已知函数已知函数f﹙x﹚=√3s
(1) f(x)=√3(sinx)^+sinxcosx
= sinx(√3sinx+cosx)=2sinxsin(x+π/6),
f(25π/6)=2sin(25π/6)sin(25π/6+π/6)
=2sin(4π+π/6)sin(4π+π/3)=2sin(π/6)sin(π/3)
=2×(1/2)×√3/2=√3/2
(2）f(α/2)=2sin(α/2)sin(α/2+π/6)=(√3/2)-(1/4), 积化差得
cos(π/6)-cos(α+π/6)=(√3/2)-(1/4), ∴ cos(α+π/6)=1/4>0,
∴ sin(α+π/6)=√[1-(1/16)=√15/4, ...全部
(1) f(x)=√3(sinx)^+sinxcosx
= sinx(√3sinx+cosx)=2sinxsin(x+π/6),
f(25π/6)=2sin(25π/6)sin(25π/6+π/6)
=2sin(4π+π/6)sin(4π+π/3)=2sin(π/6)sin(π/3)
=2×(1/2)×√3/2=√3/2
(2）f(α/2)=2sin(α/2)sin(α/2+π/6)=(√3/2)-(1/4), 积化差得
cos(π/6)-cos(α+π/6)=(√3/2)-(1/4), ∴ cos(α+π/6)=1/4>0,
∴ sin(α+π/6)=√[1-(1/16)=√15/4, 又 α=(α+π/6)-(π/6),
∴ sinα=sin(α+π/6)cos(π/6)-cos(α+π/6)sin(π/6)
=(√15/4)×(√3/2)-(1/4)×(1/2)=(3√5-1)/8
说明: 我改为f(α/2)=(√3/2)-(1/4)计算的。
∵ 若f(α/2)=(1/4)-(√3/2), 则可得cos(α+π/6)=√3-(1/4)>1,这是不可能的。 。收起
