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

Let n be a fixed positive integer such that  sin((Ο€/(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

cool

$$\mathrm{cool} \\ $$

Answered by Frix last updated on 15/Jun/23

n=6

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

Answered by witcher3 last updated on 15/Jun/23

∣cos(a)+sin(a)∣=(√(2⟨∣))sin(a+(Ο€/4))βˆ£β‰€(√2)  β‡’((√n)/2)≀(√2)β‡’n≀8  n∈{1,....8}  (sin((Ο€/(2n)))+cos((Ο€/(2n))))^2 =1+sin((Ο€/n))=(n/4)  sin((Ο€/n))=((nβˆ’4)/4)..E  sin(x)≀xβ‡’((nβˆ’4)/4)≀(Ο€/n)≀(4/n)β‡’n^2 βˆ’4nβˆ’16≀0  n∈{((4βˆ’(√(80)))/2),((4+(√(80)))/2)}β‡’n≀((13)/2)β‡’n≀6  sin n∈{1....8}β‡’sin((Ο€/n))>0..Eβ‡’nβˆ’4>0β‡’nβ‰₯5  n∈{5,6}  simply check n=6 unique solution

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

Commented by York12 last updated on 16/Jun/23

Witcher3 are you using discord

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

Answered by MM42 last updated on 15/Jun/23

1+sin((Ο€/n))=(n/4)β‡’sin((Ο€/n))=((nβˆ’4)/4)  0≀(n/4)βˆ’1≀1β‡’4≀n≀8  only   n=6  there is a tie

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

Commented by aba last updated on 15/Jun/23

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