Noelanijd

2022-07-23

$f\left(\frac{2\pi }{7}\right)+f\left(\frac{4\pi }{7}\right)+f\left(\frac{6\pi }{7}\right)=1$
let
$f\left(x\right)=\frac{1}{1+2\mathrm{cos}x}$
prove that :
$f\left(\frac{2\pi }{7}\right)+f\left(\frac{4\pi }{7}\right)+f\left(\frac{6\pi }{7}\right)=1$
My Try :
$f\left(\frac{2\pi }{7}\right)=\frac{1}{1+2\mathrm{cos}\left(\frac{2\pi }{7}\right)}$
$f\left(\frac{2\pi }{7}\right)=\frac{1}{1+2\mathrm{cos}\left(\frac{4\pi }{7}\right)}$
$f\left(\frac{2\pi }{7}\right)=\frac{1}{1+2\mathrm{cos}\left(\frac{6\pi }{7}\right)}$
$L=\frac{1}{1+2\mathrm{cos}\left(\frac{6\pi }{7}\right)}+\frac{1}{1+2\mathrm{cos}\left(\frac{4\pi }{7}\right)}+\frac{1}{1+2\mathrm{cos}\left(\frac{2\pi }{7}\right)}$
what now ?

Do you have a similar question?

Expert

Use factor ${z}^{7}-1$ into linear and quadratic factors and prove that $\mathrm{cos}\left(\pi /7\right)\cdot \mathrm{cos}\left(2\pi /7\right)\cdot \mathrm{cos}\left(3\pi /7\right)=1/8$
or if $7x=2m\pi$ where $m$ is any integer
$\mathrm{sin}4x=\mathrm{sin}\left(2m\pi -3x\right)=-\mathrm{sin}3x$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}4\mathrm{sin}x\mathrm{cos}x\mathrm{cos}2x=\mathrm{sin}x\left(4{\mathrm{sin}}^{2}x-3\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}4\mathrm{sin}x\mathrm{cos}x\left(2{\mathrm{cos}}^{2}x-1\right)=\mathrm{sin}x\left\{4\left(1-{\mathrm{cos}}^{2}x\right)-1\right\}$
So, the roots of
are $7x=2m\pi$ where $m\equiv ±1,±2,±3\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}7\right)$
Now if $u=\frac{1}{1+2\mathrm{cos}x}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}x=?$
Replace the values of $\mathrm{cos}x$ in $\left(1\right)$

Still Have Questions?

Levi Rasmussen

Expert

Let $2\mathrm{cos}\frac{2\pi }{7}=a$, $2\mathrm{cos}\frac{4\pi }{7}=b$ and $2\mathrm{cos}\frac{6\pi }{7}=c$
Hence,
$a+b+c=\frac{2\mathrm{sin}\frac{\pi }{7}\mathrm{cos}\frac{2\pi }{7}+2\mathrm{sin}\frac{\pi }{7}\mathrm{cos}\frac{4\pi }{7}+2\mathrm{sin}\frac{\pi }{7}\mathrm{cos}\frac{6\pi }{7}}{\mathrm{sin}\frac{\pi }{7}}=$
$=\frac{\mathrm{sin}\frac{3\pi }{7}-\mathrm{sin}\frac{\pi }{7}+\mathrm{sin}\frac{5\pi }{7}-\mathrm{sin}\frac{3\pi }{7}+\mathrm{sin}\frac{7\pi }{7}-\mathrm{sin}\frac{5\pi }{7}}{\mathrm{sin}\frac{\pi }{7}}=-1;$
$ab+ac+bc=4\left(\mathrm{cos}\frac{2\pi }{7}\mathrm{cos}\frac{4\pi }{7}+\mathrm{cos}\frac{2\pi }{7}\mathrm{cos}\frac{6\pi }{7}+\mathrm{cos}\frac{4\pi }{7}\mathrm{cos}\frac{6\pi }{7}\right)=$
$=2\left(\mathrm{cos}\frac{6\pi }{7}+\mathrm{cos}\frac{2\pi }{7}+\mathrm{cos}\frac{6\pi }{7}+\mathrm{cos}\frac{4\pi }{7}+\mathrm{cos}\frac{4\pi }{7}+\mathrm{cos}\frac{2\pi }{7}\right)=-2$
and
$abc=8\mathrm{cos}\frac{2\pi }{7}\mathrm{cos}\frac{4\pi }{7}\mathrm{cos}\frac{6\pi }{7}=\frac{8\mathrm{sin}\frac{2\pi }{7}\mathrm{cos}\frac{2\pi }{7}\mathrm{cos}\frac{4\pi }{7}\mathrm{cos}\frac{8\pi }{7}}{\mathrm{sin}\frac{2\pi }{7}}=\frac{\mathrm{sin}\frac{16\pi }{7}}{\mathrm{sin}\frac{2\pi }{7}}=1.$
Id est,
$f\left(\frac{2\pi }{7}\right)+f\left(\frac{4\pi }{7}\right)+f\left(\frac{6\pi }{7}\right)=\sum _{cyc}\frac{1}{1+a}=\frac{\sum _{cyc}\left(a+1\right)\left(b+1\right)}{\prod _{cyc}\left(1+a\right)}=$
$=\frac{ab+ac+bc+2\left(a+b+c\right)+3}{1+a+b+c+ab+ac+bc+abc}=\frac{-2-2+3}{1-1-2+1}=1.$
Done!

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