple of $2\pi$.
I could get the first result by transforming above two equations to $2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=1$ and $2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=1$, which implies $\tan\frac{x+y}{2}=1$.
To get the second result, I tried several things, but could not come up with it. First, I squared the given two equations and added them up. It gave $x-y=m\pi+\pi/2$ where $m$ is an integer, but I later realised it was wrong.
Can anyone find a good way of doing this?
Best Answer
When taking the inverse tangent of both sides, you should have
$$\frac{x+y}2 = \frac\pi4 + m\pi$$
since $\tan$ is $\pi$-periodic, not $2\pi$-periodic. The first intended result follows.
Substitute this into either the $\sin$-$\cos$ or $\cos$-$\cos$ product.
$$\cos\frac{x+y}{2} = \cos\frac\pi4 \cos(m\pi) = \frac{(-1)^m}{\sqrt2} \\ \implies (-1)^m \sqrt2 \cos\frac{x-y}{2}=1 \\ \implies \cos\frac{x-y}2 = \pm\cos\frac\pi4$$
This should be enough to arrive at the second required result.