题目内容
已知函数f(x)=sinx+cosx,
(1)若f(x)=2f(-x),求
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.
(1)若f(x)=2f(-x),求
| cos2x-sinxcosx |
| 1+sin2x |
(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=
,
则
=
=
=
=
,
(Ⅱ)由题意知,F(x)=cos2x-sin2x+1+2sinxcosx
=cos2x+sin2x+1=
sin(2x+
)+1,
由-
+2kπ≤2x+
≤
+2kπ(k∈z)得
-
+kπ≤x≤
+kπ(k∈z),
由
+2kπ≤2x+
≤
+2kπ(k∈z)得,
+kπ≤x≤
+kπ(k∈z),
∴函数F(x)的单调递增区间为 [-
+kπ,
+kπ](k∈z),
单调递减区间为[
+kπ,
+kπ] (k∈z).
又∵f(x)=2f(-x),
∴sinx+cosx=2(cosx-sinx)且cosx≠0∴tanx=
| 1 |
| 3 |
则
| cos2x-sinxcosx |
| 1+sin2x |
| cos2x-sinxcosx |
| cos2x+2sin2x |
=
| 1-tanx |
| 1+2tan2x |
1-
| ||
1+2×(
|
| 6 |
| 11 |
(Ⅱ)由题意知,F(x)=cos2x-sin2x+1+2sinxcosx
=cos2x+sin2x+1=
| 2 |
| π |
| 4 |
由-
| π |
| 2 |
| π |
| 4 |
| π |
| 2 |
-
| 3π |
| 8 |
| π |
| 8 |
由
| π |
| 2 |
| π |
| 4 |
| 3π |
| 2 |
| π |
| 8 |
| 5π |
| 8 |
∴函数F(x)的单调递增区间为 [-
| 3π |
| 8 |
| π |
| 8 |
单调递减区间为[
| π |
| 8 |
| 5π |
| 8 |
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