已知 A+B+C=π,sinA+sinB+sinC=cosA+cosB+cosC.求 ( cos2A+cos2B+cos
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已知 A+B+C=π,sinA+sinB+sinC=cosA+cosB+cosC.求 ( cos2A+cos2B+cos2C )÷(cosA+cosB+cosC)的值
(sinA+sinB+sinC)² = sin²A+sin²B+sin²C+2(sinAsinB+sinAsinC+sinBsinC)
(cosA+cosB+cosC)²
=cos²A+cos²B+cos²C+2(cosAcosB+cosAcosC+cosBcosC)
(cosA+cosB+cosC)²-(sinA+sinB+sinC)²
=[(cos²A+cos²B+cos²C)-(sin²A+sin²B+sin²C)]
+[2(cosAcosB+cosAcosC+cosBcosC)-2(sinAsinB+sinAsinC+sinBsinC)]
(利用倍角公式和cos的和角公式)
=(cos2A+cos2B+cos2C)+2[cos(A+B)+cos(A+C)+cos(B+C)]
=(cos2A+cos2B+cos2C)-2cosC-2cosB-2cosA=0
(cos2A+cos2B+cos2C)/(cosA+cosB+cosC))=2
(cosA+cosB+cosC)²
=cos²A+cos²B+cos²C+2(cosAcosB+cosAcosC+cosBcosC)
(cosA+cosB+cosC)²-(sinA+sinB+sinC)²
=[(cos²A+cos²B+cos²C)-(sin²A+sin²B+sin²C)]
+[2(cosAcosB+cosAcosC+cosBcosC)-2(sinAsinB+sinAsinC+sinBsinC)]
(利用倍角公式和cos的和角公式)
=(cos2A+cos2B+cos2C)+2[cos(A+B)+cos(A+C)+cos(B+C)]
=(cos2A+cos2B+cos2C)-2cosC-2cosB-2cosA=0
(cos2A+cos2B+cos2C)/(cosA+cosB+cosC))=2
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