计算sinπ/12*sin5π/12我没有分了,知道的人就帮帮忙,

sinπ/12*sin5π/12
=sin(3π/12-2π/12)*sin(3π/12+2π/12)
=sin(π/4-π/6)*sin(π/4+π/6)
=(sin(π/4)*cos(π/6)-sin(π/6)*cos(π/4))*(sin(π/4)*cos(π/6)+sin(π/6)*cos(π/4))
=(√2/2*√3/2-1/2*√2/2）*(√2/2*√3/2+1/2*√2/2）
=（√6/4-√2/4）*（√6/4+√2/4）
=6/16-2/16
=1/4

sinπ/12*sin5π/12
=(1/2)[cos(π/12-5π/12)-cos(π/12+5π/12)]
=(1/2)[cos(-π/3)-cos(π/2)]
=(1/2)(1/2-0)
=1/4

sinπ/12*sin5π/12
=-1/2[cos(π/12+5π/12)-cos(π/12-5π/12)]
=-1/2[cos(π/2)-cos(-π/3)]
=-1/2[0-1/2]
=1/4
公式:
sinαsinβ = [cos(α-β)-cos(α+β)] /2
cosαcosβ = [cos(α+β)+cos(α-β)]/2
sinαcosβ = [sin(α+β)+sin(α-β)]/2
cosαsinβ = [sin(α+β)-sin(α-β)]/2