题目内容
已知函数f(x)=2sin2x•cos2x+cos22x-sin22x.,
(I)求函数f(x)的最小正周期;
(II)若0<x<
,当f(x)=
时,求
的值.
(I)求函数f(x)的最小正周期;
(II)若0<x<
| π |
| 16 |
| ||
| 2 |
| 1+tan4x |
| 1-tan4x |
(I)f(x)=2sin2x•cos2x+cos22x-sin22x=sin4x+cos4x=
sin(4x+
)
∴T=
=
函数f(x)的最小正周期是
(II)由已知f(x)=
得?f(x)=
sin(4x+
)=
?sin(4x+
)=
而0<x<
,??
<4x+
<
?4x+
=
???
所以
=tan(4x+
)=tan
=
????
| 2 |
| π |
| 4 |
∴T=
| 2π |
| 4 |
| π |
| 2 |
函数f(x)的最小正周期是
| π |
| 2 |
(II)由已知f(x)=
| ||
| 2 |
| 2 |
| π |
| 4 |
| ||
| 2 |
| π |
| 4 |
| ||
| 2 |
| π |
| 16 |
| π |
| 4 |
| π |
| 4 |
| π |
| 2 |
| π |
| 4 |
| π |
| 3 |
所以
| 1+tan4x |
| 1-tan4x |
| π |
| 4 |
| π |
| 3 |
| 3 |
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