已知函数f(x)=sin(三分之pai+4x)+cos(4x-六分之pai)一、求f(x)的最小正周期和单调

f(x)=sin(π/3+4x)+cos(4x-π/6)
=sin(π/3+4x)+sin[π/2-(4x-π/6)]
=sin(π/3+4x)+sin(π/3-4x)
=2sin(π/3)cos4x
=√3cos4x
最小正周期=2π/4=π/2
2kπ<4x<π+2kπ减
kπ/2
π+2kπ<4x<2π+2kπ 增
π/4+kπ/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)]

f(x)=sin(π/3+4x)+cos(4x-π/6)
=sin(π/3+4x)-sin[π/2+(4x-π/6)]
=sin(π/3+4x)-sin(π/3+4x)
=0

把两个三角函数分别用两角和公式展开得
f(x)=sin4x/2+cos4x*根号3/2+cos4x*根号3/2+sin4x/2
=2*(sin4xcos60+cos4xsin60)
然后积化和差公式