{3/[sin(160)]^2-1/[cos(160)]^2}*1/sin50=?

3 /(sin 160°)^2 -1 /(cos 160°)^2
=3 /(sin 20°)^2 -1 /(-cos 20°)^2
=(√3 /sin 20°+1 /cos 20°)(√3 /sin 20°+1 /cos 20°),
因为 √3 /sin 20°±1 /cos 20°
=(√3 cos 20°±sin 20°) /(sin 20°cos 20°)
=4 [ (√3 /2) cos 20°±(1/2) sin 20°] /(2 sin 20°cos 20°)
=4 (sin 60°cos 20°±cos 60°sin 20°) /sin 40°
=4 sin (60°±20°) /sin 40°,
所以 3 /(sin 160°)^2 -1 /(cos 160°)^2
=16 sin 80°sin 40°/(sin 40°)^2
=16 *(2 sin 40°cos 40°) / sin 40°
=32 cos 40°
=32 sin 50°.
所以 原式 =32.
= = = = = = = = =
1.先用平方差公式.
2.a /sin x ±b /cos x 是常考类型,要掌握.
3.辅助角公式.
4.注意分子分母前的系数.