数学题在线解答已知tana+cota=5/2,a∈（π/4,π/2),求cos2a和sin(2a+π

数学题在线解答已知tana+cota=5/2,a∈（π/4,π/2),求cos2a和sin(2a+π

sina/cosa + cosa/sina = 5/2通分得1/(cosasina) = 5/22/sin2a = 5/2sin2a = 4/5由a∈（π/4,π/2）得2a∈（π/2,π）,所以cos2a = -根号下(1-sin2a的平方) = -3/5sin(2a+π/4) = sin2acosπ/4 + cos2asinπ/4 = 根号2/10======以下答案可供参考======供参考答案1:tana+cota=sina/cosa+cosa/sina=[(sina)^2+(cosa)^2]/sinacosa=1/(1/2sin2a)=5/2sin2a=4/5a∈（π/4,π/2)2a∈（π/2,π)cos2a=-3/5sin(2a+π/4)=sin2acosπ/4+sinπ/4cos2a=√2/10供参考答案2:tana+cota=5/2tana+1/tana=5/2因为a∈（π/4,π/2)，tana=2设tana=tcos2a=(1-t^2)/(1+t^2)=-3/5sin(α+β)=sin αcos β+cos αsin βsin(2a+π/4)=（sin2a+cos2a）*根号2/2=(4/5-3/5)*根号2/2=根号2/10供参考答案3:易知，tana=2,[1]cos2a=(cos²a-sin²a)/(sin²a+cos²a)=(1-tan²a)/(1+tan²a)=-3/5[2]sin2a=(2sinacosa)/(sin²a+cos²a)=(2tana)/(1+tan²a)=4/5∴sin[2a+(π/4)]=(√2/2)[sin2a+cos2a]=(√2/2)[(4/5)-(3/5)]=(√2)/10供参考答案4:tana+cota=5/2tana+1/tana = 5/22(tana)^2 - 5tana +2 =0(2tana-1)(tana-2) =0tana=1/2( rejected) or tana =2sina = 2√5/5 , cosa=√5/5cos2a = (cosa)^2 -(sina)^2 = 1/5 - 4/5 = -3/5sin(2a+π/4)= (√2/2)(sin2a + cos2a)= (√2/2)(2sinacosa + cos2a)= (√2/2)(4/5-3/5)= √2/10

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