我们知道:sin18°=(5~(1/2)-1)/4,通常它是通过sin(2×18°)=cos(3×18°)利用二倍角、三倍角公式展开后解方程求得的。以下我们介绍sin18°值的另外三种求法。方法一(几何法) 如图等腰三角形ABC,两底角为72°,顶角为36°,并设腰长为a,底边长为c,过C作底边AB的垂线,则∠BCB=180°在Rt△BCD中,就有 We know that sin18°=(5~(1/2)-1)/4, usually it is solved by using sin(2×18°)=cos(3×18°) to solve the equation by using the double angle and triple angle formulas. Obtained. In the following we introduce three other methods of sin18°. Method one (geometric method) As the isosceles triangle ABC, the two base angles are 72°, the vertex angle is 36°, and the waist length is a, the base length is c, and C is the vertical line of the base AB. ∠BCB=180° In Rt △BCD, there is