1.三角形ABC中,sinA^2+sinB^2=6sinC^2,则(1/tanA+1/tanB)ta

1.三角形ABC中,sinA^2+sinB^2=6sinC^2,则(1/tanA+1/tanB)ta

1.(1/tanB+1/tanA)*tanC=tanC*(tanB+tanA)/(tanBtanA)=tanC*(sinBcosA+sinAcosB)/(sinBsinA)[分子分母同时乘以cosBcosA]=sinC*sin(B+A)/(sinBsinAcosC)=(sinC)^2/(sinBsinAcosC) [B+C=180度-A]所以 (1/tanB+1/tanA)*tanC=c^2/(ab*cosC)[由正弦定理可得].1式6sinC^2＝sinB^2＋sinA^2可由正弦定理推出,6C^2=b^2+a^2 .2式再根据余弦定理,c^2=b^2+a^2-2ab*cosC .3式将2式代入3式,得5c^2=2ab*cosCc^2=2/5ab*cosC .4式最后将4式代入1式,(1/tanB+1/tanA)*tanC=c^2/(ab*cosC)=(2/5ab*cosC)/(ab*cosC)=2/52.sinα+cosα=asinα*cosα=a(sinα+cosα)^2=a^2sin^2α+2sinα*cosα+cos^2α=a^21+2a=a^2a^2-2a-1=0a^2-2a+1-1-1=0(a-1)^2-2=0(a-1-√2)(a-1+√2)=0a=1-√2 或 a=1+√2(舍去,)所以：a=1-√2sinα^3+cosα^3=(sinα+cosα)(sin^2α-sinα*cosα+cos^2α)=(sinα+cosα)(1-sinα*cosα)=a(1-a)=a-a^2=1-√2-（1-√2）^2=1-√2-3+2√2=√2-2

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