题目

1:函数y=sin^x+根号3sin(π+x)cosx+1/2 求最小正周期
2:y=-acos2x-√3asin2x+2a+b
＝-2asin(2x+π/6)+2a+b
0

提问时间：2021-01-01

答案

y=sin^x+√3sin(π+x)cosx+1/2
=(1-cos2x)/2-√3sinxcosx+1/2
=-1/2*cos2x-√3/2*sin2x+1
=-(1/2*cos2x+√3/2*sin2x)+1
=-(sinπ/6*cos2x+√cosπ/6*sin2x)+1
=-sin(2x+π/6)+1
T=2π/2=π
sin7π/6=sin(π+π/6)=-sinπ/6=-1/2
y=-acos2x-√3asin2x+2a+b
=-2a(1/2*cos2x+√3/2*sin2x)+2a+b
=-2a(sinπ/6cos2x+cosπ/6*sin2x)+2a+b(两角和公式)
=-2asin(2x+π/6)+2a+b