設(shè)f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α)
設(shè)f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α)
1.若α=-17/6π,求f(α)的值
2.若α是銳角,則sin(α-3/2π)=3/5,求f(α)的值
1.若α=-17/6π,求f(α)的值
2.若α是銳角,則sin(α-3/2π)=3/5,求f(α)的值
數(shù)學人氣：603 ℃時間：2020-06-13 10:25:30
優(yōu)質(zhì)解答
1) f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α) = [2(-sinα)(-cosα)-(-cosα)]/[(1+sin^2α+sinα-cos^2α] = [2sinαcosα+cosα]/[2sin^2α+sinα] = cosα(2sinα+1)/sinα...
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