Question

已知函数f(x)=2(cosx)^2+cos(2x+π/3).若f(α)=√(3)/3+1,0<α<π/6,求sin2α的值

Answer 1

f(α)=2cos²α+cos(2α+π/3)
=cos2α+1+cos2αcosπ/3-sin2αsinπ/3
=3/2cos2α-√3/2 sin2α +1
=√3（sinπ/3 cos2α-cosπ/3 sin2α +1
=√3sin（π/3-2α）+1
√3sin（π/3-2α）+1=√3/3+1
sin（π/3-2α）=1/3
令A=argsin1/3
则sinA=1/3
cosA=√（1-sin²A）=2√2/3
∵ 0<α<π/6 ,0<2α<π/3 ∴ π/3＞π/3-2α＞0
sin（π/3-2α）=sinA
π/3-2α=A
2α=π/3-A
sin2α=sin（π/3-A）=sinπ/3 cos A-cosπ/3 sin A
=√3/2 * 2√2/3- 1/2 *1/3
=2√6/6 -1/6
=(2√6-1)/6

Answer 2

带入，等式两边化解求解，其中用到
(cosx)^2=（1+(cos2x)），余弦展开公式，到√3cos2α-sin2α=2/3,的时候要用到cos2α=√（1-(sin2α)^2）,带入解一元二次方程组即可，注意0<2α<π/3,最后计算出结果为（√6-1）/3.