已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值
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已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值
![已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值](/uploads/image/z/6530428-28-8.jpg?t=%E5%B7%B2%E7%9F%A5tan%28%CF%80%2F4%2B%CE%B1%29%3D2%2C%E6%B1%821%2F%282sin%CE%B1cos%CE%B1%2Bcos%5E2%CE%B1%29%E7%9A%84%E5%80%BC)
tan(π/4+α)=sin(π/4+α)/cos(π/4+α)
=[sin(π/4)cosα+cos(π/4)sinα]/[cos(π/4)cosα-sin(π/4)sinα]
=[cosα+sinα]/[cosα-sinα]
=2
cosα+sinα=2cosα-2sinα
cosα=3sinα
tanα=1/3
1/(2sinαcosα+cos^2α)=1/[6(sinα)^2+9(sinα)^2]
=1/[15(sinα)^2]
2(sinα)^2=1-cos(2α)
cos(2α)=[1-(tanα)^2]/[1+(tanα)^2]
=[1-1/9]/[1+1/9]
=4/5
(sinα)^2=[1-cos(2α)]/2=1/10
1/(2sinαcosα+cos^2α)=1/[15(sinα)^2]=1/[15/10]=2/3
=[sin(π/4)cosα+cos(π/4)sinα]/[cos(π/4)cosα-sin(π/4)sinα]
=[cosα+sinα]/[cosα-sinα]
=2
cosα+sinα=2cosα-2sinα
cosα=3sinα
tanα=1/3
1/(2sinαcosα+cos^2α)=1/[6(sinα)^2+9(sinα)^2]
=1/[15(sinα)^2]
2(sinα)^2=1-cos(2α)
cos(2α)=[1-(tanα)^2]/[1+(tanα)^2]
=[1-1/9]/[1+1/9]
=4/5
(sinα)^2=[1-cos(2α)]/2=1/10
1/(2sinαcosα+cos^2α)=1/[15(sinα)^2]=1/[15/10]=2/3
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