Isolating y in the following formula

Question

How can I isolate $x$ or $y$?

$\sin(x)= \sin(y) $

$\ x\geq 0, y\geq 0, x+y \leq \pi $

Answer 1

Use Prosthaphaeresis Formulas,

$$0=\sin x-\sin y=2\sin\dfrac{x-y}2\cos\dfrac{x+y}2$$

If $\sin\dfrac{x-y}2=0,\dfrac{x-y}2=m\pi$

$\implies x-y=2m\pi$

As $0\le x+y\le\pi, -\pi\le x-y\le\pi\implies m=0$

If $\cos\dfrac{x+y}2=0,\dfrac{x+y}2=(2r+1)\dfrac\pi2\implies x+y=(2r+1)\pi$

But $0\le x+y\le\pi,r=0$

Answer 2

Suppose $x,y\ge 0$ are such that

• $\sin(x)=\sin(y)$$\\[4pt]$
• $x+y\le \pi$

Consider two cases . . .

Case$\;(1)$:$\;y=x$.

The $\sin(x)=\sin(y)$ is automatic.

Given that $y=x$, the remaining condition$\;x+y\le \pi\;$is equivalent to$\;x\le {\large{\frac{\pi}{2}}}$.

Thus, for case $(1)$, the pair $(x,y)$ satisfies the given conditions if and only if $y=x$ and $x\in [0,\frac{\pi}{2}]$.

Case$\;(2)$:$\;y\ne x$.

Since $x,y\ge 0$ and $x+y\le \pi$, it follows that $x,y\in [0,\pi]$.

Since the map$\;t\mapsto \sin(t)\;$is strictly increasing on the interval $A=[0,\frac{\pi}{2}]$, and strictly decreasing on the interval $B=[\frac{\pi}{2},\pi]$, it follows that $x,y$ can't both be in $A$, and can't both be in $B$.

First assume $x < y$.

Then $x\in A,\;y\in B$.

Identically, we have$\;\sin(\pi-x)=\sin(x)$,$\;$hence$\;\sin(y)=\sin(\pi-x)$.

But from $x\in A$, we get $\pi-x\in B$.

Since $x$ and $\pi-x$ are both in $B$, the relation $\sin(y)=\sin(\pi-x)$ yields $y=\pi-x$.

Also, from $x < y$, we get $x<{\large{\frac{\pi}{2}}}$.

Thus, assuming $x < y$, the pair $(x,y)$ satisfies the given conditions if and only if $y=\pi-x$ and $x\in [0,\frac{\pi}{2})$.

Analogously, assuming $y < x$, the pair $(x,y)$ satisfies the given conditions if and only if $x=\pi-y$ and $y\in [0,\frac{\pi}{2})$, or equivalently, $y=\pi-x$ and $x\in (\frac{\pi}{2},\pi]$.

Combining the results for the two cases, it follows that the pair $(x,y)$ satisfies the given conditions if and only if one of

• $y=x$ and $x\in [0,\frac{\pi}{2}]$.$\\[4pt]$
• $y=\pi-x$ and $x\in [0,\frac{\pi}{2})\cup (\frac{\pi}{2},\pi]$.

holds.