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  • 已知tan(π/4-a)=1/3,a屬于(0,π/4).(1)求(a)=(sin2a-2cos^2a)/(1+tana)的值.

    已知tan(π/4-a)=1/3,a屬于(0,π/4).(1)求(a)=(sin2a-2cos^2a)/(1+tana)的值.
    (2)若b屬于(0,π/2),且sin(3π/4+b)=-根號5/5,求a+b的值
    數(shù)學人氣:580 ℃時間:2020-04-07 14:15:35
    優(yōu)質解答
    tan(π/4-a)=1/3
    -tan(a-π/4)=1/3
    tan(a-π/4)=-1/3
    (tana-tanπ/4)/(1+tanπ/4tana)=-1/3
    (tana-1)/(1+tana)=-1/3
    3(tana-1)=-1-tana
    3tana-3=1+tana
    4tana=2
    tana=1/2
    (sin2a-2cos^2a)/(1+tana)
    =(sin2a-2cos^2a)/(1+1/2)
    =(2sinacosa-2cos^2a)/(3/2)
    =(2sinacosa-2cos^2a)/[3/2*(sin^2a+cos^2a)]分子分母同時除以cos^2a
    =(2sinacosa/cos^2a-2cos^2a/cos^2a)/[3/2*(sin^2a/cos^2a+cos^2a/cos^2a)]
    =(2sinacosa/cos^2a-2cos^2a/cos^2a)/[3/2*(sin^2a/cos^2a+cos^2a/cos^2a)]
    =(2tana-2)/[3/2*(tan^2a+1)]
    =(2*1/2-2)/{3/2*[(1/2)^2+1]}
    =-1/{3/2*5/4}
    =-1/{15/8}
    =-8/15
    a∈(0,π/4)
    a-π/4∈(-π/4,0)
    tan(a-π/4)=-1/3,
    sin(a-π/4)=-√10/10
    cos(a-π/4)=3√10/10
    b∈(0,π/2)
    b+3π/4∈(3π/4,5π/4)
    sin(3π/4+b)=-√5/5
    b+3π/4∈(π,5π/4)
    cos(3π/4+b)=-2√5/5
    a+b∈(0,3π/4)
    sin[(3π/4+b)+(a-π/4)]
    =sin(3π/4+b+a-π/4)
    =sin(π/2+a+b)
    =cos(a+b)
    sin[(3π/4+b)-(a-π/4)]
    =sin(3π/4+b)cos(a-π/4)-cos(3π/4+b)sin(a-π/4)
    =-√5/5*3√10/10-(-2√5/5)*(-√10/10)
    =-15√2/50-10√2/50
    =-1/2
    cos(a+b)=-1/2
    a+b=2π/3
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