已知sina+cosa=-1/5,a属于(0,π)求tana,(sinacosa)/(sin^2a-sinacosa-2cos^2a)的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/21 00:32:27
已知sina+cosa=-1/5,a属于(0,π)求tana,(sinacosa)/(sin^2a-sinacosa-2cos^2a)的值
xSJA~Bi2;Κ5EWI-آ-*MoltwP'WysvF7Mf}g̉-{ktx{1w=Q+qy_3S Q%s)4#">h"^^ NGeV@" Tx \|Cf2^2feD2*OJثv,Au\+9̠}0Ra׏o>|s#ejw%穜L&|SUkB["LP~N 뎨LM}?q|7?Ϗ*t 5~0oİ.Z^cxa0X\x)?V .F

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...

全部展开

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

收起