(2009•丰台区二模)已知函数f(x)=2sin2x•cos2x+cos22x-sin22x.,
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(2009•丰台区二模)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<
(I)求函数f(x)的最小正周期;
(II)若0<x<
π |
16 |
(I)f(x)=2sin2x•cos2x+cos22x-sin22x=sin4x+cos4x=
2sin(4x+
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)=
6
2得f(x)=
2sin(4x+
π
4)=
6
2⇒sin(4x+
π
4)=
3
2而0<x<
π
16,
π
4<4x+
π
4<
π
2⇒4x+
π
4=
π
3
所以
1+tan4x
1−tan4x=tan(4x+
π
4)=tan
π
3=
2sin(4x+
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)=
6
2得f(x)=
2sin(4x+
π
4)=
6
2⇒sin(4x+
π
4)=
3
2而0<x<
π
16,
π
4<4x+
π
4<
π
2⇒4x+
π
4=
π
3
所以
1+tan4x
1−tan4x=tan(4x+
π
4)=tan
π
3=
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