题目内容
已知f(x)=2sin(| π |
| 4 |
| x |
| 2 |
| π |
| 4 |
| x |
| 2 |
| sin2x |
| 2cosx |
(I)若f(α)=
| ||
| 2 |
| π |
| 2 |
(II)若sin
| x |
| 2 |
| 4 |
| 5 |
| π |
| 2 |
分析:(1)先对f(x)进行化简得到f(x)=
sin(x+
),再由f(α)的值及α的范围求出α的值.
(2)先由x的范围确定
的范围,进而可得cos
和sinx的值,最终求出答案.
| 2 |
| π |
| 4 |
(2)先由x的范围确定
| x |
| 2 |
| x |
| 2 |
解答:解:(I)f(x)=2sin(
+
)cos(
+
)+
=sin(
+x)+sinx=sinx+cosx
=
sin(x+
)
由f(α)=
,得
sin(α+
)=
∴sin(α+
)=
∵α∈(-
,0)
∴α+
∈(-
,
)
∴α+
=
,∴α=-
(7分)
(II)∵x∈(
,π),∴
∈(
,
)
又sin
=
,∴cos
=
∴sinx=2sin
cos
=
,cosx=-
=-
∴f(x)=sinx+cosx=
-
=
| π |
| 4 |
| x |
| 2 |
| π |
| 4 |
| x |
| 2 |
| sin2x |
| 2cosx |
=sin(
| π |
| 2 |
=
| 2 |
| π |
| 4 |
由f(α)=
| ||
| 2 |
| 2 |
| π |
| 4 |
| ||
| 2 |
∴sin(α+
| π |
| 4 |
| 1 |
| 2 |
∵α∈(-
| π |
| 2 |
∴α+
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
∴α+
| π |
| 4 |
| π |
| 6 |
| π |
| 12 |
(II)∵x∈(
| π |
| 2 |
| x |
| 2 |
| π |
| 4 |
| π |
| 2 |
又sin
| x |
| 2 |
| 4 |
| 5 |
| x |
| 2 |
| 3 |
| 5 |
∴sinx=2sin
| x |
| 2 |
| x |
| 2 |
| 24 |
| 25 |
| 1-sin2x |
| 7 |
| 25 |
∴f(x)=sinx+cosx=
| 24 |
| 25 |
| 7 |
| 25 |
| 17 |
| 25 |
点评:本题主要考查已知三角函数值和范围来确定角的值的问题.这种凑角的思想在高考中也经常被考到.
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