sinπ/18sin5π/18 sin（- 65π/18 sin之间为乘法

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/17 00:56:47

sinπ/18sin5π/18 sin（- 65π/18 sin之间为乘法
sinπ/18sin5π/18 sin（- 65π/18
sin之间为乘法

sinπ/18sin5π/18 sin（- 65π/18)
=cos(π/2-π/18)*cos(π/2-5π/18) sin（- 72π/18+7π/18)
=cos(4π/9)cos(2π/9)*sin(7π/18)
=cos(2π/9)*cos(4π/9)cos(π/2-7π/18)
=cos(π/9)*cos(2π/9)cos(4π/9)
=2sin(π/9)cos(π/9)*cos(2π/9)cos(4π/9)/[2sin(π/9)]
=2sin(2π/9)*cos(2π/9)cos(4π/9)/[4sin(π/9)]
=2sin(4π/9)*cos(4π/9)/[8sin(π/9)]
=sin(8π/9)/[8sin(π/9)]
=sin(π/9)/[8sin(π/9)]
=1/8
再问：多谢了
再答： OK

sinπ18sin5π18sin（-65π18）=（　　） (sin5π12−sinπ12)(sin5π12+sinπ12) sinπ /14*sin3π /14*sin5π /14 比较大小sin(-π /18)与sin(-π /10) (sin5π/12+cos5π/12)(sinπ/12-cosπ/12) (sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=? 已知角α终边上一点坐标为(sin5π/6,cos5π/6)求sinα和cosα的值计算 sin5π/8*cos（7/8）π+sinπ/8-cos（3/8）π 数学极限的运算lim α→π/2 （sinα-sin5α）/（cosα+cos5α） cos sin π √ 计算：sinπ12 若sin（π3