Question

求(2cosπ/9+1)*tan2π/9-2sinπ/9的值,具体过程!!!

Answer 1

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