Perpendicular tangents

We start by our given that the tangents to $\displaystyle S$ at points $\displaystyle (\alpha_k,\beta_k)$ are perpendicular to line $\displaystyle l : (\sqrt{2}-1)x+y=0$

To find $\displaystyle y'$ , we proceed by implicit differentiation :

$\displaystyle y'=\cos(x+y)(1+y')$

Rearranging:

$\displaystyle y'=\frac{\cos(x+y)}{1-\cos(x+y)}$

Equating this to the negative of the reciprocal of the slope of $\displaystyle l$ :

$\displaystyle \frac{\cos(\alpha_k+\beta_k)}{1-\cos(\alpha_k+\beta_k)}=\frac{1}{\sqrt{2}-1}$

Rearranging our equation:

$\displaystyle \cos(\alpha_k+\beta_k)=\frac{\sqrt{2}}{2}$

Therefore:

$\displaystyle \alpha_k+\beta_k= \frac{\pi}{4} + 2n\pi ---> (1)$

or,

$\displaystyle \alpha_k+\beta_k= -\frac{\pi}{4} + 2n\pi ---> (2)$

Working on $\displaystyle (1)$ :

We have $\displaystyle \beta_k=\sin(\alpha_k+\beta_k) => \beta_k=\sin( \frac{\pi}{4} + 2n\pi)=\frac{\sqrt{2}}{2}$

Hence. $\displaystyle \alpha_k=\frac{\pi}{4} + 2n\pi -\frac{\sqrt{2}}{2}$

Using Trial and Error, we find out that for $\displaystyle \alpha_k \in [-4\pi,4\pi ] , n = -2,-1,0,1$

Hence our solution pairs are :

$\displaystyle (\frac{\pi}{4} + 2n\pi -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ for $n= -2,-1,0,1$

Working on $\displaystyle (2)$ :

We have $\displaystyle \beta_k=\sin(\alpha_k+\beta_k) => \beta_k=\sin( -\frac{\pi}{4} + 2n\pi)=-\frac{\sqrt{2}}{2}$

Hence. $\displaystyle \alpha_k=-\frac{\pi}{4} + 2n\pi +\frac{\sqrt{2}}{2}$

Using Trial and Error, we find out that for $\displaystyle \alpha_k \in [-4\pi,4\pi ] , n = -1,0,1, 2$

Hence our solution pairs are :

$\displaystyle (-\frac{\pi}{4} + 2n\pi +\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$ for $n= -1,0,1,2$

Now nothing but substitution in our sum to get $\displaystyle 44.608$ then $\displaystyle 44$ is our desired solution.