Question

Explain The Graphed Solution Set As Shown In The Picture Below Of sin(x^2+2x+1)+sin(y^2+2y+1)=1

Answer 1

recall \(\sin a\cos b+\cos a\sin b=\sin(a+b)\)

Answer 2

no, that's not what I'm interested in.

we can examine \(\sin a+\sin b=1\) as a statement that \(\sin a\cos b+\cos a\sin b=k\) where \(\sin(a+b)=\cos a=\cos b=k\) so we know from \(\cos a=\cos b\) that \(a=2\pi m\pm b\). we also know from \(\sin(a+b)=k\) that \(a+b=\arcsin(k)+2\pi n\) so that \(b=\arcsin(k)+2\pi n-a\) ergo$$a=2\pi m\pm b=2\pi m\pm(\arcsin(k)+2\pi n-a)=2\pi j\pm(\arcsin(k)-a)$$ which gives two possible solutions in general:$$a=2\pi j+\arcsin(k)-a\\2a=2\pi j+\arcsin(k)\\a=\pi j+\frac12\arcsin(k)\\a=\pi j+a+b\\b=\pi j\\\quad\text{OR}\\a=2\pi j-\arcsin(k)+a\\2\pi j=\arcsin(k)\\k=0$$i.e. \(a=b=0\)

Answer 3

oops I meant to say \(a=\pi j+\frac12a+\frac12b\) ergo \(\frac12a-\frac12b=\pi j\)
so \(a-b=2\pi j\)

from \(a=(x+1)^2,y=(y+1)^2\) we get $$(x+1)^2-(y+1)^2=2\pi j$$ we get hyperbolas centered at \((-1,-1)\)

Answer 4

hmmm I'm not done though

Answer 5

Understand a little.