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Author |
hbghlyj
Posted 2020-6-18 22:54
Last edited by hbghlyj 2020-6-18 23:05$3\csc^2 40°-\sec^2 40°=\frac{3\cos^240°-\sin^240°}{\sin^240°\cos^240°}=\frac{4\cos^240°-1}{\sin^240°\cos^240°}=\frac{2(2\cos^240°-1)+1}{\sin^240°\cos^240°}=\frac{2\cos80°+1}{\frac14\sin^280°}=\frac{2\sin10°+1}{\frac14\cos^210}=\sec^210°(8\sin10°+4)$
$3\csc^2 40°-\sec^2 40°=\frac{(\sqrt3\cos40°+\sin40°)(\sqrt3\cos40°-\sin40°)}{\sin^240°\cos^240°}=\frac{4\cos10°\sin20°}{\frac14\sin^280°}=\frac{16\cos10°2\sin10°\cos10°}{\cos^210°}=32\sin10°$
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：
\[8 \sin (10 {}^{\circ}) \cos ^2(10 {}^{\circ})=4 \sin (20 {}^{\circ}) \cos (10 {}^{\circ})=2(\sin30°+\sin10°)=1+2\sin10°\] |
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