浪里白迢
花朵
共回答了19个问题采纳率:94.7% 举报
f(x)=5sin xcos x-5√3cos²x+5√3 /2
=(5/2)sin2x-(5√3/2)(1+cos2x)+5√3 /2
=√[(5/2)²+(5√3/2)]sin(2x+ψ)
=5sin(2x+ψ)(tanψ=√3,ψ=π/3)
1,∴T(+)min=2π/2=π
2,因为正弦函数在区间(-π/2+2kπ,π/2+2kπ)是增函数
所以由-π/2+2kπ≤2x+π/3≤π/2+2kπ得:
增区间为:kπ-5π/12≤x≤kπ+π/12
又正弦函数在区间(π/2+2kπ,3π/2+2kπ)是减函数
所以由π/2+2kπ≤2x+π/3≤3π/2+2kπ得:
增区间为:kπ+π/12≤x≤kπ+7π/12
3,由2x+π/3=0得:x=-π/6
故对称轴为:x=-π/6
对称中心为:(-π/6,0)
1年前
7