问题
解答题
设α∈(
|
答案
∵α∈(
,2π),3π 2
∴cosα>0,cos
<0,α 2
则原式=
=
+1 2 1 2
(1+cosα)1 2
=
+1 2 1 2 cos2α
=
(1+cosα)1 2
=|coscos2 α 2
|=-cosα 2
.α 2
设α∈(
|
∵α∈(
,2π),3π 2
∴cosα>0,cos
<0,α 2
则原式=
=
+1 2 1 2
(1+cosα)1 2
=
+1 2 1 2 cos2α
=
(1+cosα)1 2
=|coscos2 α 2
|=-cosα 2
.α 2