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

问题描述:

证明:sinπ/9[(sinπ/9)+sin(3π/9)]=sin2(2π/9)
等号后边是sin平方(2π/9)
1个回答 分类:数学 2014-09-30

问题解答:

我来补答
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)
 
 
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