Question

设函数f(x)=cos(2x-π/3)-cos2x,x属于R,求f(x)在(0,π/2)上的值域

Answer 1

1) f(x)=cos(2x-π/3)-cos2x=-2sin(2x-π/6)sin(-π/6)=sin(2x-π/6) 和差化积

x属于(0,π/2),(2x-π/6）属于(-π/6,5π/6 ) f(x)属于( -sinπ/6 sinπ/6),即 (-1/2 1/2)

2）若f(A)=1,即sin(2A-π/6)=1,2A-π/6= π/2 A= π/3

余弦定理 a^2=b^2+c^2+2bccosA c^2+3c+2=0 c=1,2