If S=x R: ^-1(x+1x^2+2x+2)- ^-1(xx^2+1)=( )/(4), then _x s( ((x^2+x+5)( )/(2))- ((x^2+x+5)…

Question

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

Answer: 4

Accepted Answer

Given equation is

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

Let's denote:

$A={\sin }^{-1}(\frac{x+1}{\sqrt{{x}^{2}+2x+2}}),$

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

So, we have the equation $A-B=\frac{\pi }{4}$.

We can also write this as $A=B+\frac{\pi }{4}$.

This gives us

$\sin (A)=\sin (B+\frac{\pi }{4}).$

We can use the identity $\sin (a+b)=\sin a\cos b+\cos a\sin b$ and rewrite this equation as:

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

After simplifying, we get:

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

Let's square both sides to remove the square roots:

On the left side, squaring gives:

${(\frac{x+1}{\sqrt{{x}^{2}+2x+2}})}^{2}=\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}.$

On the right side, squaring gives:

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

$	herefore $ $\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}$ = $\frac{1}{2}(\frac{{x}^{2}}{{x}^{2}+1}+2\frac{x}{\sqrt{{x}^{2}+1}}\sqrt{1-\frac{{x}^{2}}{{x}^{2}+1}}+1-\frac{{x}^{2}}{{x}^{2}+1})$

$\Rightarrow$ $\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}$ = $\frac{1}{2}(2	imes \frac{x}{\sqrt{{x}^{2}+1}}\sqrt{\frac{{x}^{2}+1-{x}^{2}}{{x}^{2}+1}}+1)$

$\Rightarrow$ $\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}$ = $\frac{1}{2}(2	imes \frac{x}{\sqrt{{x}^{2}+1}}\frac{1}{\sqrt{{x}^{2}+1}}+1)$

$\Rightarrow$ $\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}$ = $\frac{1}{2}(2	imes \frac{x}{{x}^{2}+1}+1)$

$\Rightarrow$ $\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}$ = $\frac{1}{2}(\frac{2x+{x}^{2}+1}{{x}^{2}+1})$

$\Rightarrow$ $\frac{(x+1{)}^{2}}{{x}^{2}+2x+2}$ = $\frac{1}{2}(\frac{{(x+1)}^{2}}{{x}^{2}+1})$

$\begin{gathered} \Rightarrow \frac{x+1}{\sqrt{{x}^{2}+2x+2}}=\frac{x+1}{\sqrt{2}\sqrt{{x}^{2}+1}} \ \Rightarrow x=-1\sqrt{{x}^{2}+2x+2}=\sqrt{2}\cdot \sqrt{{x}^{2}+1} \ \Rightarrow x=0,x=2 \ S={0,-1} \end{gathered}$

$\begin{gathered} \sum _{x\in R}(\sin (({x}^{2}+x+5)\frac{\pi }{2})-\cos (({x}^{2}+x+5)\pi )) \ =[\sin (\frac{5\pi }{2})-\cos (5\pi )]+[\sin (\frac{5\pi }{2})-\cos (5\pi )] \ =(1-(-1))+(1-(-1)) \ =2+2 \ =4 \end{gathered}$

Solutions · 1