If S=xR:^-1x+1(x+1)^2+1-^-1xx^2+1=4, then _x S[(((x^2+x+5))/(2))-((x^2+x+5))] is equal to …

Question

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

Answer: 4

Accepted Answer

Since $\dfrac{t}{\sqrt{t^2+1}}\in(-1,1)$, write the equation as

$$\sin^{-1}\frac{x+1}{\sqrt{(x+1)^2+1}}=\sin^{-1}\frac{x}{\sqrt{x^2+1}}+\frac{\pi}{4}.$$

Solving this yields $x=-1$ or $x=0$, i.e. $S=\{-1,0\}$.

At both $x=-1$ and $x=0$: $x^2+x+5=5$, so each term of the sum equals

$$\sin\left(\frac{5\pi}{2}\right)-\cos(5\pi)=1-(-1)=2.$$

$$\sum_{x\in S}\left[\cdots\right]=2+2=4.$$

Solutions · 1