Question

求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间

Answer 1

sin(π/3+4x)=cos[π/2-(π/3+4x)]=cos(π/6-4x)

f(x)=sin(π/3+4x)+cos(4x-π/6)=2cos(4x-π/6)
最小正周期是：T=2π/4=π/2

2kπ=<4x-π/6<=2kπ+π
2kπ+π/6=<4x<=2kπ+7π/6
(kπ/2)+(π/24)=<x<=(kπ/2)+(7π/24)
递减区间是[(kπ/2)+(π/24)，(kπ/2)+(7π/24)]

Answer 2

展开=sin60cos4x+sin4xcos60+cos4xcos30+sin4xsin30
=/3/2cos4x+1/2sin4x+/3/2cos4x+1/2sin4x
=/3cos4x+sin4x
=2(sin60cos4x+cos60sin4x)
=2sin(4x+60)

则，最小正周期 T=2π/w=π/2
递减区间是[(kπ/2)+(π/24)，(kπ/2)+(7π/24)]

Answer 3

f(x)=sin(π/3+4x)+sin(4x-π/6)
=sin(π/3+4x)+cos(2x+π/3)
=√2sin(4x+7π/12)
=√2cos(4x+π/12)
最小正周期T=2π/4=π/2
2kπ<=4x+π/12<=2kπ+π
kπ/2-π/48<=x<=kπ/2+11π/48
递减区间 [kπ/2-π/48，kπ/2+11π/48] k∈Z

Answer 4

展开得,?3cos4x+sin4x
=2(?3/2cos4x+1/2sin4x)
=2sin(4x+π/6 )
T=π/2