Let n be a positive integer such that sin⁡π2n+cos⁡π2n=n2 then

:sin⁡π2n+cos⁡π2n=2sin⁡π2n+π4⇒sin⁡π2n+π4=n22 Since π4<π2n+π4<3π4 for n≥1⇒12<n22≤1⇒2<n≤22⇒ 4<n≤8 For n=8,sin⁡π16+π4≠1∴ 4<n<8

Let n be a positive integer such that sin⁡π2n+cos⁡π2n=n2 then

:sin⁡π2n+cos⁡π2n=2sin⁡π2n+π4⇒sin⁡π2n+π4=n22 Since π4<π2n+π4<3π4 for n≥1⇒12<n22≤1⇒2<n≤22⇒ 4<n≤8 For n=8,sin⁡π16+π4≠1∴ 4<n<8

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Let n be a positive integer such that $\sin \frac{π}{2 n} + \cos \frac{π}{2 n} = \frac{\sqrt{n}}{2}$ then

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$4 \leq n < 8$

$6 \leq n \leq 8$

$4 < n < 8$

$4 < n \leq 8$

answer is D.

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Detailed Solution

$: \sin \frac{π}{2 n} + \cos \frac{π}{2 n} = \sqrt{2} \sin (\frac{π}{2 n} + \frac{π}{4})$
$\begin{array}{l} \Rightarrow \sin (\frac{π}{2 n} + \frac{π}{4}) = \frac{\sqrt{n}}{2 \sqrt{2}} \\ Since \frac{π}{4} < \frac{π}{2 n} + \frac{π}{4} < \frac{3 π}{4} for n \geq 1 \\ \Rightarrow \frac{1}{\sqrt{2}} < \frac{\sqrt{n}}{2 \sqrt{2}} \leq 1 \Rightarrow 2 < \sqrt{n} \leq 2 \sqrt{2} \end{array}$
$\begin{array}{l} \Rightarrow 4 < n \leq 8 \\ For n = 8, \sin (\frac{π}{16} + \frac{π}{4}) \neq 1 \\ ∴ 4 < n < 8 \end{array}$