题目内容：

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
1) 求f(x)在(0,π/2)上的值域
2) 记三角形ABC的内角A,B,C的对边长分部为a,b,c,若f(A)=1,a=根号7,b=3,求c的值

优质解答

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