sinx+siny=-sinz (1)
cosx+cosy=-cosz (2)
(1)^2+(2)^2
=>2+2sinxcosx+2cosxcosy=1
=>cos(x-y)=cosxcosy+sinxsiny=-1/2
已知sinx+siny+sinz等于0,cosx+coxy+cosz等于0,则cos(x-y)的值为?
已知sinx+siny+sinz等于0,cosx+coxy+cosz等于0,则cos(x-y)的值为?
数学人气:307 ℃时间:2020-04-11 07:28:41
优质解答
我来回答
类似推荐