Question

高中数学，三角函数求值（在线等待）

求值：sinA*sinB*sinC(其中A=∏/14,B=3∏/14,C=5∏/14)
答案为1/8
请较为详细的写出解答过程，十分感谢。

Answer 1

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

Answer 2

sinA*sinB*sinC
=cos6π/14*cos4π/14*cos2π/14
=[(cos6π/14*cos4π/14*cos2π/14)*2sin2π/14]/2sin2π/14
=[cos6π/14*cos4π/14*cos2π/14*2sin2π/14]/2sin2π/14
=[(cos6π/14*cos4π/14*sin4π/14]/2sin2π/14
=1/2*[cos6π/14*sin8π/14]/2sin2π/14
=1/2*[cos6π/14*sin(π-8π/14)]/2sin2π/14
=1/2*[cos6π/14*sin6π/14]/2sin2π/14
=1/4sin12π/14/2sin2π/14
=1/8sin(π-12π/14)/sin2π/14
=1/8sin2π/14/sin2π/14
=1/8