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

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(B) Given $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.Multiply both sides by $\frac{1}{\sqrt{2}}$:$\frac{1}{\sqrt{2}} \sin \f

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$4 < n \le 8$

Vedclass · Answer

(B) Given $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.Multiply both sides by $\frac{1}{\sqrt{2}}$:$\frac{1}{\sqrt{2}} \sin \frac{\pi }{2^n} + \frac{1}{\sqrt{2}} \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2\sqrt{2}}$.$\sin \left( \frac{\pi }{2^n} + \frac{\pi }{4} ight) = \frac{\sqrt{n}}{2\sqrt{2}}$.Since the maximum value of $\sin 	heta$ is $1$,we have $\frac{\sqrt{n}}{2\sqrt{2}} \le 1$ $\Rightarrow \sqrt{n} \le 2\sqrt{2}$ $\Rightarrow n \le 8$.Also,for $n=1$,$\sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$,while $\frac{\sqrt{1}}{2} = 0.5$. $1 
eq 0.5$.For $n=2$,$\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$,while $\frac{\sqrt{2}}{2} \approx 0.707$.For $n=3$,$\sin \frac{\pi}{8} + \cos \frac{\pi}{8} = \sqrt{1 + \sin \frac{\pi}{4}} = \sqrt{1 + \frac{1}{\sqrt{2}}} \approx 1.306$,while $\frac{\sqrt{3}}{2} \approx 0.866$.For $n=4$,$\sin \frac{\pi}{16} + \cos \frac{\pi}{16} > 1$,while $\frac{\sqrt{4}}{2} = 1$.Since the function $f(n) = \sin \frac{\pi}{2^n} + \cos \frac{\pi}{2^n}$ is decreasing for $n \ge 2$ and $g(n) = \frac{\sqrt{n}}{2}$ is increasing,the equality holds for $n > 4$ and $n \le 8$.

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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