Let n be a positive integer such that

Question

$\sin {\frac {π}{2n}}+\cos {\frac {π}{2n}}=\frac{\sqrt n}{2}$. Then

A) $6\le n\le8$, B) $4\lt n\le8$,

C) $4\le n\le 8$, D) $4 \lt n\lt8$,

I couldn't get started in solving it. That's why I asked help

Answer 1

Given $\sin\dfrac{\pi}{2n}+\cos\dfrac{\pi}{2n}=\dfrac{\sqrt n}{2}$

$\left(\sin\dfrac{\pi}{2n}+\cos\dfrac{\pi}{2n}\right)^2 = \left(\dfrac{\sqrt n}{2}\right)^2$

$\sin^2\dfrac{\pi}{2n}+\cos^2\dfrac{\pi}{2n}+2\sin\dfrac{\pi}{2n}\cos\dfrac{\pi}{2n}=\dfrac{n}{4}$

$1+\sin\dfrac{\pi}{n}=\dfrac{n}{4}$

$\sin\dfrac{\pi}{n}=\dfrac{n-4}{4}$

Since $-1\le\sin\theta\le1$

$\dfrac{n-4}{4}\le1$

$n\le8$

Also, $\dfrac{n-4}{4}\ge0$

$n\ge4$

So, we have $4\le n\le8$

Now lets check whether it satisfies $\sin\dfrac{\pi}{n}=\dfrac{n-4}{4}$

Lets take $n=4$ and we get that $\sin\dfrac{\pi}{4}\ne0$

and for $n=8$ we get that $$\sin\dfrac{\pi}{8}\ne1$$

Therefore, the answer is $4<n<8$

Answer 2

By C-S $$\frac{\sqrt{n}}{2}=\sin\frac{\pi}{2n}+\cos\frac{\pi}{2n}\leq\sqrt{(1+1)(\sin^2\frac{\pi}{2n}+\cos^2\frac{\pi}{2n})}=\sqrt2,$$ which gives $n\leq8.$

The equality does not occur, which says $n<8.$

Also, by squaring easy to see that $n>4,$ which gives the answer: $$4<n<8.$$

Answer 3

Let $$f(x)=\sin x+\cos x=2\sin\tfrac\pi 4\cos (x-\tfrac\pi 4)=\sqrt 2 \cos(x-\tfrac\pi 4).$$ (Brush up on trigonometric addition theorems!) and $$ g(n)=f(\tfrac\pi{2n}).$$

We are lookoing for positive integer solutions of $$\tag1g(n)=\frac{\sqrt n}2.$$ As $g(1)=\sqrt 2\cos\frac\pi4=1\ne\frac{\sqrt 2}2$ and $g(2)=\sqrt 2\cos 0=\sqrt 2\ne\frac{\sqrt 2}2$, we need only cosider $n$ with $n>2$. In that range, $g$ is a strictly decreasing function of $n$ because $f$ is strictly increasing on $(0,\frac{\pi}{4})$. As the right hand side of $(1)$ is strictly increasing as a function of $n$, we conclude that $(1)$ has at most one solution in positive integers. Incidentally, $$g(6)=\sqrt 2\cos\left(\frac\pi{12}-\frac\pi4\right)=\sqrt 2\cos\frac\pi6=\sqrt 2\cdot\frac{\sqrt 3}2=\frac{\sqrt 6}2$$ so that we can conclude:

The equation $\sin\frac\pi{2n}+\cos\frac \pi{2n}=\frac{\sqrt n}2$ has exactly one solution in positive integers, namely $n=6$.

The condition thus found is strictly stronger than each of the given options A,B,C,D.

Remark: Even if we allow $n$ to be any positive real, this would not introduce any additional solutions.