已知函数f(x)=2sin x cos x +cos 2x(1)求f(π/2)的值(2)设α∈(O,π),f(α/2)=√2/2,求sinα的值

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已知函数f(x)=2sin x cos x +cos 2x(1)求f(π/2)的值(2)设α∈(O,π),f(α/2)=√2/2,求sinα的值

已知函数f(x)=2sin x cos x +cos 2x(1)求f(π/2)的值(2)设α∈(O,π),f(α/2)=√2/2,求sinα的值
已知函数f(x)=2sin x cos x +cos 2x
(1)求f(π/2)的值
(2)设α∈(O,π),f(α/2)=√2/2,求sinα的值

已知函数f(x)=2sin x cos x +cos 2x(1)求f(π/2)的值(2)设α∈(O,π),f(α/2)=√2/2,求sinα的值
(1)f(x)=2sin x cos x +cos 2x
=sin2x+cos2x
=√2sin(2x+π/4)
∴f(π/2)=√2sin(π+π/4)=-√2sinπ/4=-1
(2)f(a/2)=√2sin(a+π/4)=√2/2
∴sin(a+π/4)=1/2
∵0<a<π
∴π/4<a+π/4<5π/4
∴a+π/4=5π/6
∴cos(a+π/4)=-√3/2
sina=sin(a+π/4-π/4)
=sin(a+π/4)cosπ/4-cos(a+π/4)sinπ/4
=√2/4+√6/4
=(√2+√6)/4
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(1)f(x)=sin2x+cos2x=√2sin(2x+π/4)
f(π/2)=√2sin(3π/4)=1
(2)因为α∈(0,π),所以α+π/4∈(π/4,5π/4)
f(α/2)=√2sin(α+π/4)=√2/2
所以sin(α+π/4)=1/2>0
α+π/4=π/6(舍)或5π/6
所以α=7π/12.
sinα=sin(7π/12)=sin(π/3+π/4)=(√6+√2)/4.

(1)
f(x)=2sinx.cosx +cos2x
f(π/2)= 0-1=-1
(2)
f(x) = sin2x+cos2x
f(α/2) = sinα + cosα
=√2(sin(α+π/4)=√2/2
sin(α+π/4)=1/2
α+π/4 = 5π/6
α = 7π/12

将公式变形化简为=√2sin(2x+π/4)
后就很容易得出

f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)

  1. f(π/2)=√2sin(2×π/2+π/4)=-√2×√2/2=-1

  2. 0<α<π
    π/4<α+π/4<5π/4
    f(α/2)=√2sin(α+π/4)=√2/2
    sin(α+π/4)=1/2
    α+π/4=5π/6
    α=7π/12=π/3+π/4
    sinα=sinπ/3cosπ/4+cosπ/3sinπ/4=(√6+√2)/4