When $\cos(x)+\cos(y)=\sin(x)+\sin(y)=1$, $x$ or $y$ must be a multiple of $2\pi$?

Question

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?

Answer 1

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.

Answer 2

As an alternative by subtracting we obtain

$$\cos x-\sin x + \cos y - \sin y =0 \iff \sqrt 2\cos\left(x+\frac \pi 4\right)+\sqrt 2\cos\left(y+\frac \pi 4\right)=0$$

$$\iff \cos\left(x+\frac \pi 4\right)=-\cos\left(y+\frac \pi 4\right)$$

which requires

$$x+\frac \pi 4=\pi - \left(y+\frac \pi 4\right)+2n\pi\;\lor\; x+\frac \pi 4=\pi + \left(y+\frac \pi 4\right)+2m\pi$$

and from the first one we obtain

$$x+y =\frac \pi 2 +2n\pi $$

using this one in the original equations we obtain

$$\cos x+\cos y =1 \iff \cos x +\cos\left(\frac \pi 2 +2n\pi-x\right)=1\iff \cos x +\sin x =1$$

$$\sin x+\sin y =1 \iff \sin x +\sin\left(\frac \pi 2 +2n\pi-x\right)=1\iff \sin x +\cos x =1$$

which leads to

$$\sin x +\cos x =\sqrt 2\sin \left(\frac \pi 4+x\right)=1$$

and therefore

$$\begin{cases}x=2m\pi \implies y=\frac \pi 2 +2(n-m)\pi \\ \\ x=\frac \pi 2+2m\pi \implies y=2(n-m)\pi \end{cases}$$

Answer 3

Using the given condition $ \ \sin x + \sin y \ = \ 1 \ \ , \ $ we may choose $ \ u \ = \ \sin x \ \ $ and $ \ 1 - u \ = \ \sin y \ \ . \ $ This leads us to $ \ \cos x \ = \ \pm \sqrt{1 - u^2} \ \ $ and $ \ \cos y \ = \ \pm \sqrt{1 - (1 - u)^2} \ = \ \pm \sqrt{2u - u^2} \ \ . \ $

But under the second condition $ \ \cos x + \cos y \ = \ 1 \ \ , \ $ we then have $ \ \pm \sqrt{1 - u^2} \pm \sqrt{2u - u^2} \ = \ 1 \ \ , \ $ for which the domain is the intersection of $ \ |u| \ \le \ 1 \ \Rightarrow \ -1 \ \le \ u \ \le \ 1 \ $ and $ \ 0 \ \le \ u·(2 - u) $ $ \Rightarrow \ 0 \ \le \ u \ \le \ 2 \ \ , \ $ which is $ \ 0 \ \le \ u \ \le \ 1 \ \ . \ $ Solving the version of this equation with both positive square roots, we obtain $$ \sqrt{1 - u^2} \ + \ \sqrt{2u - u^2} \ = \ 1 \ \ \Rightarrow \ \ 2u \ - \ u^2 \ \ = \ \ 1 \ - \ 2 · \sqrt{1 - u^2} \ + \ (1 - u^2) $$ $$ \Rightarrow \ \ 2u \ \ = \ \ 2 \ - \ 2 · \sqrt{1 - u^2} \ \ \Rightarrow \ \ 1 \ - \ u^2 \ \ = \ \ 1 \ - \ 2u \ + \ u^2 $$ $$ \Rightarrow \ \ 2u^2 \ - \ 2u \ \ = \ \ 2u·(u - 1) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ u \ = \ \sin x \ = \ 0 \ , \ 1 \ \ , \ \ \sin y \ \ = \ \ 1 \ , \ 0 \ \ , \ \ \cos x \ \ = \ \ 1 \ , \ 0 \ \ , \ \ \cos y \ \ = \ \ 0 \ , \ 1 \ \ . $$ Only one angle per "cycle" satisfies these results, hence $$ \sin x \ = \ 0 \ \ , \ \ \cos x \ = \ 1 \ \ , \ \ \sin y \ = \ 1 \ \ , \ \ \cos y \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ = \ 2k \pi \ \ , \ \ y \ = \ \frac{\pi}{2} + 2m \pi $$ or $$ \sin x \ = \ 1 \ \ , \ \ \cos x \ = \ 0 \ \ , \ \ \sin y \ = \ 0 \ \ , \ \ \cos y \ \ = \ \ 1 \ \ \Rightarrow \ \ x \ = \ \frac{\pi}{2} + 2k \pi \ \ , \ \ y \ = \ 2m \pi \ \ , $$ with $ \ k \ $ and $ \ m \ $ being integers. This demonstrates the second proposition in the problem statement and $ \ x + y \ = \ \frac{\pi}{2} + 2·(k + m) \pi \ \ $ demonstrates the first.

[If we solve the versions of the equation with one negative square-root term, we obtain only one of the solutions shown above, with the other being "spurious". The version with both negative square-root terms has no real solutions. So only the equation with two positive square-root terms is applicable.]

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We might also treat the problem as a question of locating the intersections of the curves described by the two equations. We could start by asking about the "principal" curve for $ \ \sin x + \sin y \ = \ 1 \ \ , \ $ where $ \ 0 \ \le \ x \ , \ y \ \le \ 2 \pi \ \ . \ $ The equation is symmetric under an exchange of the variables $ \ x \ $ and $ \ y \ \ , \ $ so the curve is symmetric about the line $ \ y \ = \ x \ \ ; \ $ since the curve equation is equivalent to $ \ \cos \left(\frac{\pi}{2} - x \right) + \cos \left(\frac{\pi}{2} - y \right) \ = \ 1 \ \ , \ $ it also has symmetry about $ \ \frac{\pi}{2} - y \ = \ - \left(\frac{\pi}{2} - x \right) \ \rightarrow \ x + y \ = \ \pi \ \ . \ $

Differentiating the curve equation implicitly produces $ \ \cos x + \cos y · y' \ = \ 0 \ \Rightarrow \ \large{ y' \ = \ -\frac{\cos x}{\cos y} } \ \ . \ $ This tells us that this "principal curve" has horizontal tangent lines at $ \ x \ = \ \frac{\pi}{2} \ $ and vertical tangents at $ \ y \ = \ \frac{\pi}{2} \ \ . \ $ Applying the curve equation gives us tangent points at $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ , \ \left(\frac{\pi}{2} \ , \ \pi \right) \ , $ $ \left(0 \ , \ \frac{\pi}{2} \right) \ , \ \left(\pi \ , \ \frac{\pi}{2} \right) \ . \ $ So this curve is bounded with the box $ \ [0 \ , \ \pi] \times [0 \ , \ \pi] \ \ ; \ $ its center is located, as would be expected, at the intersection of the symmetry lines, $ \left(\frac{\pi}{2} \ , \ \frac{\pi}{2} \right) \ \ . $

If we perform an investigation of the extrema of the two-variable functions $ \ y + x \ $ and $ \ y - x \ $ under the constraint $ \ \sin x + \sin y \ = \ 1 \ \ $ (for which we will not show the calculations here), we also find that the "principal curve" is bounded by the lines $ \ x + y \ = \ \frac{\pi}{3} \ \ , \ \ x + y \ = \ \frac{5 \pi}{3} \ \ , \ \ y - x \ = \ \frac{2 \pi}{3} \ \ , \ \ $ $ y - x \ = \ -\frac{2 \pi}{3} \ \ . \ $ A graph of this curve is shown below.

The periodicity of the sine function produces an infinite family of such "loops" [in green in the graph below] with their centers spaced in the horizontal and vertical directions by multiples of $ \ 2 \pi \ \ $ (the "principal curve" is simply the loop that is tangent to both coordinate axes). Having characterized the "curve" for $ \ \sin x + \sin y \ = \ 1 \ \ , \ $ we can now readily describe that for $ \ \cos x + \cos y \ = \ 1 \ \ . \ $ As this latter equation is equivalent to $ \ \sin \left(\frac{\pi}{2} - x \right) + \sin \left(\frac{\pi}{2} - y \right) \ = \ 1 \ \ , \ $ it represents a second infinite family of loops [in blue], effectively translated "to the left and downward" relative to those of $ \ \sin x + \sin y \ = \ 1 \ \ , \ $ placing the centers of these loops at $ \ (2k \pi \ , \ 2m \pi) \ \ . $ It can then be argued that the intersections of these families occur on the lines equidistant from their loop centers $$ y \ - \ \left(\frac{\pi}{4} + 2m \pi \right) \ \ = \ \ -\left[ \ x \ - \ \left(\frac{\pi}{4} + 2k \pi \right) \ \right] \ \ \rightarrow \ \ x + y \ = \ \frac{\pi}{2} + \ 2·(k + m)·\pi \ \ . \ $$ By the argument presented earlier, the intersections of the principal loops for the two curve equations occur at $ \ \left(0 \ , \ \frac{\pi}{2} \right) \ $ and $ \ \left( \frac{\pi}{2} \ , \ 0 \right) \ \ $ (in fact, the loops intersect orthogonally at those points!), so all of the loop intersections occur periodically at $ \ \left(2k \pi \ , \ \frac{\pi}{2} + 2m \pi \right) \ $ and $ \ \left( \frac{\pi}{2} + 2k \pi \ , \ 2m \pi \right) \ \ . $