1.三角形ABC中,sinA^2+sinB^2=6sinC^2,则(1/tanA+1/tanB)tanC=?
1个回答
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*cosC
c^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)
=2/5
2.
sinα+cosα=a
sinα*cosα=a
(sinα+cosα)^2=a^2
sin^2α+2sinα*cosα+cos^2α=a^2
1+2a=a^2
a^2-2a-1=0
a^2-2a+1-1-1=0
(a-1)^2-2=0
(a-1-√2)(a-1+√2)=0
a=1-√2 或 a=1+√2(舍去,)
所以：a=1-√2
sinα^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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