sin3B
=sin(B+2B)
=sinBcos2B+cosBsin2B
=sinB(1-2sin²B)+cosB*2sinBcosB
=sinB-2sin³B+2sinBcos²B
=sinB-2sin³B+2sinB(1-sin²B)
=3sinB-4sin³B
所以原式=3-4sin²B