How would you sketch $\sin y = - \sin x$?

Question

I at first did $\sin^{-1}$ both sides to get $y = -x$.

This is wrong, can someone explain why?

The graph is made up of a lot of crosses and looks like a grid on desmos. I need a way to get the answer without using a graphing calculator, any help?

This should be tagged [tag:graphing-functions], not [tag:graph-theory]. The $\sin$ function on $\mathbb R$ is not one-to-one — J. W. Tanner, Dec 16 '20 at 17:41
Note that $\sin y=-\sin x=\sin (-x)=\sin (-x+2\pi)=\sin(-x-4\pi)$ — mfl, Dec 16 '20 at 17:48

score 1 · Answer 1 · answered Dec 16 '20 at 17:52

1

As $\sin y = -\sin x = \sin ( \pi + x)$, so must have $$ y = 2n \pi + (\pi + x) = (2n+1)\pi + x $$ for some integer $n$.

Now as $n$ varies over the set of integers, we obtain different straight lines, each of slope $1$, with $y$-intercept equal to $(2n+1) \pi$.

Hence the grid you got.

answered Dec 16 '20 at 17:52

Saaqib Mahmood

26,762

Also, $\sin(2\pi-x) = -\sin(x)$ – 19aksh Dec 16 '20 at 17:54
1

To get the grid, you'll also need a slope of $-1$, otherwise, they'll be just parallel lines of slope $1$ – 19aksh Dec 16 '20 at 17:58

score 0 · Answer 2 · answered Dec 16 '20 at 17:49

You have,

$\sin x + \sin y = 0 \Rightarrow 2\sin\frac{x+y}2 \cos\frac{x-y}2 = 0 \Rightarrow \sin\frac{x+y}{2} = 0$ or $\cos\frac{x-y}{2} = 0$

So,

$\frac{x+y}{2} = n\pi$ or $\frac{x-y}{2} = (m+\frac12)\pi$ where $m,n \in Z$

$\Rightarrow x = 2n\pi-y$ ($2n$ is always even)
or $x = (2m+1)\pi +y$ ($2m+1$ is always odd)

So,

$x = k\pi + (-1)^{k-1}y ,\ k \in \Bbb Z$

or

$y = k\pi + (-1)^{k-1}x ,\ k \in \Bbb Z$

How would you sketch $\sin y = - \sin x$?

2 Answers2