四道三角函数的题,1.用cosα表示(sinα)^4-(sinα)^2+(cosα)^22.化简:(根号下(1-2sin

6个回答

  • (一)

    (sinα)^4-(sinα)^2+(cosα)^2

    =(1-cosα^2)^2-(1-cosα^2)+(cosα)^2

    =1-2cosα^2+cosα^4-1+cosα^2+cosα^2

    =cosα^4

    (二)化简分子

    =√[sin^2(10°)+cos^2(10°)-2sin10°cos10°]

    =√(sin10°-cos10°)^2=sin10°-cos10°;

    化简分母

    =cos10°-√[1-cos^2(180°-10°)]

    =cos10°-√[1-cos^2(10°)]

    =cos10°-sin10°

    原式=(sin10°-cos10°)/(cos10°-sin10°)=-1

    (三)第一项化简

    =cosα√{[(1-sinα)(1-sinα)]/[(1+sinα)(1-sinα)]}

    =cosα√[(1-sinα)^2]/[1-sin^2(α)]

    =cosα√[(1-sinα)^2]/cos^2(α)

    =cos(1-sinα)/(-cosα) =sinα-1

    因为α在第二象限,所以cosα是负数.开方出来要加负号.

    第二项化简

    =sinα√{[(1-cosα)(1-cosα)]/[(1+cosα)(1-cosα)]}

    =sinα√[(1-cosα)^2]/[1-cos^2(α)]

    =sinα√[(1-cosα)^2]/sin^2(α)

    =sinα(1-cosα)/sinα =1-cosα

    所以原式=sinα-1+1-cosα=sinα-cosα

    (四)化简分子

    =(sinx)^2+(cosx)^2-(sinx)^4(sinx)^2-(cosx)^4(cosx)^2

    =(sinx)^2[1-(sinx)^4]+(cosx)^2[1-(cosx)^4]

    =(sinx)^2{[1+(sinx)^2][1-(sinx)^2]}+(cosx)^2{[(1+(cosx)^2)][(1-(cosx)^2]}

    =(sinx)^2[1+(sinx)^2](cosx)^2+(cosx)^2{[(1+(cosx)^2)](sinx)^2

    =(sinx)^2(cosx)^2{1+(sin)^2+1+(cosx)^2)}

    =3(sinx)^2(cosx)^2

    化简分母

    =(sinx)^2+(cosx)^2-(sinx)^2(sinx)^2-(cosx)^2(cosx)^2

    =(sinx)^2[1-(sinx)^2]+(cosx)^2[1-(cosx)^2]

    =(sinx)^2(cosx)^2+(cosx)^2(sinx)^2

    =2(sinx)^2(cosx)^2

    原式=[3(sinx)^2(cosx)^2]/[2(sinx)^2(cosx)^2]=3/2