设A+B+C=180°,求证:sin2A+sin2B+sin2C-2cosAcosBcosC=2设A+B+C=180°,求证:(sinA)^2+(sinB)^2+(sinC)^2-2cosAcosBcosC=2
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![设A+B+C=180°,求证:sin2A+sin2B+sin2C-2cosAcosBcosC=2设A+B+C=180°,求证:(sinA)^2+(sinB)^2+(sinC)^2-2cosAcosBcosC=2](/uploads/image/z/8586577-1-7.jpg?t=%E8%AE%BEA%2BB%2BC%3D180%C2%B0%2C%E6%B1%82%E8%AF%81%3Asin2A%2Bsin2B%2Bsin2C-2cosAcosBcosC%3D2%E8%AE%BEA%2BB%2BC%EF%BC%9D180%C2%B0%2C%E6%B1%82%E8%AF%81%EF%BC%9A%28sinA%29%5E2%2B%28sinB%29%5E2%2B%28sinC%29%5E2-2cosAcosBcosC%3D2)
设A+B+C=180°,求证:sin2A+sin2B+sin2C-2cosAcosBcosC=2设A+B+C=180°,求证:(sinA)^2+(sinB)^2+(sinC)^2-2cosAcosBcosC=2
设A+B+C=180°,求证:sin2A+sin2B+sin2C-2cosAcosBcosC=2
设A+B+C=180°,求证:(sinA)^2+(sinB)^2+(sinC)^2-2cosAcosBcosC=2
设A+B+C=180°,求证:sin2A+sin2B+sin2C-2cosAcosBcosC=2设A+B+C=180°,求证:(sinA)^2+(sinB)^2+(sinC)^2-2cosAcosBcosC=2
好象没有太简单的方法.
由A+B+C=180°得sinA=sin(180°-B-C)=sin(B+C),cosA=-cos(B+C),
(sinA)^2+(sinB)^2+(sinC)^2-2cosAcosBcosC
=(sinCcosB+sinBcosC)^2+(sinB)^2+(sinC)^2+2(cosBcosC-sinBsinC)cosBcosC
=(sinCcosB)^2+(sinBcosC)^2+(sinB)^2+(sinC)^2+2(cosBcosC)^2(有两项抵消)
=[(sinC)^2+(cosC)^2](cosB)^2+(sinB)^2+[(sinB)^2+(cosB)^2](cosC)^2+(sinC)^2 (将2(cosBcosC)^2拆开,分别与(sinCcosB)^2、(sinBcosC)^2合并即得此式)
=(sinB)^2+(cosB)^2+(sinC)^2+(cosC)^2=2