Questions about these Trigonometry Graphs involving sin() and cos()

  • #1
pairofstrings
411
7
TL;DR Summary
a sin(x) - b cos(y) = y
a sin(x) + b cos(y) = 1
Hi.
I have two trigonometric equations whose graphs I am trying to understand.
Here are the equations:
1. a sin(x) - b cos(y) = y; a = 2, b = 2

Web capture_20-8-2023_152359_www.desmos.com.jpeg

2. a sin(x) + b cos(y) = 1; a = 1, b = 1

Web capture_20-8-2023_15261_www.desmos.com.jpeg

My question is why the graphs are the way they are.
What should I do to understand them?
Can anyone explain these graphs?

Thanks for the help.
 
Last edited:
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  • #2
When you consider level sets ##\{(x,y)\mid f(x,y)=const\}## it is important to find critical points of the function ##f## and understand which kind these critical points are.
So first find the points such that ##df=0##.
It is like drawing a phase portrait of a Hamiltonian system with the Hamiltonian f.
 
Last edited:
  • #3
Thanks. So, I need to do Analysis first?
 
  • #4
pairofstrings said:
Thanks. So, I need to do Analysis first?
The second graph looks off to me. You have
$$\cos y = 1 - \sin x$$If ##\sin x <0##, then there are no solutions for ##y##. You have solutions for ##0 \le x \le \pi##, with symmetry about ##x = \frac \pi 2##. Whatever solutions you have in this range are repeated every ##2\pi## units along the x-axis.

It would be better have units of ##\pi## along both axes.

Does that get you started?
 

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