Question

For $x\in \left(0,\frac{\pi }{4}\right)$
$R_n=\sum _{r=1}^{2n}\sin \left({\sin }^{-1}{x}^{4r-3}\right),S_n=\sum _{r=1}^{2n}\cos \left({\cos }^{-1}{x}^{4r-2}\right),$
$T_n=\sum _{r=1}^{2n}\tan \left({\tan }^{-1}{x}^{4r-1}\right),U_n=\sum _{r=1}^{2n}\cot \left({\cot }^{-1}{x}^{4r}\right)$
where $n\in N$ and $n\ge 4$

• A. $R_n<S_n<T_n<U_n$
• B. $R_n>S_n>T_n>U_n$
• C. $\underset{n\to \infty }{\mathrm{Lim}}\left(R_n+S_n+T_n+U_n\right)$ is equal to $\frac{x}{1-x}$
• D. The value of $x$ for which $R_n+S_n=T_n+U_n$ is $2\sin \frac{\pi }{10}$

Answer 1

Answer: (B), (C)

$\Theta R_n=\sum _{r=1}^{2n}{x}^{4r-3}=x+x^5+x^9\dots \dots \dots .$.
$S_n=\sum _{r=1}^{2n}{x}^{4r-2}=x^2+x^6+x^{10}\dots \dots \dots .$
$T_n=\sum _{r=1}^{2n}{x}^{4r-1}=x^3+x^7+x^{11}\dots \dots \dots .$
$U_n=\sum _{r=1}^{2n}{x}^{4-r}=x^4+x^8+x^{12}\dots \dots \dots .$
clearly, $R_n>S_n>T_n>U_n$ as $x\in \left(0,\frac{\pi }{4}\right)$
and $\underset{n\to \infty }{\mathrm{Lim}}\left(R_n+S_n+T_n+U_n\right)=x+x^2+\dots \dots \dots \dots ..=\frac{x}{1-x}$
$\Theta R_n+S_n=T_n+U_n$
$\Rightarrow \frac{x\left(1-x^{8n}\right)}{1-x^4}+\frac{x^2\left(1-x^{8n}\right)}{1-x^4}=\frac{x^3\left(1-x^{8n}\right)}{1-x^4}+\frac{x^4\left(1-x^{8n}\right)}{1-x^4}$
$\Rightarrow x+x^2=x^3+x^4\Rightarrow \left(x+x^2\right)\left(x^2-1\right)=0\Rightarrow \text{ no value of }x$