解
m(-2sinx,cosx),n=(√3cox,2cosx)
f(x)=1-mn
=1-(-2√3sinxcosx+2cosxcosx)
=2√3sinxcosx-2cosxcosx+1
=√3sin2x-1-cos2x+1
=2sin(2x-π/6)
令2kπ-π/2≤2x-π/6≤2kπ+π/2 得 kπ-π/6≤x≤kπ+π/3
f(x)单调递增区间为( kπ-π/6,kπ+π/3)
f(x)的图像可以由g(x)=sinx向右平移π/6得到y=sinx-π/6,再令纵坐不变,横坐标变成原来的1/2倍得到
y=sin(2x-π/6,),再令横坐标不变,纵坐标变成原来的2倍得到y=2sin(2x-π/6)