已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)

已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
⑴求函数f(x)的最小正周期和图像的对称轴方程
⑵求函数f(x)在区间[-π/12,π/2]上的值域
pinglitan 1年前 已收到3个回答 举报

温柔小美 幼苗

共回答了12个问题采纳率:83.3% 举报

2sin(x-π/4)sin(x+π/4)
=2sin(x-π/4)cos[π/2-(x-π/4)]
=2sin(x-π/4) cos(π/4-x)
=2sin(x-π/4) cos(x-π/4)
=sin(2x-π/2)
=-sin(π/2-2x)
=-cos2x
∴ f(x)=cos2xcosπ/3+sin2xsinπ/3-cos2x
=-(cos2xcosπ/3-sin2xsinπ/3)
=-cos(2x+π/3)
T=2π/2=π
对称轴2x+π/3=kπ
x=(kπ-π/3)/2
做到这一步,相信下面的你会做了,
建议楼主多给点分,还有你的三角函数要好好的学习,高考肯定是免不了的

1年前

4

与天相依 幼苗

共回答了36个问题 举报

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos2xcosπ/3+sin2xsinπ/3+2(sinxcosπ/4-cosxsinπ/4)(sinxcosπ/4+cosxsinπ/4)
可以这样打开,不用记太多公式,但计算复杂。

1年前

2

matforce 幼苗

共回答了3个问题 举报

2sin(x-π/4)sin(x+π/4)
=2sin(x-π/4)cos[π/2-(x-π/4)]
=2sin(x-π/4) cos(π/4-x)
=2sin(x-π/4) cos(x-π/4)
=sin(2x-π/2)
=-sin(π/2-2x)
=-cos2x
∴ f(x)=cos2xcosπ/3+sin2xsinπ/3-cos2x
=-(cos2xcosπ/3-sin2xsinπ/3)
=-cos(2x+π/3)
T=2π/2=π

1年前

0
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