高一数学

三角形中,求证cosa^2+co

三角形中,求证（cosA）^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1 证明：（cosA）^2+(cosB)^2+(cosC)^2+2cosAcosBcosC =cosA*[cosA+2cosB*cosC]+(cosB)^2+(cosC)^2 =-cos(B+C)*[-cos(B+C)+2cosB*cosC]+(cosB)^2+(cosC)^2 =-cos(B+C)*cos(B-C)+(cosB)^2+(cosC)^2 =-(1/2)*[cos(2B)+cos(2C)]+(cosB)^2+(cosC)^2 =-(1/2)*[2(cosB)^2－1＋2（cosC）...全部

  三角形中,求证（cosA）^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1 证明：（cosA）^2+(cosB)^2+(cosC)^2+2cosAcosBcosC =cosA*[cosA+2cosB*cosC]+(cosB)^2+(cosC)^2 =-cos(B+C)*[-cos(B+C)+2cosB*cosC]+(cosB)^2+(cosC)^2 =-cos(B+C)*cos(B-C)+(cosB)^2+(cosC)^2 =-(1/2)*[cos(2B)+cos(2C)]+(cosB)^2+(cosC)^2 =-(1/2)*[2(cosB)^2－1＋2（cosC）^2-1]+(cosB)^2+(cosC)^2 =1 三角形中,求证cosa^2+cosb^2+cosc^2+2cosacosbcosc=1 cosa^2+cosb^2+cosc^2+2cosacosbcosc =cosa*[cosa+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)+2] =-cos(b+c)*[-cos(b+c)+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)]+1 =-cos(b+c)*cos(b-c)+cos(b+c)*cos(b-c)+1 =1 。
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