已知x属于[0.π/2],求y=3sinx+4cosx的值域

Question

已知x属于[0.π/2],求y=3sinx+4cosx的值域

Answer 1

【解】y=3sinx+4cosx=5sin(x+ω)
这里tanω=4/3,ω∈[0.π/2],
所以5sin(π/2+ω)≤ 5sin(x+ω)≤ 5
即:5cosω≤ 5sin(x+ω)≤ 5
3≤ 5sin(x+ω)≤ 5
所以y=3sinx+4cosx的值域是[3,5]