一道数学正/余弦定理的题在三角形ABC中,若b^2*sin^2C+c^2*sin^2B=2bcosBcosC,判断三角形形状.过程尽量表述清楚.
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![一道数学正/余弦定理的题在三角形ABC中,若b^2*sin^2C+c^2*sin^2B=2bcosBcosC,判断三角形形状.过程尽量表述清楚.](/uploads/image/z/8579772-36-2.jpg?t=%E4%B8%80%E9%81%93%E6%95%B0%E5%AD%A6%E6%AD%A3%2F%E4%BD%99%E5%BC%A6%E5%AE%9A%E7%90%86%E7%9A%84%E9%A2%98%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%8B%A5b%5E2%2Asin%5E2C%2Bc%5E2%2Asin%5E2B%3D2bcosBcosC%2C%E5%88%A4%E6%96%AD%E4%B8%89%E8%A7%92%E5%BD%A2%E5%BD%A2%E7%8A%B6.%E8%BF%87%E7%A8%8B%E5%B0%BD%E9%87%8F%E8%A1%A8%E8%BF%B0%E6%B8%85%E6%A5%9A.)
一道数学正/余弦定理的题在三角形ABC中,若b^2*sin^2C+c^2*sin^2B=2bcosBcosC,判断三角形形状.过程尽量表述清楚.
一道数学正/余弦定理的题
在三角形ABC中,若b^2*sin^2C+c^2*sin^2B=2bcosBcosC,判断三角形形状.
过程尽量表述清楚.
一道数学正/余弦定理的题在三角形ABC中,若b^2*sin^2C+c^2*sin^2B=2bcosBcosC,判断三角形形状.过程尽量表述清楚.
代入正弦定理
a/sinA=b/sinB=c/sinC=2R
=>(2RsinB)^2(sinC)^2+(2RsinC)^2(sinB)^2=2*(2RsinB*2RsinC)cosBcosC
两边除以4R^2sinBsinC
=>2sinBsinC=2cosBcosC
=>cos(B+C)=0
=>B+C=π/2
三角形是直角三角形.
作BC边上的高,则b*cosC+c*cosB=a=>a^2=b^2*cos^2C+c^2*cos^2B+2bc*cosBcosC=>
a^2=b^2+c^2-(b^2*sin^2C+c^2*sin^2B)+2bc*cosBcosC=>
a^2=b^2+c^2-2b*cosBcosC+2bc*cosBcosC
又a^2=b^2+c^2-2bc*cosA=b^2+c^2+2bc...
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作BC边上的高,则b*cosC+c*cosB=a=>a^2=b^2*cos^2C+c^2*cos^2B+2bc*cosBcosC=>
a^2=b^2+c^2-(b^2*sin^2C+c^2*sin^2B)+2bc*cosBcosC=>
a^2=b^2+c^2-2b*cosBcosC+2bc*cosBcosC
又a^2=b^2+c^2-2bc*cosA=b^2+c^2+2bc*cos(B+C)【余弦定理】=>
-2b*cosBcosC+2bc*cosBcosC=2bc*cos(B+C)=>cosBcosC=c*sinBsinC
又b/sinB=c/sinC则b*sinC=c*sinB【正弦定理】,则b^2*sin^2C+c^2*sin^2B=2bcosBcosC=>
2c^2*sin^2B=2bcosBcosC则c^2*sin^2B=bc*sinBsinC恒成立=>△为任意△
注意题目条件!b^2*sin^2C+c^2*sin^2B=2bcosBcosC!而不是b^2*sin^2C+c^2*sin^2B=2bc*cosBcosC
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