If $$S=\left\{x \in \mathbb{R}: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}$$, then $$\sum_\limits{x \in s}\left(\sin \left(\left(x^{2}+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^{2}+x+5\right) \pi\right)\right)$$ is equal to ____________.

Given equation is <br/><br/>$$ \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}. $$ <br/><br/>Let's denote: <br/><br/>$A = \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right),$ <br/><br/>$B = \sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right).$ <br/><br/>So, we have the equation $A - B = \frac{\pi}{4}$. <br/><br/>We can also write this as $A = B + \frac{\pi}{4}$. <br/><br/>This gives us <br/><br/>$\sin(A) = \sin\left(B + \frac{\pi}{4}\right).$ <br/><br/>We can use the identity $\sin(a + b) = \sin a \cos b + \cos a \sin b$ and rewrite this equation as: <br/><br/>$$\frac{x+1}{\sqrt{(x+1)^2+1}} = \frac{x}{\sqrt{x^2+1}} \cos\left(\frac{\pi}{4}\right) + \sqrt{1-\left(\frac{x}{\sqrt{x^2+1}}\right)^2} \sin\left(\frac{\pi}{4}\right).$$ <br/><br/>After simplifying, we get: <br/><br/>$$\frac{x+1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right).$$ <br/><br/>Let's square both sides to remove the square roots: <br/><br/>On the left side, squaring gives: <br/><br/>$\left(\frac{x+1}{\sqrt{x^2 + 2x + 2}}\right)^2 = \frac{(x+1)^2}{x^2 + 2x + 2}.$ <br/><br/>On the right side, squaring gives: <br/><br/>$$\left(\frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right)\right)^2 = \frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right).$$ <br/><br/>$\therefore$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = $$\frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right)$$ <br/><br/>$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = $${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}\sqrt {{{{x^2} + 1 - {x^2}} \over {{x^2} + 1}}} + 1} \right)$$ <br/><br/>$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = $${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}{1 \over {\sqrt {{x^2} + 1} }} + 1} \right)$$ <br/><br/>$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {{x^2} + 1}} + 1} \right)$ <br/><br/>$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {{{2x + {x^2} + 1} \over {{x^2} + 1}}} \right)$ <br/><br/>$\Rightarrow$ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {{{{{\left( {x + 1} \right)}^2}} \over {{x^2} + 1}}} \right)$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{x+1}{\sqrt{x^2+2x+2}}=\frac{x+1}{\sqrt{2} \sqrt{x^2+1}} \\\\ & \Rightarrow x=-1 \text { OR } \sqrt{x^2+2x+2}=\sqrt{2} \cdot \sqrt{x^2+1} \\\\ & \Rightarrow x=0, x=2 \text { (Rejected) } \\\\ & S=\{0,-1\} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \sum_{x \in R}\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right) \\\\ & =\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]+\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right] \\\\ & = (1 -(-1)) + (1 -(-1))\\\\ & = 2 + 2 \\\\ & = 4 \end{aligned} $$