作业帮sinπ/18sin5π/18 sin（- 65π/18 sin之间为乘法sinπ/18sin5π/18 sin（- 65π/18sin之间为乘法
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sinπ/18sin5π/18 sin（- 65π/18 sin之间为乘法sinπ/18sin5π/18 sin（- 65π/18sin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 sin之间为乘法sinπ/18sin5π/18 sin（- 65π/18sin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
sinπ/18sin5π/18 sin（- 65π/18 sin之间为乘法sinπ/18sin5π/18 sin（- 65π/18sin之间为乘法
计算：sinπ／12 sin5π／12
(sin5π/12-sinπ/12)(sin5π/12+sinπ/12)的值 急!
sinπ /14*sin3π /14*sin5π /14
sin5π/3=sin(2π-π/3)
(sinπ/12)-(sin5π/12)+(2sinπ/8)(sin3π/8)
sinπ/12-sin5π/12+2sinπ/8sin3π/8
sin(-5),sin5比较大小
(sin5π/12+cos5π/12)(sinπ/12-cosπ/12)
(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=?
sinπ /14*sin3π /14*sin5π /14答案是1/8
化简sin5π/4＋sin(﹣π)＋cos9π/4
sin5π/4cos4π/5tan11π/6sin乘cos乘tan
计算sinπ/12*sin5π/12我没有分了,知道的人就帮帮忙,
sin（-π/3）+2sin5/3π+3sin2/3的值等于?
sin（-5）＜sin3＜sin5 ,
sin3＜sin（-5）＜sin5,
sin 1’ ,sin5’ ,sin10' ,sin20' 分别是多少?