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已知0<α<π/2,sinα=4/5
(1)求(sin²α+sin2α)/(cos²α+cos2α)的值
(2)求tan(α-5π/4)的值
人气:444 ℃ 时间:2020-03-28 15:19:53
解答
∵0<α<π/2,sinα=4/5∴cosα>0∵(sinα)^2+(cosα)^2=1∴(cosα)^2=1-(sinα)^2=1-(4/5)^2=9/25从而 cosα=3/5又 sin2α=2sinα*cosα=2*4/5*3/5=24/25cos2α=1-2(sinα)^2=1-2*(4/5)^2=1-2*16/25=-7/25∴(sin...
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