一条三角函数的值问题!急f(x)=√2sin(2x-π/4)x∈[π/8,3π/4]2x∈[π/4,3π/2]2x-π/4∈[0,5π/4]sin(2x-π/4)∈[-√2/2,1]f(x)=sin(2x-π/4)∈[-1,√2]2x-π/4∈[0,5π/4]sin(2x-π/4)∈[-√2/2,1]晕死…还没打完…2x-π/4∈[0,5π/4]

Question

一条三角函数的值问题!急
f(x)=√2sin(2x-π/4)
x∈[π/8,3π/4]
2x∈[π/4,3π/2]
2x-π/4∈[0,5π/4]
sin(2x-π/4)∈[-√2/2,1]
f(x)=sin(2x-π/4)∈[-1,√2]
2x-π/4∈[0,5π/4]
sin(2x-π/4)∈[-√2/2,1]
晕死…还没打完…
2x-π/4∈[0,5π/4] 怎么变成 sin(2x-π/4)∈[-√2/2,1]的？
他的值域和最大最小值又是什么？（其实我知道最大最小值…）
关键是加入sin后[0,5π/4]怎么变成[-√2/2,1]！还有值域阿！

Answer 1

2x-π/4∈[0,π]时 值域为[0,1]
2x-π/4∈[π,5π/4] 值域为[-√2/2,0]
两个并起来就行
画个图不就出来了...估计您是概念理解偏差了...
[0,5π/4]求正弦 就是[-√2/2,1]