I'm sure almost all of you have seen this well-known identity:

\[\displaystyle \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } =1\]

But what if this is made into something you can graph? Consider

\[\displaystyle \sin ^{ 2 }{ x } +\sin ^{ 2 }{ (x+y) } =1\]

In this case, we are looking for \(y\) values such that \(\left| \cos{ x } \right| =\left| \sin{ (x+y) } \right| \). For now, lets say that \((x,y)\in [ -\frac { \pi }{ 2 } , \frac { \pi }{ 2 } ] \).

The solutions we have can be found from:

\[\displaystyle \left| \cos { \left( \pm \frac {\pi}{2} \pm x \right)} \right|= \left| \sin {x} \right|\]

This gives \(y=\{\pm \frac{\pi}{2}, \pm \frac{\pi}{2}-2x\}\). Here's what that looks like:

Since \(\sin^{2}{x}\) is periodic (that is, we can add \(\pi\) to x and get the same results for \(y\)), the graph is this parallelogram tessellated on the plane! Cool, huh?

But wait... what if we didn't square the terms? It gets harder then...

\[\displaystyle \sin{x}+\sin{\left(x+y\right)}=1\]

Intuitively, we're going to get some shape that is tiled on the plane but half as densely - due to the period being twice as long.

I'll update this note with the entire process later, but essentially the graph of this can be found by the union of \[\displaystyle y=2\pi k-\arcsin{\left( 1-\sin{x}\right)}-x+\pi\] \[\displaystyle y=\arcsin{\left( 1-\sin{x}\right)}-x-2\pi k\] for \(k \in Z\). This results in a shape somewhat like a bean.

## Comments

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TopNewestAgain, some interesting work!

Good luck investigating \(\sin{x}+\sin(x+y)=1.\) If I had to study that curve, I would look at \(\sin{x}+\sin{y}=1\) first... you can always "shear" your result to get the original curve. Also, it may help to study \(\sin{x}+\sin{y}=b\) for various values of \(b\) between \(-2\) and \(2\), including \(b=0\)... that way you get a feel for the problem.

Keep us posted! – Otto Bretscher · 2 years, 1 month ago

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Thanks!@Otto Bretscher – Swapnil Das · 2 years, 1 month ago

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It's not about comparing yourself to others, but enjoying your studies and doing meaningful work (whether it's learning, teaching, or doing research). There is a big portion of luck involved in becoming a really great mathematician (I'm surrounded by them here in Cambridge, MA): You have to be at the right place at the right time (Goettingen before WWII, Paris in the 60s, Moscow in the 80s), finding the right teachers, focus on the right questions at a very early age. Every honest biography of a great mathematician contains a sentence like "I have been incredibly lucky with my teachers". But even those of us who don't make it to that level (almost all of us) can have a fun-filled and meaningful professional life... in my case I have dedicated myself mostly to teaching.

In my opinion, Brilliant is actually a great way to "think about math"... I always tell my students to "do the problems they can't do". If you know in advance which technique to use, a problem is no longer interesting. Try the most challenging problems...

I also tell my ambitious students to read biographies of great mathematicians. Right now I read Edward Frenkel's "Love & Math", and I'm very impressed (I taught with Prof. Frenkel for a few years at Harvard, mostly Linear Algebra)... maybe you can get a copy!

Good luck, take care, and stay in touch! – Otto Bretscher · 2 years, 1 month ago

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Good note! Keep posting! – Swapnil Das · 2 years, 1 month ago

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