三角恒等式20°、40°

hbghlyj

$\sqrt3=\tan20°+4\sin20°=\cot 10°-4\cos10°$
$=\cot20°-\sec10°=\csc40°+\tan10°=4\sin40°-\tan40°$

hbghlyj

$3\csc^2 40°-\sec^2 40°=\frac{3\cos^240°-\sin^240°}{\sin^240°\cos^240°}=\frac{4\cos^240°-1}{\sin^240°\cos^240°}=\frac{2(2\cos^240°-1)+1}{\sin^240°\cos^240°}=\frac{2\cos80°+1}{\frac14\sin^280°}=\frac{2\sin10°+1}{\frac14\cos^210}=\sec^210°(8\sin10°+4)$
$3\csc^2 40°-\sec^2 40°=\frac{(\sqrt3\cos40°+\sin40°)(\sqrt3\cos40°-\sin40°)}{\sin^240°\cos^240°}=\frac{4\cos10°\sin20°}{\frac14\sin^280°}=\frac{16\cos10°2\sin10°\cos10°}{\cos^210°}=32\sin10°$
由上面导出一个恒等式：\[8 \sin (10 {}^{\circ}) \cos ^2(10 {}^{\circ})=2 \sin (10 {}^{\circ})+1\]
直接证明也很简单：
\[8 \sin (10 {}^{\circ}) \cos ^2(10 {}^{\circ})=4 \sin (20 {}^{\circ}) \cos (10 {}^{\circ})=2(\sin30°+\sin10°)=1+2\sin10°\]