(1)
定义域:x∈R
值域:-1≤sin(π/4-x)≤1-2≤2sin(π/4-x)≤2
值域为[-2,2]
y=2sin(π/4-x)=-2sin(x-π/4)
x-π/4∈[π/2+2kπ,3π/2+2kπ]x∈[3π/4+2kπ,7π/4+2kπ]k∈z
单调递增区间[3π/4+2kπ,7π/4+2kπ]k∈z
(2)y=log1/2底sinx
定义域sinx>0x∈(2kπ,π+2kπ)
值域:0
x-π/4∈[π/2+2kπ,3π/2+2kπ]??为啥啊
y=-2sin(x-π/4),当sin(x-π/4)单调减的时候整个复合函数单调递增(因为前面的负号)也就是要求sin(x-π/4)的减区间了,正弦函数的减区间为[π/2+2kπ,3π/2+2kπ](可以看一下正弦函数的图像帮助一下)也就是x-π/4∈[π/2+2kπ,3π/2+2kπ]为复合函数的增区间,解出x为[3π/4+2kπ,7π/4+2kπ]k∈z有帮助请采纳,祝你学习进步,谢谢