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tan2Θ=-2√2,π/4
人气:318 ℃ 时间:2020-05-06 12:26:03
解答
π/4tana<0
tan2a=-2√2
=(2tana)/(1-tan^2 a)
tana=-√2/2
=sina/cosa
sina=-√2/2cosa
(2cos^2a/2-sina-1)/√2sin(π/4+a)
=(1+cosa-sina-1)/(sina+cosa)
=[-(sina-cosa)]/(sina+cosa)
=3+2√2
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