Let-n-be-a-fixed-positive-integer-such-that-sin-pi-2n-cos-2n-n-2-Then-find-n-

Question Number 193494 by aba last updated on 15/Jun/23

Let n be a fixed positive integer such that

\sin (\frac{π}{2 n}) + \cos (\frac{π}{2 n}) = \frac{\sqrt{n}}{2}

Then find n

Commented by maths_plus last updated on 15/Jun/23

cool

Answered by Frix last updated on 15/Jun/23

n = 6

Answered by witcher3 last updated on 15/Jun/23

∣ \cos (a) + \sin (a) ∣= \sqrt{2 ⟨ ∣} \sin (a + \frac{π}{4}) ∣⩽ \sqrt{2}

\Rightarrow \frac{\sqrt{n}}{2} ⩽ \sqrt{2} \Rightarrow n ⩽ 8

n \in {1, \dots . 8}

{(\sin (\frac{π}{2 n}) + \cos (\frac{π}{2 n}))}^{2} = 1 + \sin (\frac{π}{n}) = \frac{n}{4}

\sin (\frac{π}{n}) = \frac{n - 4}{4} . . E

\sin (x) ⩽ x \Rightarrow \frac{n - 4}{4} ⩽ \frac{π}{n} ⩽ \frac{4}{n} \Rightarrow n^{2} - 4 n - 16 ⩽ 0

n \in {\frac{4 - \sqrt{80}}{2}, \frac{4 + \sqrt{80}}{2}} \Rightarrow n ⩽ \frac{13}{2} \Rightarrow n ⩽ 6

\sin n \in {1 \dots . 8} \Rightarrow \sin (\frac{π}{n}) > 0 . . E \Rightarrow n - 4 > 0 \Rightarrow n ⩾ 5

n \in {5, 6}

simply check n = 6 unique solution

Commented by York12 last updated on 16/Jun/23

W i t c h e r 3 a r e y o u u s i n g d i s c o r d

Answered by MM42 last updated on 15/Jun/23

1 + s i n (\frac{π}{n}) = \frac{n}{4} \Rightarrow s i n (\frac{π}{n}) = \frac{n - 4}{4}

0 ⩽ \frac{n}{4} - 1 ⩽ 1 \Rightarrow 4 ⩽ n ⩽ 8

o n l y n = 6 t h e r e i s a t i e

Commented by aba last updated on 15/Jun/23

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