已知sina+cosa=-1/5,a属于(0,π)求tana,(sinacosa)/(sin^2a-sinacosa-2cos^2a)的值
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已知sina+cosa=-1/5,a属于(0,π)求tana,(sinacosa)/(sin^2a-sinacosa-2cos^2a)的值
已知sina+cosa=-1/5,a属于(0,π)
求tana,(sinacosa)/(sin^2a-sinacosa-2cos^2a)的值
已知sina+cosa=-1/5,a属于(0,π)求tana,(sinacosa)/(sin^2a-sinacosa-2cos^2a)的值
sina=-1/5-cosa
sin²a+cos²a=1=(-1/5-cosa)²+cos²a
cosa=-4/5或3/5
a∈(0,π)
cosa=-4/5,sina=3/5
tana=sina/cosa=-3/4
(sinacosa)/(sin²a-sinacosa-2cos²a)
=(3/5*(-4/5))/((3/5)²-(3/5)*(-4/5)-2*(-4/5)²)
=(-12/25)/(-12/25)=1
由sinα+cosα=-1/5,得(√2)cos(α-π/4)=-1/5,cos(α-π/4)=-1/(5√2)=-√2/10
故α-π/4=arccos(-√2/10)=π-arccos(√2/10),α=5π/4-arccos(√2/10);tanarccos(√2/10)=7.
故tanα=tan[5π/4-arccos(√2/10)]=[tan(5π/4)-tanarcco...
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由sinα+cosα=-1/5,得(√2)cos(α-π/4)=-1/5,cos(α-π/4)=-1/(5√2)=-√2/10
故α-π/4=arccos(-√2/10)=π-arccos(√2/10),α=5π/4-arccos(√2/10);tanarccos(√2/10)=7.
故tanα=tan[5π/4-arccos(√2/10)]=[tan(5π/4)-tanarccos(√2/10)]/[1+tan(5π/4)tanarccos(√2/10)]
=[1-tanarccos(√2/10)]/[1+tanarccos(√2/10)]=(1-7)/(1+7)=-6/8=-3/4.
(sinαcosα)/(sin²α-sinαcosα-2cos²α)【分子分母同除以cos²α】
=(tanα)/(tan²α-tanα-2)=(-3/4)/[(9/16)+(3/4)-2]=(-3/4)/(-1)=3/4
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