[Mastermander]

$\sin\theta=\sin\frac{2\pi}{13},\sin\frac{4\pi}{13},\sin\frac{6\pi}{13},\dots$

$\sin7\theta=-\sin6\theta$

$\sin4\theta\cos3\theta+\cos4\theta\sin3\theta=-2\sin3\theta\cos3\theta$

$2\sin2\theta\cos2\theta(4\cos^3\theta-3\cos\theta)+(1-2\sin^22\theta)(3\sin\theta-4\sin^3\theta)=-2(3\sin\theta-4\sin^3\theta)(\cos\theta)(4\cos^2\theta-3)$

$4\sin\theta\cos^2\theta(1-2\sin^2\theta)(1-4\sin^2\theta)+(1-8\sin^2\theta\cos^2\theta)(3\sin\theta-4\sin^3\theta)=-2\cos\theta(3\sin\theta-4\sin^3\theta)(1-4\sin^2\theta)$

Let $x=\sin\theta$

$4x(1-x^2)(1-2x^2)(1-4x^2)+(1-8x^2+8x^4)(3x-4x^3)=-2\sqrt{1-x^2}(3x-4x^3)(1-4x^2)$

$-64x^7+80x^5+32x^4-32x^3-24x^2+7x=-2\sqrt{1-x^2}(16x^5-16x^3+3x)$

$4096x^{14}-10240x^{12}-4096x^{11}+10496x^{10}+8192x^9-4992x^8-5888x^7+608x^6+1984x^5-336x^3+49x^2=-1024x^{12}+3072x^{10}-3456x^8+1792x^6-420x^4+36x^2$

$4096x^{14}-9216x^{12}-4096x^{11}+7424x^{10}+8192x^9-8448x^8-5888x^7-1184x^6+1984x^5+548x^4-336x^3+13x^2=0$

divided by $x^2=0$

$4096x^{12}-9216x^{10}-4096x^{9}+7424x^8+8192x^7-8448x^6-5888x^5-1184x^4+1984x^3+548x^2-336x^1+13=0$

เนื่องจาก $x=\sin\theta$ เป็นคำตอบและจากทฤษฎีสมการจึงได้

$\sin\frac{2\pi}{13}\sin\frac{4\pi}{13}\dots\sin\frac{24\pi}{13}=\frac{13}{4096}$

$\Big(\sin\frac{2\pi}{13}\sin\frac{4\pi}{13}\dots\sin\frac{12\pi}{13}\Big)^2=\frac{13}{4096}$

$\sin\frac{2\pi}{13}\sin\frac{4\pi}{13}\dots\sin\frac{12\pi}{13}=\frac{\sqrt{13}}{64}$