求sin^2 40°+cos^2 10°-sin40°sin80°的值.

2个回答

  • 用到积化和差公式和倍角公式

    sinαsinβ=-[cos(α+β)-cos(α-β)]/2

    cos^2 α=1/2(1+cos2α)

    sin^2 40°+cos^2 10°-sin40°sin80°

    =cos^2 50°+cos^2 10°+1/2[cos (40°+80°)-cos(40°-80°)

    =1/2(1+cos100°)+1/2(1+cos20°)+1/2[cos (40°+80°)-cos(40°-80°)

    =1+1/2(cos100°+cos20°)+1/2(cos120°-cos40°)

    =1+1/2(cos100°+cos20°)-1/4-1/2cos40°

    =3/4+cos60°cos40°-1/2cos40°

    =3/4