Question

The smallest positive integer $n$ such that $\sin \left(\frac{\pi }{2n}\right)+\cos \left(\frac{\pi }{2n}\right)=\frac{\sqrt{n}}{2}$

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is B $6$

$\sin \left(\frac{\pi }{2n}\right)+\cos \left(\frac{\pi }{2n}\right)=\frac{\sqrt{n}}{2}\Rightarrow \frac{\sqrt{2}}{2}\sin \left(\frac{\pi }{2n}\right)+\frac{\sqrt{2}}{2}\cos \left(\frac{\pi }{2n}\right)=\frac{\sqrt{2}n}{4}\Rightarrow \cos \left(\frac{\pi }{4}\right)\sin \left(\frac{\pi }{2n}\right)+\sin \left(\frac{\pi }{4}\right)\cos \left(\frac{\pi }{2n}\right)=\frac{\sqrt{2}n}{4}\Rightarrow \sin (\frac{\pi }{4}+\frac{\pi }{2n})=\frac{\sqrt{2}n}{4}$

The left side is sine if an angle $(\frac{\pi }{4},\frac{3\pi }{4})$. So that sine is in $(\frac{1}{\sqrt{2}},1)$. Actually its easy to see that$n=2$ does not satisfy the original equation so that sine cannot be $1$

$\frac{\sqrt{2}}{2}<\frac{\sqrt{2}n}{4}<14<n<8$

If there are any solutions and knowing that all angles are in first quadrant original equation is equivalent to

$\sqrt{\frac{1-\cos \left(\pi /n\right)}{2}}+\sqrt{\frac{1+\cos \left(\pi /2\right)}{2}}=\frac{\sqrt{n}}{2}=\sqrt{1-\cos \left(\frac{\pi }{n}\right)}+\sqrt{1+\cos \left(\frac{\pi }{n}\right)}=\sqrt{\frac{n}{2}}2+2\sqrt{1-\cos \left(\frac{\pi }{n}\right)(1+\cos (\frac{\pi }{n}\left)\right)}=\frac{n}{2}\sqrt{(1-\cos \frac{\pi }{n})(1+\cos \frac{\pi }{n})}=\frac{n}{4}-1\sqrt{1-{\cos }^2\left(\frac{\pi }{n}\right)}=\frac{n}{4}-1\sin \left(\frac{\pi }{n}\right)=\frac{n}{4}-1$

We know that $\sin \left(\frac{\pi }{n}\right)$ is irrational

$So,n=6$ is any solution.