sin(x+π/3)=-sin(x+π/3-π)=-sin(x-2π/3)
f(x)=cos(x+π/6)cos(x-2π/3)+sin(x+π/6)sin(x+π/3)=cos(x+π/6)cos(x-2π/3)-sin(x+π/6)sin(x-2π/3)
=cos[(x+π/6)+(x-2π/3)]
=cos(2x-π/2)
cosx单调递减区间为[2kπ,2kπ+π]
∴f(x)单调递减区间为[kπ+π/4,kπ+3π/4]
f(x)=cos(x+π/6）cos(x-2π/3)+sin(x+π/6)sin(x+π/3) 求函数F（X）的单调递减
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