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If $\displaystyle a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx, $ then find the value of

$$\begin{vmatrix} a_1 & a_{51} & a_{101} \\ a_2 &a_{52} & a_{102}\\ a_3 & a_{53}&a_{103}\\ \end{vmatrix}.$$

My approach :

We know that $S_{n+1} - S_{n} = T_n$ where $S_{n+1} $ is sum of $n+1$ term and $S_n $ is sum of $n$ terms and $T_n $ is $n$th term.

Can we use this here somehow ..... as I used :

$\displaystyle \frac{\sin^2(n+1)x}{\sin^2x}- \frac{\sin^2nx}{\sin^2x} = \frac{\sin^2(n+1)x - \sin^2{nx}}{\sin^2x} $.... now what to do further... please suggest... thanks.

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Set $x=y/2$, then $$a_n=\frac{1}{2}\int_0^{\pi}\frac{\sin^2\frac{ny}{2}}{\sin^2\frac{y}{2}}dy.$$ If $y=-z$, you get that $$a_n=\frac{1}{2}\int_{-\pi}^0\frac{\sin^2\frac{nz}{2}}{\sin^2\frac{z}{2}}dz\Rightarrow a_n=\frac{1}{4}\int_{-\pi}^{\pi}\frac{\sin^2\frac{ny}{2}}{\sin^2\frac{y}{2}}dy=\frac{n}{4}\int_{-\pi}^{\pi}F_n(y)\,dy=\frac{2\pi n}{4},$$ where $F_n$ is the Fejer kernel, which has the property that its integral from $-\pi$ to $\pi$ is $2\pi$. So, your determinant is equal to $$\frac{(2\pi)^3}{4^3}\left|\begin{array}{c c c}1 & 51 & 101\\ 2 & 52 & 102\\ 3 & 53 & 103\end{array}\right|=\frac{(2\pi)^3}{4^3}\left|\begin{array}{c c c}1 & 50 & 101\\ 2 & 50 & 102\\ 3 & 50 & 103\end{array}\right|=\frac{(2\pi)^3}{4^3}\left|\begin{array}{c c c}1 & 50 & 100\\ 2 & 50 & 100\\ 3 & 50 & 100\end{array}\right|=0.$$

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As $\displaystyle a_n =\int^{\frac\pi2}_0 \frac{\sin^2nx}{\sin^2x}dx, $

If $n=1, \displaystyle a_1 =\int^{\frac\pi2}_0 dx=\frac\pi2$

Else we are interested in $n>1$

$\displaystyle a_{n+1}-a_n=\int^{\frac\pi2}_0 \frac{\sin^2nx-\sin^2(n-1)x}{\sin^2x}dx$

Using $\sin^2A-\sin^2B=\sin(A-B)\sin(A+B),$ $\displaystyle a_{n+1}-a_n=\int^{\frac\pi2}_0 \frac{\sin (2n-1)x}{\sin x}dx=b_n$(say)

$\implies b_1=\displaystyle a_2-a_1=\int^{\frac\pi2}_0 \frac{\sin x}{\sin x}dx=\frac\pi2$

Now, $\displaystyle b_n-b_{n-1}=\int^{\frac\pi2}_0 \frac{\sin (2n-1)x-\sin(2n-3)x}{\sin x}dx$

Using $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2, \displaystyle b_n-b_{n-1}=\int^{\frac\pi2}_02\cos(2n-2)dx$

As $n\ne1,b_n-b_{n-1}=\frac{2\sin(2n-2)}{2n-2}|_0^{\frac\pi2}=0$ as $\sin m\pi=0$ for integer $m$

$\implies b_n=b_{n-1}$

So if $n>1, b_n=\cdots=b_1=\frac\pi2$

$\implies$ if $n>1, a_{n+1}-a_n=b_n=\frac\pi2$ and we have $a_1=\frac\pi2$

So, $a_n=??$ for $n>1$

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@sultan, how about this method? – lab bhattacharjee Aug 1 '13 at 17:10
At line $5$ you mention that $\sin^2 A-\sin^2 B=\sin(A-B)\sin(A+B)$, but this is not true. – detnvvp Aug 2 '13 at 0:39
Oh! Sorry, you are right. – detnvvp Aug 2 '13 at 14:26

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