(1)已知函數(shù)f(x)= -(√3)sin²x+sinxcosx,求f(25π/6).
f(x)= -(√3)sin²x+sinxcosx=(√3/2)(cos2x-1)+(1/2)sin2x
=cos2xcos(π/6)+sin2xsin(π/6)-(√3/2)=cos(2x-π/6)-(√3/2)；
故f(25π/6)=cos(50π/6-π/6)-(√3)/2=cos(49π/6)-(√3)/2=cos(8π+π/6)-(√3)/2
=cos(π/6)-(√3)/2=(√3)/2-(√3)/2=0
(2)設(shè)α∈(0,π）,f(α/2)=1/4-(√3)/2,求sinα的值
f(α/2)=cos(α-π/6)-(√3/2)=1/4-(√3)/2,故cos(α-π/6)=1/4；sin(α-π/6)=√(1-1/16)=(1/4)√15；
cos(α-π/6)=(√3/2)cosα+(1/2)sinα=1/4.(1)
sin(α-π/6)=(√3/2)sinα-(1/2)cosα=(1/4)√15
即sin(α-π/6)=-(1/2)cosα+(√3/2)sinα=(1/4)√15.(2)
(1)+(√3)×(2)得2sinα=1/4+(1/4)√45=1/4+(3/4)√5=(1/4)(1+3√5)
故sinα=(1/8)(1+3√5)