求(2cosπ/9+1)*tan2π/9-2sinπ/9的值,
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求(2cosπ/9+1)*tan2π/9-2sinπ/9的值,
求(2cosπ/9+1)*tan2π/9-2sinπ/9的值,
求(2cosπ/9+1)*tan2π/9-2sinπ/9的值,
显然tan2π/9=(sin2π/9) / (cos2π/9),
而sin2π/9= 2sinπ/9 * cosπ/9,
所以
(2cosπ/9+1)*tan2π/9 -2sinπ/9
=(2cosπ/9+1)* 2sinπ/9 * cosπ/9 / (cos2π/9) - 2sinπ/9
= 2sinπ/9 * [(2cosπ/9+1) * cosπ/9 / (cos2π/9) - 1]
= 2sinπ/9 * (2cos²π/9+cosπ/9 -cos2π/9) / (cos2π/9)
由公式cos2x= 2cos²x -1可以知道,2cos²π/9 -cos2π/9=1
故
原式可以化简为：2sinπ/9 * (1+cosπ/9) / (cos2π/9)
而1+cosπ/9=2cos²π/18=2cosπ/18 *sin4π/9=4cosπ/18 *sin2π/9 *cos2π/9,
故原式
=2sinπ/9 * (1+cosπ/9) / (cos2π/9)
=8sinπ/9 *cosπ/18 *sin2π/9
而由积化和差公式sinαsinβ= -[cos(α+β) -cos(α-β)]/2,
可以得到2sinπ/9 *sin2π/9= cosπ/9 -cosπ/3,故
8sinπ/9 *cosπ/18 *sin2π/9
=4(cosπ/9 -cosπ/3) *cosπ/18 代入cosπ/3=1/2
= -2cosπ/18 +4cosπ/9 *cosπ/18
而cosαcosβ=[cos(α+β)+cos(α-β)]/2,
故2cosπ/9 *cosπ/18= cosπ/6 +cosπ/18
即
-2cosπ/18 +4cosπ/9 *cosπ/18
= -2cosπ/18 + 2cosπ/6 + 2cosπ/18
= 2cosπ/6
= √3
所以
计算得到
(2cosπ/9+1)*tan2π/9-2sinπ/9
= √3