I just thought a way to find $\sin\frac{2π}{7}$.
Considering the equation $x^7=1$
$⇒(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0$
$⇒(x-1)[(x+\frac1 x)^3+(x+\frac1 x)^2-2(x+\frac1 x)-1]=0$
We can then get the 7 solutions of x, but the steps will be very complicated, especially when solving cubic equation, and expressing x as a+bi. The imaginary part of the second root of x will be $\sin\frac{2π}{7}$.
Besides this troublesome way, are any other approach? Thank you.
Just for laughs, we can at least in principle compute $\cos{(\pi/7)}$ by observing that
$$\sin{\frac{3 \pi}{7}} = \sin{\frac{4 \pi}{7}}$$
Using a combination of double-angle forumlae, we end up with a cubic equation for $\cos{(\pi/7)}$:
$$8 \cos^3{\frac{\pi}{7}} - 4 \cos^2{\frac{\pi}{7}} - 4 \cos{\frac{\pi}{7}}+1=0$$
This equation has one real solution which is $\cos{(\pi/7)}$. The bad news is that the expression is unwieldy at best:
$$\cos{\frac{\pi}{7}}=\frac{1}{6} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 i\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}\right)$$
The imaginary part of this expression is of course zero. The real part, however, ends up being expressed in terms of a sine and cosine of another angle, and I think the point of an exercise like this is to not do that. Anyway, I hope this adds to the discussion above.
Target=> $\sin\frac{2\pi}{7}$
set $$\sqrt[7]{1}=ζ^{k}_{7}$$ $$ζ^{k}_{7}=e^{\frac{2ki\pi}{7}}$$ as seventh roots of unity so, the statements applies $$ζ^{1}_{7}+ζ^{2}_{7}+ζ^{3}_{7}+ζ^{4}_{7}+ζ^{5}_{7}+ζ^{6}_{7}+1=0$$ $$ζ^{k}_{7}=\cos\frac{2k\pi}{7}\pm i\sin\frac{2k\pi}{7}$$ $$ζ^{-k}_{7}=\cos\frac{2k\pi}{7}\mp i\sin\frac{2k\pi}{7}$$ $$ζ^{k}_{7}+ζ^{-k}_{7}=2\cos\frac{2k\pi}{7}$$ $$ζ^{k}_{7}-ζ^{-k}_{7}=\pm2i\sin\frac{2k\pi}{7}$$ $$ζ^{2k}_{7}-2+ζ^{-2k}_{7}=-4\sin^2\frac{2k\pi}{7}$$ $$2\cos\frac{4k\pi}{7}-2=-4\sin^2\frac{2k\pi}{7}$$ $$\sqrt{2\cos\frac{4k\pi}{7}-2}=\pm2i\sin\frac{2k\pi}{7}$$ the $$ζ^{1}_{7}+ζ^{2}_{7}+ζ^{3}_{7}+ζ^{4}_{7}+ζ^{5}_{7}+ζ^{6}_{7}+1=0$$ gives us irreducible cubic polynomial $$t^3+t^2-2t-1=0$$ where $$t_{k}=2\cos\frac{2k\pi}{7}$$ applying lagrange resolvent for cubic $$-1=t_{1}+t_{2}+t_{3}$$ $$r_1=t_{1}+ζ^{1}_{3}t_{2}+ζ^{2}_{3}t_{3}$$ $$r_2=t_{1}+ζ^{2}_{3}t_{2}+ζ^{1}_{3}t_{3}$$ where $ζ^{2}_{3}$ is the cuberoot of unity, Applying Discrete Fourier Transform inverse $$3t_{1}=-1+r_1+r_2$$ $$3t_{2}=-1+ζ^{2}_{3}r_1+ζ^{1}_{3}r_2$$ $$3t_{3}=-1+ζ^{1}_{3}r_1+ζ^{2}_{3}r_2$$ on the action of lagrange method, lagrange resolvent gives $$r^3_1=7\left(3ζ^{1}_{3}-1\right)$$ $$r^3_2=7\left(3ζ^{2}_{3}-1\right)$$ now... $$r_1=\sqrt[3]{7\left(3ζ^{1}_{3}-1\right)}$$ $$r_2=\sqrt[3]{7\left(3ζ^{2}_{3}-1\right)}$$ $$ζ^{1}_{3}=\frac{-1+\sqrt{-3}}{2}$$ $$ζ^{2}_{3}=\frac{-1-\sqrt{-3}}{2}$$ $$r_1=\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}$$ $$r_1=\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}$$ $$3t_{1}=-1+r_1+r_2$$ $$3t_{2}=-1+ζ^{2}_{3}r_1+ζ^{1}_{3}r_2$$ $$3t_{3}=-1+ζ^{1}_{3}r_1+ζ^{2}_{3}r_2$$ $$3t_{1}=-1+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}$$ $$3t_{2}=-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}$$ $$3t_{3}=-1+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}$$ $$2\cos\frac{2\pi}{7}=\frac{-1+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}}{3}$$ $$2\cos\frac{4\pi}{7}=\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}}{3}$$ $$2\cos\frac{6\pi}{7}=\frac{-1+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}}{3}$$ $$\sqrt{2\cos\frac{4k\pi}{7}-2}=\pm2i\sin\frac{2k\pi}{7}$$ $$\sqrt{t_{2k}-2}=2i\sin\frac{2k\pi}{7}$$ $$\sqrt{t_{2}-2}=2i\sin\frac{2\pi}{7}$$ $$\sqrt{t_{4}-2}=2i\sin\frac{4\pi}{7}$$ $$\sqrt{t_{6}-2}=2i\sin\frac{6\pi}{7}$$ $$t_{4}=t_{3}$$ $$t_{6}=t_{1}$$ $$\sqrt{t_{2}-2}=2i\sin\frac{2\pi}{7}$$ $$\sqrt{\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}}{3} -2}=2i\sin\frac{2\pi}{7}$$ $$\sqrt{\frac{-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}}{3}}=2i\sin\frac{2\pi}{7}$$ $$\sqrt{\frac{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}\right)}{9}}=2i\sin\frac{2\pi}{7}$$ $$\frac{\sqrt{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}\right)}}{3}=2i\sin\frac{2\pi}{7}$$ so, $$\sin\frac{2\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1-3\sqrt{-3}\right)}\right)}}{6}$$
Using this or Point#$24$ of this ,
$\sin 7x=7s-56s^3+112s^5-64s^7$ where $s=\sin x$
If $\sin 7x=0,7x=n\pi$ where $n$ is any integer.
So, $x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$
So, the roots of $7s-56s^3+112s^5-64s^7=0--->(1)$ are $\sin\frac{n\pi}7$ where $n=0,1,2,\cdots 5,6$
So, the roots of $64s^6-112s^4+56s^2-7=0--->(2)$ are $\sin\frac{n\pi}7$ where $n=1,2,\cdots 5,6$
So, the roots of $64t^3-112t^2+56t-7=0 --->(3)$ are $\sin^2\frac{n\pi}7$ where $n=1,2,4$ or $3,5,6$ as $\sin \frac{(7-r)\pi}7=\sin(\pi-\frac{r\pi}7)=\sin\frac{r\pi}7$
If we choose $n=1,2,4$ observe that $\sin^2\frac{4\pi}7-\sin^2\frac{2\pi}7=2\sin\frac{\pi}7\cos\frac{3\pi}7>0$ (Using $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$)
Similarly, $\sin^2\frac{2\pi}7-\sin^2\frac{\pi}7>0$
So, $\sin^2\frac{4\pi}7>\sin^2\frac{2\pi}7>\sin^2\frac{\pi}7$
Using Cardano's method, we can solve the Cubic equation $(3)$
