If $A = {\sin(2\pi/7)} + \sin(4\pi/7) +\sin(8\pi/7)$ and $B = \cos(2\pi/7) + \cos(4\pi/7) + \cos(8\pi/7)$ ,Then $A^{2}+B^{2}$ is equal to $?$

Question

If $$A=\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$$ $$B=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}.$$ then $A^2+B^2$ is equal to?

I solve got that $A^{2}+B^{2} = 3 + 2(\cos(2\pi/7)+\cos(4\pi/7)+\cos(6\pi/7))$ From here I am not able to solve. Please help me to solve this problem.

Answer 1

$$B+Ai=e^{2i\pi/7}+e^{4i\pi/7}+e^{8i\pi/7}$$

$$B-Ai=e^{-2i\pi/7}+e^{-4i\pi/7}+e^{-8i\pi/7}$$

So $$B^2+A^2=(B+Ai)(B-Ai)=3+2\cos 2\pi/7 + 2\cos 4\pi/7+2\cos 6\pi/7$$

But $\cos (2k\pi/7) = \cos(-2k\pi/7)=\cos(2(7-k)\pi/7)$.

So this means that $B^2+A^2=3+\sum_{k=1}^{6}\cos(2k\pi/7)$.

But that sum $\sum_{k=0}^{6}\cos(2k\pi/7) = 0$, so you get that:

$$B^2+A^2=3+(-1)=2$$

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More generally, for $p\equiv 3\pmod{4}$ a prime, let:

$$A=\sum_{k=1}^{(p-1)/2} \sin\frac{2\pi k^2 i}{p}\\ B=\sum_{k=1}^{(p-1)/2} \cos\frac{2\pi k^2 i}{p}$$

Then $A+Bi=\sum_{k=1}^{(p-1)/2} e^{2\pi k^2 i/p}$, and:

$$\begin{align}A^2+B^2&=\frac{p-1}{2}+\sum_{1\leq j\neq k\leq \frac{p-1}{2}}\cos\left(\frac{2\pi(k^2-j^2)}{p}\right)\\ &=\frac{p-1}{2}+\frac{1}{4}\sum_{1\leq j,k<p;\,j\not\equiv \pm k}\cos\left(\frac{2\pi(k^2-j^2)}{p}\right)\end{align}\tag{1}$$

For any factorization $n\equiv ab\pmod{p}$ with $a\not\equiv \pm b$, you have a pair $j,k$ so that $j+k=a,j-k=b$. For every $n$, since either $n$ or $-n$ is a square, there are exactly $2$ factorizations $n=ab$ ith $a\equiv \pm b$. There are $p-1$ total factorizations, there are $p-3$ factorizations with $a\not\equiv \pm b$.

So you get:

$$A^2+B^2=\frac{p-1}{2}+\frac{p-3}{4}\sum_{n=1}^{p-1}\cos(2\pi ni/p)=\frac{p+1}{4}$$

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There's also some elementary Galois theory going on here. If $\zeta_p=e^{2\pi i/p}$. then the Galois group of $\mathbb Q[\zeta_p]$ over $p$ is $\mathbb Z/p\mathbb Z^{\times}\cong\mathbb Z/(p-1)\mathbb Z$.

The subgroup of squares modulo $p$ is of index $2$ in the Galois group. But $B+Ai$ is in the field fixed by this subgroup, so this means that $B+Ai$ be quadratic over $\mathbb Q$, and it is an algebraic integer. This means that $A^2+B^2$ must be an integer when $p\equiv 3\pmod{4}$.

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If $p\equiv 1\pmod 4$, then $A=0$ and $B+Ai=B$ is a real algebraic integer of degree $2$ over $\mathbb Q$. You can use (1) to get $B^2$ in terms of $B$.

For $n$ not a square modulo $p$, you get that there are $p-1$ distinct factorizations $n=ab$ with $a\not\equiv -b$. For $n$ a square modulo $p$, you get that there are $p-1$ solutions to $n=ab$, but $4$ of them have $a\equiv \pm b\pmod p$, so there are $p-5$ terms for $n$. This means:

$$B^2=\frac{p-1}{2}+\frac{p-1}{4}\sum_{n=1}^{p-1}\cos\frac{2\pi ni}{p} - \frac{4}{4}B$$

or:

$$B^2=\frac{p-1}{4}-B$$

This gives that $B=\frac{-1\pm\sqrt{p}}{2}$.

Not sure which root $B$ is in general, but it is $\frac{-1+\sqrt{p}}{2}$ for small $p\equiv 1\pmod{4}$.

Answer 2

hint

If $$z=e^{\frac {2i\pi}{7}} =e^{2ia}$$

then $A^2+B^2$ is the real part of the sum

$$3+2 (z+z^2+z^3)=3+2z\frac {1-z^3}{1-z} $$

which is

$$3+2\cos (4a)\frac {\sin (3a)}{\sin (a)} $$ $$=3+\frac {1}{\sin (a)}\Bigl(\sin (7a)-\sin (a )\Bigr)$$ $$=3-1=2.$$

Answer 3

$$A^2+B^2=3+2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)=$$ $$=3+\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}=$$

$$=3+\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}=2.$$ Done!