已知cos(x+y)=1/3,cos(x-y)+2/3,且0
问题描述：
已知cos(x+y)=1/3,cos(x-y)+2/3,且0
答
cos(x+y)=1/3,所以x+ycos(x+y)=1/3,cos(x-y)+2/3 可知sin(x+y)=2√2/3,sin(x-y)=√5/3.
cos2x=cos[(x+y)-(x-y)]=cos(x+y)cos(x-y)+sin(x+y)sin(x-y)=1/3x2/3+2√2/3x√5/3=(2+2√10)/9
cos(x-y)/cos(x+y)=2，用和差化积公式展开分子分母，再同时除以cosx*cosy有：
(1+tanx*tany)/(1-tanx*tany)=2.
最后得tanx*tany=1/3
答
cos(x+y)=1/3,cos(x-y)=2/3.0＜x＜90º,60º＜y＜90º∴可得:0＜x＜30＜60＜y＜90ºsin(x+y)=(2√2)/3.cos(x+y)=1/3.sin(y-x)=(√5)/3,cos(y-x)=2/3.[[[1]]]cos2x=cos[(y+x)-(y-x)]=cos(x+y)cos(y-x...