KVPY #21

Similar solution with Md Zuhair 's

$\begin{aligned} \sin \frac \pi{2n} + \cos \frac \pi{2n} & = \frac {\sqrt n}2 & \small \color{#3D99F6} \text{Squaring both sides} \\ {\color{#3D99F6}\sin^2 \frac \pi{2n}} + {\color{#D61F06}2 \sin \frac \pi{2n} \cos \frac \pi{2n}} + {\color{#3D99F6}\cos^2 \frac \pi{2n}} & = \frac n4 & \small \color{#3D99F6} \text {Note that } \sin^2 x + \cos^2 x = 1 \\ {\color{#3D99F6}1} + {\color{#D61F06}\sin \frac \pi n} & = \frac n4 & \small \color{#D61F06} \text {Note that } 2\sin x \cos x = \sin 2x \\ \implies \sin \frac \pi n & = \frac n4 - 1 \end{aligned}$

Note that the RHS is rational, the LHS must be rational too. There are only three $n$ 's when the LHS is rational:

$\begin{cases} n = 1 & \implies \sin \pi = 0 & \implies \dfrac 14 - 1 = - \dfrac 34 & \implies \color{#D61F06} LHS \ne RHS \\ n = 2 & \implies \sin \dfrac \pi 2 = 1 & \implies \dfrac 24 - 1 = - \dfrac 12 & \implies \color{#D61F06} LHS \ne RHS \\ n = 6 & \implies \sin \dfrac \pi 6 = \dfrac 12 & \implies \dfrac 64 - 1 = \dfrac 12 & \implies \color{#3D99F6} LHS = RHS \end{cases}$

$\implies n = \boxed{6}$

$Sin\dfrac \pi {2n}+Cos\dfrac \pi {2n}=\dfrac{\sqrt n} 2 .\\ \therefore~Sin\dfrac \pi 2 *\dfrac 1 n+Sin\dfrac \pi 2*(1-\dfrac1 n)=\dfrac{\sqrt n} 2.\\ \implies~2*Sin\frac \pi 2 *(\frac 1 n+1-\frac1 n)*Cos\frac \pi 2 *(\frac 1 n -1+\frac 1 n)= \dfrac{\sqrt n} 2.\\ \therefore~ 2*Sin\frac \pi 2*Cos\frac \pi 2 *(\frac 2 n -1)=\dfrac{\sqrt n} 2,~~squaring~ both~sides , ~we~have:-\\ Cos^2\frac \pi 2 *(\frac 2 n -1)=\dfrac n {4*4}. \\ Since~n~is~an~integer~LHS~,a~Cos~"square"~must ~also~be~rational. \\ Only~possible~angles~ are~0, ~\frac \pi 6,~\frac \pi 4,~\frac \pi 3,~\frac \pi 4,~their~negative values..\\$

Expression : $\sin \dfrac{π}{2n}$ + $\cos \dfrac{π}{2n} =\dfrac{\sqrt{n}}{2}$

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To Find $n$

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Solution

Squaring out expression $\sin \dfrac{π}{2n}$ + $\cos \dfrac{π}{2n} =\dfrac{\sqrt{n}}{2}$ on both sides we get

$\implies$ $\sin^2 \dfrac{π}{2n}$ + $\cos^2 \dfrac{π}{2n} + 2\sin \dfrac {\pi}{2n} \cos \dfrac{\pi}{2n}=\dfrac{n}{4}$ $\implies 1 + sin \dfrac{\pi}{n} = \dfrac{n}{4}$

$\implies sin \dfrac{\pi}{n} = \dfrac{n}{4} - 1$

$\implies sin \dfrac{\pi}{n} = \dfrac{n-4}{4}$

This equation is satisfied by only $n=6$ [Determined by graph]

So $\boxed{n=6}$

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$Q.E.D$