Question

求y=sinxcosx+cosx的平方在x∈【－3分之π,3分之π】上的值域

Answer 1

【参考答案】
y=sinxcosx+(cosx)^2
=(1/2)sin2x+(1/2)(1+cos2x)
=(1/2)+(1/2)sin2x+(1/2)cos2x
=(1/2)+(√2
/2)[(√2
/2)sin2x+(√2
/2)cos2x]
=(1/2)+(√2
/2)sin(2x+
π/4)
当
x∈[-π/3,
π/3]时，
2x+
π/4∈[-5π/12,
11π/12]
sin(2x+
π/4)∈[-(√6+√2)/4,
1]
所以
y∈[(1-√3)/4,
(1+√2)/2]
满意敬请采纳，不理解欢迎追问