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Question

(C) Given,$\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$.Squaring both sides:$\left(\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}\right)

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given,$\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$.Squaring both sides:$\left(\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}ight)^2 = \frac{n}{4}$$1 + \sin \frac{\pi}{n} = \frac{n}{4}$$\sin \frac{\pi}{n} = \frac{n-4}{4}$.Since $n > 2$,$\frac{\pi}{n} \in (0, \frac{\pi}{2}]$,so $\sin \frac{\pi}{n} > 0$,which implies $n-4 > 0$,so $n > 4$.Also,$\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \sqrt{2} \sin \left(\frac{\pi}{4} + \frac{\pi}{2n}ight) = \frac{\sqrt{n}}{2}$.$\sin \left(\frac{\pi}{4} + \frac{\pi}{2n}ight) = \frac{\sqrt{n}}{2\sqrt{2}} = \sqrt{\frac{n}{8}}$.Since $\sin 	heta \le 1$,we have $\sqrt{\frac{n}{8}} \le 1 \Rightarrow n \le 8$.For $n=8$,$\sin \frac{\pi}{16} = \frac{8-4}{4} = 1$,which is impossible as $\frac{\pi}{16} 
eq \frac{\pi}{2}$.Thus,$4 The number of integral values is $3$.

The number of all possible integral values of $n > 2$ such that $\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$ is:

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