Question

设函数f(x)=cos(x+2派/3)+2cos方x/2(x属于r)

求f(x)的最大值 设三角形ABC的内角ABC的对边长分别为abc若f(B)=1b=1c=根号3求a的值

Answer 1

f(x)=-1/2*cosx-√3/2*sinx+cosx+1=1/2*cosx-√3/2*sinx+1=cos(x+π/3)+1
0<=f(x)<=2
f(x)最大 2

f(B)=cos(B+π/3)+1=1
B=π/6
sinB=1/2
sinC=c*sinB/b=√3/2
C=2π/3 或π/3
A=π/6或π/2
a=1或2

Answer 2

f(B)=cos(B+2π/3)+2cos²B/2=1推出cos(B+2π/3)= -cosB所以B+2π/3+B=π推出B=π/6

cosB=(a²+c²-b²)/2*a*c=（a²+3-1)/2*a*根号3=根号3/2推出a=1或2~~

Answer 3

①2cos