Question

函数y=√3sinx+cosx,x∈[-π/2,π/2]的最大值是

请写出过程，谢谢

Answer 1

y=√3sinx+cosx
=2sin(x+A)，(其中tanA=1/√3，即A=π/6)
=2sin(x+π/6)

因为x∈[-π/2,π/2]
所以(x+π/6)∈[-π/3,(2π)/3]
所以y(max)=2

Answer 2

y=2sin(x+π/6) x=π/3时有最大值2。x=-π/2有最小值-√3