Question

If $A_n=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin (2n-1)x}{\sin x}dx,B_n=\underset{0}{\overset{\pi /2}{\int }}{\left(\frac{\sin nx}{\sin x}\right)}^{2}dx,$ for $n\in N,$ then

• A. $A_{n+1}=A_n$
• B. $B_{n+1}=B_n$
• C. $A_{n+1}-A_n=B_{n+1}$
• D. $B_{n+1}-B_n=A_{n+1}$

Answer 1

Answer: (A), (D)

The correct options are
A $A_{n+1}=A_n$
D $B_{n+1}-B_n=A_{n+1}$

$A_{n+1}-A_n=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx=\underset{0}{\overset{\pi /2}{\int }}2\cos 2nxdx=0\Rightarrow A_{n+1}=A_n$

Now,
$B_{n+1}-B_n=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^2(n+1)x-{\sin }^{2}nx}{{\sin }^{2}x}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\sin \right(n+1)x-\sin nx)\left(\sin \right(n+1)x+\sin nx)}{{\sin }^{2}x}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{2\cos \left(\frac{2nx+x}{2}\right)\sin \left(\frac{x}{2}\right)\times 2\sin \left(\frac{2nx+x}{2}\right)\cos \left(\frac{x}{2}\right)}{{\sin }^{2}x}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin (2nx+x)\sin x}{{\sin }^{2}x}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin (2n+1)x}{\sin x}dx=A_{n+1}\Rightarrow B_{n+1}-B_n=A_{n+1}$