问题
填空题
已知0<y<x<π,且tanxtany=2,sinxsiny=
|
答案
由题意可得tanxtany=
=2,sinxsiny cosxcosy
解得cosxcosy=
,故cos(x-y)=cosxcosy+sinxsiny=1 6
+1 6
=1 3 1 2
故x-y=2kπ±
,k∈Z,π 3
又0<y<x<π,所以-π<x-y<π.
所以x-y=π 3
故答案为:π 3
已知0<y<x<π,且tanxtany=2,sinxsiny=
|
由题意可得tanxtany=
=2,sinxsiny cosxcosy
解得cosxcosy=
,故cos(x-y)=cosxcosy+sinxsiny=1 6
+1 6
=1 3 1 2
故x-y=2kπ±
,k∈Z,π 3
又0<y<x<π,所以-π<x-y<π.
所以x-y=π 3
故答案为:π 3