Question

已知向量a=（4cosx，-1），向量b=（sin（x+π/3），根号3），函数f（x）=1/2*a*b

求f（x）的解析式和单调递增区间

Answer 1

f（x）=1/2*a*b
=(1/2)*[4sin(x+π/3)cosx-√3]
=2sin(x+π/3)cosx-√3/2
=sin(2x+π/3)+sin(π/3)-√3/2
=sin(2x+π/3)+√3/2-√3/2
所以解析式y=sin(2x+π/3)
单调递增区间2x+π/3∈[2kπ-π/2, 2kπ+π/2]
x∈[kπ-5π/12, kπ+π/12]
希望能帮到你，祝学习进步O(∩_∩)O