Question

已知函数f(x)=-根号3sin平方x+sinxcosx,求f(25π/6)，

(2)设a属于（0，π），f(a/2)=1/4-根号3/2,求sina的值

Answer 1

(1)已知函数f(x)= -(√3)sin²x+sinxcosx，求f(25π/6)。
解：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)设α∈(0，π），f(α/2)=1/4-(√3)/2，求sinα的值
解：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)

Answer 2

f(x)=-√3sin²x+sinxcosx
=-√3(1-cos2x)/2+(1/2)sin2x
=(1/2)sin2x+(√3/2)cos2x-√3/2
=sin(2x+π/3)-√3/2
(1)
f(25π/6)=sin(2*25π/6+π/3)-√3/2=sin(26π/3)-√3/2=sin(2π/3)-√3/2=sin(π/3)-√3/2=√3/2-√3/2=0
(2)
f(a/2)=1/4-√3/2
所以sin(2*a/2+π/3)-√3/2=1/4-√3/2
所以sin(a+π/3)=1/4
因为a∈(0,π)
所以a+π/3∈(π/3,4π/3)
又sin(a+π/3)=1/4
可以知道a+π/3必然是钝角
所以cos(a+π/3)=-√[1-(1/4)²]=-√15/4

所以sina=sin[(a+π/3)-π/3]=sin(a+π/3)cos(π/3)-cos(a+π/3)sin(π/3)=(1/4)*(1/2)-(-√15/4)*(√3/2)=(1+3√5)/8