题目内容

已知函数f(x)=sinx+cosx,
(1)若f(x)=2f(-x),求
cos2x-sinxcosx
1+sin2x
的值;
(2)设函数F(x)=f(x)•f(-x)+f2(x),试讨论函数F(x)的单调性.
(Ⅰ)∵f(x)=sinx+cosx,∴f(-x)=cosx-sinx.
又∵f(x)=2f(-x),
∴sinx+cosx=2(cosx-sinx)且cosx≠0∴tanx=
1
3

cos2x-sinxcosx
1+sin2x
=
cos2x-sinxcosx
cos2x+2sin2x

=
1-tanx
1+2tan2x
=
1-
1
3
1+2×(
1
3
)2
=
6
11

(Ⅱ)由题意知,F(x)=cos2x-sin2x+1+2sinxcosx
=cos2x+sin2x+1=
2
sin(2x+
π
4
)+1

-
π
2
+2kπ≤2x+
π
4
π
2
+2kπ
(k∈z)得
-
8
+kπ≤x≤
π
8
+kπ
(k∈z),
π
2
+2kπ≤2x+
π
4
2
+2kπ
(k∈z)得,
π
8
+kπ≤x≤
8
+kπ
(k∈z),
∴函数F(x)的单调递增区间为 [-
8
+kπ,
π
8
+kπ]
(k∈z),
单调递减区间为[
π
8
+kπ,
8
+kπ]
  (k∈z).
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