f（x）=3sin（ωx+
（1）求f（0）； （2）求f（x）的解析式； （3）已知f（
|
（1）f（0）=3sin（ω•0+
）=3×π 6
=1 2
，3 2
（2）∵T=
=2π ω
∴ω=4π 2
所以f（x）=3sin（4x+
）．π 6
（3）f（
+α 4
）=3sin[4（π 12
+α 4
）+π 12
]=3sin（α+π 6
）=π 2 9 5
∴cosα=3 5
∴sinα=±
=±1-cos2α 4 5
| 完形填空。 | |||
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