# Solving the system $\cos^2x+\sin^2y=1$, $\cos y\sin y=\cos x\sin x$

I have a set of trigonometric equations as follows:

$$\cos^2(x)+\sin^2(y)=1$$ $$\cos(y)\sin(y)=\cos(x)\sin(x)$$

I have tried to plot these two graphs on desmos and it seems that two functions agree on the line $$x=y+n\pi$$. However, I don't see any clue in getting this relation and I am hard stuck right now.

Could anyone give me some hint on this? Thanks!

• Hint: Move one of the trig functions in the first equation to the right-hand side, and simplify.
– Blue
Aug 29 '20 at 20:06
• You can use a double angle formula to convert the second to a different form if you wish ... Aug 29 '20 at 20:15
• What if I just have the second function? Can I solve it as well? edit:Nevermind--I can use double angle formula as @MarkBennet said
– Edi
Aug 29 '20 at 20:28

$$\cos^2x+\sin^2y=1 \iff \cos^2x=\cos^2 y \iff y=\pm x+2k\pi \, \lor\, y=\pi \pm x+2k\pi$$
From the 1st equation $$(2\cos^2 x - 1) - (1 - 2\sin^2 y) = \cos 2x - \cos 2y = -2\sin (x+y) \sin (x-y) = 0$$ From the 2nd equation $$2\cos x \sin x - 2 \cos y \sin y = \sin 2x - \sin 2y = 2\cos (x+y) \sin (x-y) = 0$$
Either $$\sin (x-y) = 0 \implies x = y+n\pi$$ or $$\sin (x+y) = \cos(x+y) = 0 \implies \text{ no solution }$$