I'm trying to find the extrema of $f(x,y)= \cos(x^2-y^2)$ constrained to $x^2+y^2=1.$ Using Lagrange Multipliers I get this far:
$-x(\sin(2x^2-1)=\lambda x$
$-y(\sin(-2y^2+1)=\lambda y$
But I don't know how to proceed after this.
Could anyone explain to me what I have to do next please?
I would advise not substituting in the relation $x^2+y^2=1$ yet, as you did in the equations you showed. Instead, just stick with $f_x(x,y)$ and $f_y(x,y)$. You get $$-x\sin(x^2-y^2)=\lambda x$$ $$y\sin(x^2-y^2)=\lambda y$$ You can divide the first equation by $x$ and the second by $y$ to get $$-\sin(x^2-y^2)=\lambda,\quad \sin(x^2-y^2)=\lambda$$ We can set the equations equal to one another: $$-\sin(x^2-y^2)=\sin(x^2-y^2)$$ Since $\sin$ is an odd function, $-\sin(x)=\sin(-x)$. Therefore, we have $$\sin(y^2-x^2)=-\sin(x^2-y^2)$$ From this, we have $$y^2-x^2=x^2-y^2$$ Therefore, we can rearrange, and then use the relation $x^2+y^2=1$.
One thing, though. Dividing out $x$ and $y$ means that we assumed that $x$ and $y$ did not equal zero. This is not necessarily the case. We should test the function to see what we get for when $x=0$ and/or $y=0$< which will produce other extrema.
I thank Thibaut Demaerel for pointing this out to me.