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((π/(2n)))+cos((𝛑/(2n)))=((√n)/2) Then find n

$$\boldsymbol{\mathrm{Let}}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{fixed}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{2n}}\right)+\mathrm{cos}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}}\right)=\frac{\sqrt{\mathrm{n}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{Then}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{n}} \\ $$

Commented by maths_plus last updated on 15/Jun/23

$$\mathrm{cool} \\ $$

Answered by Frix last updated on 15/Jun/23

$${n}=\mathrm{6} \\ $$

Answered by witcher3 last updated on 15/Jun/23

∣cos(a)+sin(a)∣=(√(2⟨∣))sin(a+(π/4))∣≤(√2) ⇒((√n)/2)≤(√2)⇒n≤8 n∈{1,....8} (sin((π/(2n)))+cos((π/(2n))))^2 =1+sin((π/n))=(n/4) sin((π/n))=((n−4)/4)..E sin(x)≤x⇒((n−4)/4)≤(π/n)≤(4/n)⇒n^2 −4n−16≤0 n∈{((4−(√(80)))/2),((4+(√(80)))/2)}⇒n≤((13)/2)⇒n≤6 sin n∈{1....8}⇒sin((π/n))>0..E⇒n−4>0⇒n≥5 n∈{5,6} simply check n=6 unique solution

$$\mid\mathrm{cos}\left(\mathrm{a}\right)+\mathrm{sin}\left(\mathrm{a}\right)\mid=\sqrt{\mathrm{2}\langle\mid}\mathrm{sin}\left(\mathrm{a}+\frac{\pi}{\mathrm{4}}\right)\mid\leqslant\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{n}}}{\mathrm{2}}\leqslant\sqrt{\mathrm{2}}\Rightarrow\mathrm{n}\leqslant\mathrm{8} \\ $$$$\mathrm{n}\in\left\{\mathrm{1},….\mathrm{8}\right\} \\ $$$$\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{2n}}\right)+\mathrm{cos}\left(\frac{\pi}{\mathrm{2n}}\right)\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{sin}\left(\frac{\pi}{\mathrm{n}}\right)=\frac{\mathrm{n}}{\mathrm{4}} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{n}}\right)=\frac{\mathrm{n}−\mathrm{4}}{\mathrm{4}}..\mathrm{E} \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\leqslant\mathrm{x}\Rightarrow\frac{\mathrm{n}−\mathrm{4}}{\mathrm{4}}\leqslant\frac{\pi}{\mathrm{n}}\leqslant\frac{\mathrm{4}}{\mathrm{n}}\Rightarrow\mathrm{n}^{\mathrm{2}} −\mathrm{4n}−\mathrm{16}\leqslant\mathrm{0} \\ $$$$\mathrm{n}\in\left\{\frac{\mathrm{4}−\sqrt{\mathrm{80}}}{\mathrm{2}},\frac{\mathrm{4}+\sqrt{\mathrm{80}}}{\mathrm{2}}\right\}\Rightarrow\mathrm{n}\leqslant\frac{\mathrm{13}}{\mathrm{2}}\Rightarrow\mathrm{n}\leqslant\mathrm{6} \\ $$$$\mathrm{sin}\:\mathrm{n}\in\left\{\mathrm{1}….\mathrm{8}\right\}\Rightarrow\mathrm{sin}\left(\frac{\pi}{\mathrm{n}}\right)>\mathrm{0}..\mathrm{E}\Rightarrow\mathrm{n}−\mathrm{4}>\mathrm{0}\Rightarrow\mathrm{n}\geqslant\mathrm{5} \\ $$$$\mathrm{n}\in\left\{\mathrm{5},\mathrm{6}\right\} \\ $$$$\mathrm{simply}\:\mathrm{check}\:\mathrm{n}=\mathrm{6}\:\mathrm{unique}\:\mathrm{solution} \\ $$$$ \\ $$

Commented by York12 last updated on 16/Jun/23

$${Witcher}\mathrm{3}\:{are}\:{you}\:{using}\:{discord} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by MM42 last updated on 15/Jun/23

1+sin((π/n))=(n/4)⇒sin((π/n))=((n−4)/4) 0≤(n/4)−1≤1⇒4≤n≤8 only n=6 there is a tie

$$\mathrm{1}+{sin}\left(\frac{\pi}{{n}}\right)=\frac{{n}}{\mathrm{4}}\Rightarrow{sin}\left(\frac{\pi}{{n}}\right)=\frac{{n}−\mathrm{4}}{\mathrm{4}} \\ $$$$\mathrm{0}\leqslant\frac{{n}}{\mathrm{4}}−\mathrm{1}\leqslant\mathrm{1}\Rightarrow\mathrm{4}\leqslant{n}\leqslant\mathrm{8} \\ $$$${only}\:\:\:{n}=\mathrm{6}\:\:{there}\:{is}\:{a}\:{tie} \\ $$$$ \\ $$

Commented by aba last updated on 15/Jun/23

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