已知tan(π/4+α)=3,计算(1)tanα;(2)(2sinαcosα+3cos2α)/5cos2α-3sin2α
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已知tan(π/4+α)=3,计算(1)tanα;(2)(2sinαcosα+3cos2α)/5cos2α-3sin2α
已知tan(π/4+α)=3,计算(1)tanα;(2)(2sinαcosα+3cos2α)/5cos2α-3sin2α
已知tan(π/4+α)=3,计算(1)tanα;(2)(2sinαcosα+3cos2α)/5cos2α-3sin2α
(1)∵tan(π/4+α)=3
==>(1+tanα)/(1-tanα)=3 (应用和角公式)
==>4tanα=2
∴tanα=1/2
(2)∵tanα=1/2
==>cosα=2sinα
==>(sinα)^2+(2sinα)^2=1
==>5(sinα)^2=1
∴(sinα)^2=1/5
故(2sinαcosα+3cos2α)/(5cos2α-3sin2α)
=(4(sinα)^2+3-6(sinα)^2)/(5-10(sinα)^2-12(sinα)^2) (应用倍角公式)
=(3-2(sinα)^2)/(5-22(sinα)^2)
=(3-2(1/5)^2)/(5-22(1/5)^2)
=73/103.