求y=sin(x+π/6),x∈[-π/6,5π/6]值域求y=cos(-2x+π/6)单调递增区间今天就要

Question

求y=sin(x+π/6),x∈[-π/6,5π/6]值域
求y=cos(-2x+π/6)单调递增区间
今天就要

Answer 1

1、
-π/6<=x<=5π/6
所以
0<=x+π/6<=π
所以0<=sin(x+π/6)<=1
所以值域[0,1]
2、
y=cos(-2x+π/6)=cos(2x-π/6)
cosx的增区间是(2kπ-π,2kπ)
所以2kπ-π<2x-π/6<2kπ
2kπ-5π/6<2x<2kπ+π/6
kπ-5π/12所以增区间(kπ-5π/12,kπ+π/12)