首先利用 sinA=cos(π/2 -A).cosA=-cos(π -A)
你的题目中
sinA*sinB*sinC=cos(3π/7)cos(2π/7)cos(π/7)
=-cos(4π/7)cos(2π/7)cos(π/7)
=-cos(4π/7)cos(2π/7)cos(π/7)*sin(π/7)/sin(π/7)
=-sin(8π/7)/8sin(π/7)
=1/8
首先利用 sinA=cos(π/2 -A).cosA=-cos(π -A)
你的题目中
sinA*sinB*sinC=cos(3π/7)cos(2π/7)cos(π/7)
=-cos(4π/7)cos(2π/7)cos(π/7)
=-cos(4π/7)cos(2π/7)cos(π/7)*sin(π/7)/sin(π/7)
=-sin(8π/7)/8sin(π/7)
=1/8