已知0<α<π/2,且sinα=4/5. (1)求(sin^2 α+sin2α)/(cos^2 α
已知0<α<π/2,且sinα=3/5.(1)求(sin^2α+sin2α)/(cos^2α+cos2α)的值(2)求tan(α-5/4π)的值谢谢...
已知0<α<π/2,且sinα=3/5. (1)求(sin^2 α+sin2α)/(cos^2 α+cos2α)的值 (2)求tan(α-5/4π)的值
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0<α<π/2,且sinα=3/5. 则
cosα=4/5,tanα=3/4。
(1),(sin^2α+sin2α)/(cos^2α+cos2α)
=(sin^2α+2sinαcosα)/(cos^2α+2cos^2α-1)
=(9/25+2*12/25)/(16/25+32/25-1)
=(9+24)/(16+32-25)
=33/23
(2)tan(α-5π/4)=tan(α-π/4)
=(tanα-tanπ/4)/(1+tanαtanπ/4)
=(3/4-1)/(1+3/4)=-1/7
cosα=4/5,tanα=3/4。
(1),(sin^2α+sin2α)/(cos^2α+cos2α)
=(sin^2α+2sinαcosα)/(cos^2α+2cos^2α-1)
=(9/25+2*12/25)/(16/25+32/25-1)
=(9+24)/(16+32-25)
=33/23
(2)tan(α-5π/4)=tan(α-π/4)
=(tanα-tanπ/4)/(1+tanαtanπ/4)
=(3/4-1)/(1+3/4)=-1/7
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