Question

已知函数f(x)=sin^2+根号3sinxcosx-1/2（1）求f(-π/12)的值（2）若x∈{0，π/2}，求函数y=f(x)的最小值及

最小值时的x值

Answer 1

解：f(x)=(1-cos2x)/2+√3sinxcosx-1/2.
=√(3/2)sin2x-(1/2)cos2x.
=sin2xcosπ-cos2xsinπ.
=sin(2x-π/6).
(1) f(-.π/12)=sin(-2*π/12-π/6).
=sin(-π/3).
=-√3/2.
(2)∵x∈[0,π/2] ∴-π/6≤2x-π/6≤5π/6.
f(x)=sin(2x--π/6)=sin(-π/6)=-sinπ/6=-1/2.
=-1/2.
∴当x=0时,f(x)min=-1/2.