Let n be a positive integer such that sin fra | vedclass

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(B) Given equation: $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.Multiply both sides by $\sqrt{2}$:$\sqrt{2} \left( \sin \frac

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$4 < n \le 8$

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(B) Given equation: $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.Multiply both sides by $\sqrt{2}$:$\sqrt{2} \left( \sin \frac{\pi }{2^n} \cdot \frac{1}{\sqrt{2}} + \cos \frac{\pi }{2^n} \cdot \frac{1}{\sqrt{2}} ight) = \frac{\sqrt{n}}{2}$.Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get:$\sqrt{2} \sin \left( \frac{\pi }{4} + \frac{\pi }{2^n} ight) = \frac{\sqrt{n}}{2}$.Since the maximum value of $\sin 	heta$ is $1$,we have $\sin \left( \frac{\pi }{4} + \frac{\pi }{2^n} ight) \le 1$.Therefore,$\frac{\sqrt{n}}{2} \le \sqrt{2} \implies \sqrt{n} \le 2\sqrt{2} \implies n \le 8$.For $n=4$,$\sin \frac{\pi}{16} + \cos \frac{\pi}{16} = \sqrt{2} \sin(\frac{\pi}{4} + \frac{\pi}{16}) = \sqrt{2} \sin(\frac{5\pi}{16}) \approx \sqrt{2} 	imes 0.831 = 1.175$,while $\frac{\sqrt{4}}{2} = 1$. Since $1.175 > 1$,the equation holds for $n > 4$.Thus,$4 < n \le 8$.

Let $n$ be a positive integer such that $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}.$ Then

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