题目内容
已知函数f(x)=sinx(sinx+
cosx),其中x∈[0,
].
(1)求f(x)的最大值和最小值;
(2)若cos(α+
)=
,求f(α)的值.
| 3 |
| π |
| 2 |
(1)求f(x)的最大值和最小值;
(2)若cos(α+
| π |
| 6 |
| 3 |
| 4 |
(1)∵已知函数f(x)=sinx(sinx+
cosx)=sin2x+
sinxcosx=
+
=sin(2x-
)+
,
∵x∈[0,
],
∴-
≤2x-
≤
,
故当2x-
=
时,f(x)的最大值为 1+
=
,
故当2x-
=-
时,f(x)的最小值为-
+
=0.
(2)若cos(α+
)=
,
则f(α)=sin(2α-
)+
=-cos[
+(2α-
)]+
=cos2(α+
)+
=2cos2(α+
)-
=2×
-
=
.
| 3 |
| 3 |
| 1-cos2x |
| 2 |
| ||
| 2 |
| π |
| 6 |
| 1 |
| 2 |
∵x∈[0,
| π |
| 2 |
∴-
| π |
| 6 |
| π |
| 6 |
| 5π |
| 6 |
故当2x-
| π |
| 6 |
| π |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
故当2x-
| π |
| 6 |
| π |
| 6 |
| 1 |
| 2 |
| 1 |
| 2 |
(2)若cos(α+
| π |
| 6 |
| 3 |
| 4 |
则f(α)=sin(2α-
| π |
| 6 |
| 1 |
| 2 |
| π |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
| 9 |
| 16 |
| 1 |
| 2 |
| 5 |
| 8 |
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