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.

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