Question

已知函数f x cos (π/3+x)cos(π/3-x)-sinxcosx+1/4 求单调递增区间

Answer 1

设F(X)=g(x)+h(x)+1/4
g(x)=cos(π/3+x)cos(π/3-x)
=[cosπ/3cosx-sinπ/3sinx][cosπ/3cosx+sinπ/3sinx]
=[1/2cosx-√3/2sinx][1/2cosx+√3/2sinx]
=1/4cos²x-3/4sin²x
=1/4cos²x-3/4(1-cos²x)
=cos²x-3/4
=(cos2x+1)/2-3/4
=1/2cos2x-1/4
h(x)=1/2(2sinxcosx)=1/2cos2x
所以F(
X)=cos2x
2kπ-π<=2x<=2kπ
kπ-π/2<=x<=kπ
k属于整数
即单调递增区间为[kπ-π/2,kπ]
k属于整数