If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$

Question

• $\sin x + \sin y = 1$
• $\cos x + \cos y = 0$

Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.

I got the question from chapter 26 of a comic called Yamada-kun.

How can I solve this equation?

Answer 1

There is an interesting trick: you may couple the two equations by writing $$ e^{ix}+e^{iy} = i \tag{1}$$ hence $e^{ix}$ and $e^{iy}$, that are two points on the unit circle, are simmetric with respect to the imaginary axis. By imposing that their sum has unit norm, we clearly get $\{x,y\}=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$:

Answer 2

There are so many different ways to solve it the real question is which way.

What reaches out to grab me is:

$\cos x + \cos y = 0$

$\cos x = - \cos y$ which means either $y = \pi - x$ (within a period of $2\pi$) or $y = x + \pi$ (within a period of $2\pi$).

If $y = x + \pi$ then $\sin y = - \sin x$ and $\sin y + \sin x = 0 \ne 1$ which is impossible.

If $y = x - \pi$ then $\sin y = \sin x$ and $\sin y + \sin x = 2 \sin x$. If this is so (and it's our only option) then $\sin x = 1/2$ which means $x = \{\pi/6, 5\pi/6\}$.

So $(x,y) = (\pi/6, 5\pi/6)$ or $(x,y)= (5\pi/6, \pi/6)$ (within periods of $2\pi$)

Answer 3

HINT

Squaring both equations you get $$ \sin^2 x + \sin^2 y + 2\sin x \sin y = 1\\ \cos^2 x + \cos^2 y + 2\cos x \cos y = 0 $$ Now add them together to get $$ 2 + 2 \sin x \sin y + 2 \cos x \cos y = 1 $$ or in other words $$ \frac{-1}{2} = \cos x \cos y + \sin x \sin y = \cos (x-y) $$

Answer 4

There are identities for sum of sin and cos:

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

Using the first equation tells us that $\cos\left(\frac{x-y}{2}\right)\neq 0.$ Therefore by the second equation, $\cos\left(\frac{x+y}{2}\right) = 0.$

This may or may not be useful in finishing the problem.

Answer 5

I will only look for solutions on $[0, 2\pi)$

We need to make $2$ observations:

$1)$ The maximum value of $\sin \theta$ is $1$. Therefore, if any of $\sin x$, $\sin y$ is negative, $\sin x + \sin y < 1$. It follows that both $x$ and $y$ must be $\in [0, \pi]$.

$2)$ $\cos x= -\cos y \iff \cos x = \cos (\pi - y)$ . (You can prove this using the cosine subtraction formula). Since we are working only on the interval $[0, \pi]$, we must have $x=\pi-y$. This makes sense, because the cosines of symmetric angles on opposite sides of the $y$ axis will cancel out to $0$.

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Substitute $x=\pi-y$ into the first equation:

$$\sin (\pi-y) + \sin y = 1$$

Using $\sin (a-b)= \sin a \cos b - \cos a \sin b$ we can deduce that $\sin (\pi-y)=\sin y$

$$\sin y + \sin y = 1$$

$$\sin y = \dfrac 12$$

$$y = \dfrac {\pi}{6}, y= \dfrac {5\pi}{6}$$

$$x=\pi-y$$

Therefore the solutions are $ \left( \dfrac {\pi}{6}, \dfrac {5\pi}{6} \right)$ and $ \left(\dfrac {5\pi}{6}, \dfrac {\pi}{6} \right)$. The symmetry of the solutions is expected, since both original equations and symmetric.

Answer 6

$$\sin x + \sin y = 1\Rightarrow 2\sin\frac{x+y}{2}\cos \frac{x-y}{2}=1\\\cos x+\cos y=0\Rightarrow 2\cos\frac{x+y}{2}\cos \frac{x-y}{2}=0$$ It follows $$ \cos \frac{x+y}{2}=0\text{ and } \cos \frac{x-y}{2}=0\iff $$ The first ($\cos \frac{x+y}{2}=0$) gives, from the first given equation,

$$\begin{cases}x+y=\pi\\x-y=\frac{2\pi}{3}\end{cases}\iff(x,y)=(\frac{\pi}{6},\frac{5\pi}{6})$$ The second ($\cos \frac{x-y}{2}=0$) is discarded by incompatibility.

Thus the solutions are given by $$(x,y)=( \frac{\pi}{6}+2m\pi,\space \frac{5\pi}{6}+2n\pi)$$ It is not worthless to see the graphic solutions: all intersections of the red lattice, with blue closed curves.

Answer 7

This is probably improper vocabulary but I think this concepts are fundamental.

Your "basic" angle is $0 \le \theta \le \pi/2$. All other angles are "essentially" equivalent "upto reflection on one or two axes". (Or up to periods of multiples of $2\pi$.)

Example. $\theta$ and $\pi - \theta$ are "equivalent up to reflection on the y-axis" and $\sin \theta = \sin (\pi - \theta)$ and $\cos \theta = - \cos (\pi - \theta)$ and "sines change signs on reflection over x axis" and "cosines change signs on reflection over y axis".

$\theta$ and $\pi + \theta$ are equivalent "up to reflection on both axes" and $\sin \theta = - \sin (\pi + \theta)$ and $\cos \theta = - \cos (\pi + \theta)$.

Similarly for $\theta$ and $-\theta$ are equivalent "up to reflection on the x axis" and $\sin \theta = - \sin (- \theta)$ and $\cos \theta = \cos (- \theta)$.

....... .......

Okay, So $\cos x + \cos y = 0\implies \cos x = - \cos y$ means $x$ and $y$ are equivalent upto but reflected over the y-axis.

So $\sin x = \pm \sin y$ but $\sin x + \sin y \ne 0$ means $\sin x = \sin y$ meaning $x$ and $y$ are not reflected over the x-axis.

So $y = \pi - x$.

And now as $2\sin x= 2\sin y = 1$, the "basic angle" is $x|y = \pi/6$ and $y|x = \pi - \pi/6 = 5\pi/6$.