题目内容
已知函数f(x)=sin(
-x)+sinx.
(1)求函数y=f(x)的单调递增区间;
(2)若f(α-
)=
,求f(2α+
)的值.
| π |
| 2 |
(1)求函数y=f(x)的单调递增区间;
(2)若f(α-
| π |
| 4 |
| ||
| 3 |
| π |
| 4 |
(1)f(x)=sin(
-x)+sinx=cosx+sinx=
sin(x+
)
∵y=sinx在[-
,
]上单调递增,
∴-
≤x+
≤
整理得:-
≤x≤
∴f(x)在2kπ-
≤x≤2kπ+
(k∈Z)上单调递增.
(2)由(1)知f(x)=
sin(x+
)
∴f(α-
)=
sinα=
∴sinα=
f(2α+
)=
sin(2α+
)=
cos2α=
(1-2sin2α)=
×(1-2×
)=
| π |
| 2 |
| 2 |
| π |
| 4 |
∵y=sinx在[-
| π |
| 2 |
| π |
| 2 |
∴-
| π |
| 2 |
| π |
| 4 |
| π |
| 2 |
整理得:-
| 3π |
| 4 |
| π |
| 4 |
∴f(x)在2kπ-
| 3π |
| 4 |
| π |
| 4 |
(2)由(1)知f(x)=
| 2 |
| π |
| 4 |
∴f(α-
| π |
| 4 |
| 2 |
| ||
| 3 |
∴sinα=
| 1 |
| 3 |
f(2α+
| π |
| 4 |
| 2 |
| π |
| 2 |
| 2 |
| 2 |
| 2 |
| 1 |
| 9 |
7
| ||
| 9 |
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