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已知⊙O的半径为R,它的内接三角形ABC满足2R(sin^2A-sin^2C)=(√2a-b),sinB,求三角形面积最大值.
人气:394 ℃ 时间:2020-01-29 05:25:24
解答
根据正弦定理
a/sinA=b/sinB=c/sinC=2R,得
sinA=a/(2R)
sinB=b/(2R)
sinC=c/(2R)
代入2R(sin^2A-sin^2C)=(√2a-b)*sinB,得
a^2+b^2-c^2=ab√2
而a^2+b^2-c^2=2abcosC,所以
2abcosC=ab√2,得
cosC=√2/2
C=45°
S=1/2*absinC
=√2R^2sinAsinB
=√2R^2/2[cos(A-B)-cos(A+B)]
=√2R^2/2[cos(A-B)-cos135°]
=√2R^2/2[cos(A-B)+√2/2]
≤√2R^2/2(1+√2/2)
=(1+√2)*R^2/2
当且仅当A=B时,S最大为:(1+√2)*R^2/2
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