题目

已知函数f(x)=2sin²(π/4+X)+√3cos2x-1,x∈R （1）求函数f（x）的最小正周期和单调增区间
（2）在三角形ABC中,若f(C)=√3,2sinB=cos(A-C)-cos(A+C),求tanA的值

提问时间：2021-01-12

答案

f(x) = 2sin²(π/4+x)+√3cos2x-1
= {1-cos[2(π/4+x)] + √3cos2x - 1
= -cos(π/2+2x) + √3cos2x
= -sin2x + √3cos2x
= -2(sin2xcosπ/3 - cos2xsinπ/3)
= -2sin(2x-π/3)
最小正周期 = 2π/2 = π
2x-π/3∈（2kπ+π/2,2kπ+3π/2）,其中k∈Z时单调增
单调增区间：（kπ+5π/12,kπ+11π/12）,其中k∈Z-2sin(2x-π/3)
C∈（0,π）
2C-π/3∈（-π/3,5π/3）
f(C)=√3
-2sin(2x-π/3)=√3
sin(2x-π/3)=-√3/2
2C-π/3=4π/3
C=5π/6
B=π-(A+C)
sinB=sin(A+C)
2sinB=cos(A-C)-cos(A+C)
2sin(A+C)=cos(A-C)-cos(A+C)
2sinAcosC+2cosAsinC = cosAcosC+sinAsinC-(cosAcosC-sinAsinC)
2sinAcosC+2cosAsinC = 2sinAsinC
-√3sinA+cosA= sinA
(√3+1)sinA=cosA
tanA = 1/(√3+1) = (√3-1)/2