问题标题:
高一数学三角函数请详细解答,谢谢!(420:57:55)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.
问题描述:
高一数学三角函数请详细解答,谢谢!(420:57:55)
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.
刘臣勇回答:
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[-...
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