sin78°-sin66°-sin42°+sin6°

1个回答

  • sin78°-sin42°=sin(60°+18°)-sin(60°-18°)= 2cos60°sin18°=sin18°

    sin66°-sin6°=sin(60°+6°)-sin6°=sin60°cos6°+cos60°sin6°-sin6°

    =sin(60°-6°)=sin54°

    ∴sin78°-sin66°-sin42°+sin6°=sin78°-sin42°-(sin66°-sin6°)

    =sin18°-sin54°=sin18°-cos(2*18°)=sin18°-1+2sin²18°

    下面求sin18°

    令x = 18°

    ∴cos3x = sin2x

    ∴4(cosx)^3 - 3cosx = 2sinxcosx

    ∵cosx≠ 0

    ∴4(cosx)^2 - 3 = 2sinx

    ∴4sinx² + 2sinx - 1 = 0,

    ∴sin18°+2sin²18°=1/2

    ∴原式=1/2-1=-1/2