(2005•浙江)已知函数f(x)=2sinxcosx+cos2x
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(2005•浙江)已知函数f(x)=2sinxcosx+cos2x
(Ⅰ)求f(
)
(Ⅰ)求f(
π |
4 |
![(2005•浙江)已知函数f(x)=2sinxcosx+cos2x](/uploads/image/z/515135-47-5.jpg?t=%EF%BC%882005%E2%80%A2%E6%B5%99%E6%B1%9F%EF%BC%89%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2sinxcosx%2Bcos2x)
(Ⅰ)∵f(x)=sin2x+cos2x
∴f(
π
4)=sin
π
2+cos
π
2=1
(Ⅱ)f(
α
2)=cosα+sinα=
2
2
∴sin(α+
π
4)=
1
2,cos(α+
π
4)=±
3
2.sinα=sin(α+
π
4−
π
4)=
1
2×
2
2∓
3
2×
2
2=
2∓
∴f(
π
4)=sin
π
2+cos
π
2=1
(Ⅱ)f(
α
2)=cosα+sinα=
2
2
∴sin(α+
π
4)=
1
2,cos(α+
π
4)=±
3
2.sinα=sin(α+
π
4−
π
4)=
1
2×
2
2∓
3
2×
2
2=
2∓
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