作业帮高一 数学 三角函数 请详细解答,谢谢! (4 20:57:55)若sin(3兀+乄)=1/4,求cos(兀+乄)/cos乄[cos(兀+乄)-1]+cos(乄-2兀)/[cos(乄+2兀).cos(乄+兀)+cos(-乄)]的值.
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![高一 数学 三角函数 请详细解答,谢谢! (4 20:57:55)若sin(3兀+乄)=1/4,求cos(兀+乄)/cos乄[cos(兀+乄)-1]+cos(乄-2兀)/[cos(乄+2兀).cos(乄+兀)+cos(-乄)]的值.](/uploads/image/z/563208-24-8.jpg?t=%E9%AB%98%E4%B8%80+%E6%95%B0%E5%AD%A6+%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0+%E8%AF%B7%E8%AF%A6%E7%BB%86%E8%A7%A3%E7%AD%94%2C%E8%B0%A2%E8%B0%A2%21++++%284+20%3A57%3A55%29%E8%8B%A5sin%283%E5%85%80%2B%E4%B9%84%29%3D1%2F4%2C%E6%B1%82cos%28%E5%85%80%2B%E4%B9%84%29%2Fcos%E4%B9%84%5Bcos%28%E5%85%80%2B%E4%B9%84%29-1%5D%2Bcos%28%E4%B9%84-2%E5%85%80%29%2F%5Bcos%28%E4%B9%84%2B2%E5%85%80%29.cos%28%E4%B9%84%2B%E5%85%80%29%2Bcos%28-%E4%B9%84%29%5D%E7%9A%84%E5%80%BC.)
高一 数学 三角函数 请详细解答,谢谢! (4 20:57:55)若sin(3兀+乄)=1/4,求cos(兀+乄)/cos乄[cos(兀+乄)-1]+cos(乄-2兀)/[cos(乄+2兀).cos(乄+兀)+cos(-乄)]的值.
高一 数学 三角函数 请详细解答,谢谢! (4 20:57:55)
若sin(3兀+乄)=1/4,求cos(兀+乄)/cos乄[cos(兀+乄)-1]+cos(乄-2兀)/[cos(乄+2兀).cos(乄+兀)+cos(-乄)]的值.
高一 数学 三角函数 请详细解答,谢谢! (4 20:57:55)若sin(3兀+乄)=1/4,求cos(兀+乄)/cos乄[cos(兀+乄)-1]+cos(乄-2兀)/[cos(乄+2兀).cos(乄+兀)+cos(-乄)]的值.
因为sin(3兀+乄)=1/4,所以sin(乄)=-1/4,因为[sin(乄)]^2+[cos(乄)]^2=1,所以cos(乄)=√15/4或-√15/4(4分之根号15)
cos(兀+乄)/cos乄[cos(兀+乄)-1]+cos(乄-2兀)/[cos(乄+2兀).cos(乄+兀)+cos(-乄)]
=-cos(乄)/cos乄[-cos(乄)-1]-cos(乄)/[-cos(乄).cos(乄)-cos(乄)]
=1/[cos(乄)+1]+1/[cos(乄)+1]
=2/[cos(乄)+1]
当cos(乄)=√15/4则
2/[cos(乄)+1]
=2/(√15/4+1)
=8/(√15+4)
=32-8√15
当cos(乄)=-√15/4则
2/[cos(乄)+1]
=2/(-√15/4+1)
=8/(-√15+4)
=32+√15