题目内容
已知函数f1(x)=3sin(2x-
),f2(x)=4sin(2x+
),则函数f(x)=f1(x)+f2(x)的振幅为( )
| π |
| 3 |
| π |
| 3 |
A.
| B.5 | C.7 | D.13 |
函数f(x)=f1(x)+f2(x)
=3sin(2x-
)+4sin(2x+
)
=3sin2xcos
-3cos2xsin
+4sin2xcos
+4cos2xsin
=7sin2xcos
+cos2xsin
=
sin2x+
cos2x
=
sin(2x+θ).其中tanθ=
.
所以函数的振幅为
.
故选A.
=3sin(2x-
| π |
| 3 |
| π |
| 3 |
=3sin2xcos
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
=7sin2xcos
| π |
| 3 |
| π |
| 3 |
=
| 7 |
| 2 |
| ||
| 2 |
=
| 13 |
| ||
| 7 |
所以函数的振幅为
| 13 |
故选A.
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