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已知tana=3.求sin2a+sinacosa+2cos2a
人气:367 ℃ 时间:2020-10-02 06:16:29
解答
(sin2a+sinacosa+2cos2a)
=(sin2a+sinacosa+2cos2a)/1
=(2sinacosa+sinacosa+2cos^2a-2sin^2a)/(sin^2a+cos^2a)
=(3sinacosa+2cos^2a-2sin^2a)/(sin^2a+cos^2a) (上下同除cos^2a)
=(3tana+2-2tan^2a)/(tan^2a+1)
=(9+2-18)/(9+1)
=-4/5
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