sinπ/18sin5π/18 sin(- 65π/18 sin之间为乘法
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/05 01:17:00
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
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
=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