求sin12*sin24*sin48*sin96=?注意是度,不是弧度!

Question

求sin12*sin24*sin48*sin96=?
注意是度,不是弧度!

Answer 1

sin12*sin24*sin48*sin96
=（sin96*sin24）（sin48*sin12）
= -1/2[cos120-cos72]*(-1/2)[cos60-cos36]
= 1/4(-1/2-cos72)(1/2-cos36)
= 1/4(cos72+1/2)(cos36-1/2)
= 1/4cos72cos36+1/8(cos36-cos72)-1/16
= 1/4cos72cos36+1/8*(-2)sin54sin(-18)-1/16
= 1/4cos72cos36+1/4cos36cos72-1/16
=1/2cos36cos72-1/16
=1/4*(2sin36cos36cos72)/sin36-1/16
=1/4*sin72cos72/sin36-1/16
=1/8*sin144/sin36-1/16
=1/8-1/16
=1/16