y=根号3sinx+cosx x∈[0,π/2] 求值域
问题描述：
y=根号3sinx+cosx x∈[0,π/2] 求值域
答
y=根号3sinx+cosx
=2(√3/2*sinx+1/2*cosx)
=2sin(x+π/6)
∵x∈[0,π/2]∴x+π/6∈[π/6,2π/3]
∴sin(x+π/6)∈[1/2,1]
∴函数值域为【1.2】
答
解
y=√3sinx+cosx
=2[sinx*√3/2+cosx*(1/2)]
=2[sinxcos(π/6)+cosxsin(π/6)]
=2sin(x+π/6)
x∈[0,π/2]
x+π/6∈[π/6,2π/3]
所以 sin(x+π/6)∈【1/2,1】
所以值域是[1, 2 ]
答
y=√3sinx+cosx
=2（√3/2*sinx+1/2*cosx ）
=2sin(2x+π/6）
因0≤x≤π/2
所以π/6≤2x+π/6≤5π/6
1/2≤sin(2x+π/6）≤1
1≤2sin(2x+π/6）≤2
所以y=√3sinx+cosx 值域
【1,2】