高数 函数y=sin&+cos&的奇偶性

1个回答

  • 是(SINX)^2+(COSX)^2= 1

    sin(A+B) = sinAcosB+cosAsinB

    sin(A-B) = sinAcosB-cosAsinB 

    cos(A+B) = cosAcosB-sinAsinB

    cos(A-B) = cosAcosB+sinAsinB

    tan(A+B) = (tanA+tanB)/(1-tanAtanB)

    tan(A-B) = (tanA-tanB)/(1+tanAtanB)

    cot(A+B) = (cotAcotB-1)/(cotB+cotA) 

    cot(A-B) = (cotAcotB+1)/(cotB-cotA)

    [编辑本段]倍角公式

    Sin2A=2SinA•CosA

    Cos2A=CosA^2-SinA^2=1-2SinA^2=2CosA^2-1

    tan2A=2tanA/(1-tanA^2)

    (注:SinA^2 是sinA的平方 sin2(A) )

    [编辑本段]三倍角公式

    sin3α=4sinα·sin(π/3+α)sin(π/3-α)

    cos3α=4cosα·cos(π/3+α)cos(π/3-α)

    tan3a = tan a · tan(π/3+a)· tan(π/3-a)

    cosα=sin(90-α)

    [编辑本段]半角公式

    tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA);

    cot(A/2)=sinA/(1-cosA)=(1+cosA)/sinA.

    [编辑本段]和差化积

    sinθ+sinφ = 2sin[(θ+φ)/2]cos[(θ-φ)/2]

    sinθ-sinφ = 2cos[(θ+φ)/2]sin[(θ-φ)/2]

    cosθ+cosφ = 2cos[(θ+φ)/2]cos[(θ-φ)/2]

    cosθ-cosφ = -2sin[(θ+φ)/2]sin[(θ-φ)/2]

    tanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1-tanAtanB)

    [编辑本段]积化和差

    sinαsinβ = -1/2*[cos(α+β)-cos(α-β)]

    cosαcosβ = 1/2*[cos(α+β)+cos(α-β)]

    sinαcosβ = 1/2*[sin(α+β)+sin(α-β)]

    cosαsinβ = 1/2*[sin(α+β)-sin(α-β)]

    [编辑本段]诱导公式

    sin(-a) = -sin(a)

    cos(-a) = cos(a)

    sin(π/2-a) = cos(a)

    cos(π/2-a) = sin(a)

    sin(π/2+a) = cos(a)

    cos(π/2+a) = -sin(a)

    sin(π-a) = sin(a)

    cos(π-a) = -cos(a)

    sin(π+a) = -sin(a)

    cos(π+a) = -cos(a)

    tanA= sinA/cosA

    tan(π/2+α)=-cotα

    tan(π/2-α)=cotα

    tan(π-α)=-tanα

    tan(π+α)=tanα