若arccosx+arccosy+arccosz=π,求证：x^2+y^2+z^2+2xyz=1

如题，详细过程

令A=arccosx,B=arccosy,C=arccosZ ,则有A+B+C=π，待证明的是cosA^2+cosB^2+cosC^2+2cosAcosBcosC=1

cos(A+B+C)=-1

又有cos(A+B+C)=cosAcos(B+C)-sinAsin(B+C)

    =cosAcos(π-A)-sinAsin(B+C)

    =-cosA^2-sinA(sinBcosC+sinCcosB)

    =-cosA^2-sinAsinBcosC-sinAsinCcosB    （1）

sinAsinBcosC=sin(B+C)sinBcosC

    =sinB^2cosC^2+sinBcosBsinCcosC

    =(1-cosB^2)cosC^2+sinBcosBsinCcosC

sinAsinCcosB=sin(B+C)sinCcosB

    =sinBcosBsinCcosC+sinC^2cosB^2

    =sinBcosBsinCcosC+(1-cosC^2)cosB^2

把上述两式代入（1）有

cos(A+B+C)=-cosA^2-sinAsinBcosC-sinAsinCcosB

    =-cosA^2-cosB^2-cosC^2-2sinBcosBsinCcosC+2cosB^2cosC^2

    =-cosA^2-cosB^2-cosC^2+2cosBcosC(cosBcosC-sinBsinC)

    =-cosA^2-cosB^2-cosC^2+2cosBcosCcos(B+C)

    =-cosA^2-cosB^2-cosC^2-2cosAcosBcosC

    =-1

即cosA^2+cosB^2+cosC^2+2cosAcosBcosC=1

证毕

设arccosx=αarccosy=βarccosz=γ，则α+β+γ=π ∴x^2+y^2+z^2+2xyz =(cosα)^2+(cosβ)^2+(cosγ)^2+2cosαcosβcosγ =(cosα)^2+(cosβ)^2+(cos(л-α-β) ^2+2cosαcosβcos(л-α-β) =(cosα)^2+(cosβ)^2+(cos(α+β))^2+2cosαcosβcos(α+β) =(cosα)^2+(cosβ)^2+（cosαcosβ-sinαsinβ）^2+2cosαcosβ（cosαcosβ-sinαsinβ） =(cosα)^2+(cosβ)^2+3(cosα)^2 (cosβ)^2+(1-(cosα)^2)（1-(cosβ)^2） =1