f(x)=cos(x+π/6)cos(x-2π/3)+sin(x+π/6)sin(x+π/3) 求函数F(X)的单调递减区间
题目
f(x)=cos(x+π/6)cos(x-2π/3)+sin(x+π/6)sin(x+π/3) 求函数F(X)的单调递减区间
答案
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[2k...
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