高一数学设1+tanθ/1-ta
设1+tanθ /1-tanθ=3+2倍的根号2,求sin2θd的值
(1+tanθ)/(1-tanθ)=3+2√2
===> 1+tanθ=(3+2√2)*(1-tanθ)
===> 1+tanθ=3+2√2-(3+2√2)tanθ
===> (4+2√2)tanθ=2+2√2
===> tanθ=(2+2√2)/(4+2√2)=(1+√2)/(2+√2)=√2/2
则,sin2θ=2sinθcosθ=2(sinθ/cosθ)*cos^2 θ
=2tanθ/sec^2 θ
=2tanθ/(1+tan^2 θ)
=2*(√2/2)/[1+(√2/2)^2]
=(2√2)/3 。 全部
设1+tanθ /1-tanθ=3+2倍的根号2,求sin2θd的值
(1+tanθ)/(1-tanθ)=3+2√2
===> 1+tanθ=(3+2√2)*(1-tanθ)
===> 1+tanθ=3+2√2-(3+2√2)tanθ
===> (4+2√2)tanθ=2+2√2
===> tanθ=(2+2√2)/(4+2√2)=(1+√2)/(2+√2)=√2/2
则,sin2θ=2sinθcosθ=2(sinθ/cosθ)*cos^2 θ
=2tanθ/sec^2 θ
=2tanθ/(1+tan^2 θ)
=2*(√2/2)/[1+(√2/2)^2]
=(2√2)/3 。
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