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

发布网友 发布时间：2024-10-23 22:30

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热心网友 时间：4分钟前

f(α)=2cos²α+cos(2α+π/3)
=c
os2
α+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

热心网友 时间：5分钟前

注：3^(1/2)为根号3
cos^2(π/12)为cos的平方π/121、f(x)=sinπ/3-cosπ/2+2cos^2(π/12)=sinπ/3-cosπ/2+2cos^2(π/3-π/4)=3^(1/2)-0+1+3^(1/2)=1+3^(1/2)
2、先化简：f(x)=sin2xcosπ/6+cos2xsinπ/6-cos2xcosπ/3+sin2xsinπ/3+1+cos2x=二分之根号三sin2x+1/2cos2x-1/2cos2x+二分之根号三sin2x+1+cos2x=根号三sin2x+cos2x+1=2sin(2x+π/6)+1
当sin(2x+π/6)等于1时取得最大值3
当
sin(2x+π/6)=1
等价于2x+π/6=2kπ+π/2
解得x=kπ+π/6
我写了这么多
怎么也得采纳吧
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