证明：sinπ／9[（sinπ／9)+sin(3π／9)]=sin2(2π／9）

此题用到和差化积公式和倍角公式
和差化积公式：sin（a+b）+sin（a-b）=2sina*cosb,也可写成
sina+sinb=22sin（a+b）/2*cos(a-b)/2
倍角公sin（2a）=2sina*cosa
套用公式即可
解题如下：
sinπ／9[（sinπ／9)+sin(3π／9)]
=sin(π/9)* 2sin [(π/9+3π/9)/2] cos[(π/9-3π/9)/2
= sin(π/9)*2sin (2π/9)cos(π/9)
=2sin(π/9)cos(π/9) sin (2π/9)
=sin (2π/9)*sin (2π/9)
=sin 2 (2π/9)

sinπ／9[（sinπ／9)+sin(3π／9)]=sin20°（sin(40°-20°）+sin(40°+20°）】
=sin20°【sin40°cos20°-cos40°sin20°+sin40°cos20°+cos40°sin20°】
=sin20°*2sin40°cos20°=[sin40°】^2

证： 左边=sinπ/9[2sin(π/9+3π/9)/2]*[cos(π/9-3π/9)/2].
=sinπ/9*[2sin(2π/9)*cosπ/9,
=sin2π/9*(2sinπ/9cosπ/9.
=sin2π/9*sin2π/9.
=sin^2(2π/9).
∴左=右。
原题得证。

证明：sin(π/9)[sin(π/9)+sin(3π/9)]
=sin(π/9) 2*sin [(π/9+3π/9)/2] cos[(π/9-3π/9)/2] 和差化积
= 2sin(π/9) sin (2π/9) cos(π/9)，cos(π/9)=cos(-π/9) 偶函数
= 2sin(π/9)cos(π/9) * sin (2π/9)
= sin^2 (2π/9) ,^2表现 sin平方(2π/9)