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若α为锐角,(cosα)^4-(sinα)^4=1/2,则sinα的值为?
人气:461 ℃ 时间:2020-06-12 10:02:10
解答
(cosα)^4-(sinα)^4=1/2
[(cosα)^2-(sinα)^2]*[(cosα)^2+(sinα)^2]=1/2
(cosα)^2-(sinα)^2=1/2 1)
(cosα)^2+(sinα)^2=1 2)
1)-2),得
2(sinα)^2=1/2
sinα=1/2
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