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在三角形里,有恒等式sin^2A+sin^2B+sin^2C-2cosAcosBcosC=2
恒等式证明:
sinA^2+sinB^2+sinB^2-2cosAcosBcosC
=3-(cosA^2+cosB^2+cosC^2+2cosAcosBcosC)
=3-{cosA*[cosA+2cosB*cosC]+(1/2)*[cos(2B)+cos(2C)+2]}
=3-{-cos(B+C)*[-cos(B+C)+2cosB*cosC]+cos(B+C)*cos(B-C)+1}
=3-{-cos(B+C)*cos(B-C)+cos(B+C)*cos(B-C)+1}
=2
当sin^2A+sin^2B+sin^2C>2时,cosAcosBcosC
1年前
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