作业帮已知sinθ+cosθ=－（根号10）/5,求tanθ的值
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已知sinθ+cosθ=－（根号10）/5,求tanθ的值
已知sinθ+cosθ=－（根号10）/5,求tanθ的值
已知sinθ+cosθ=－（根号10）/5,求tanθ的值
sinθ+cosθ=－（根号10）/5
两边平方
(sinθ)^2+2sinθcosθ+(cosθ)^2=2/5
1+2sinθcosθ=2/5
sinθcosθ=-3/10
[(sinθ)^2+(cosθ)^2]/(sinθcosθ)=1/(sinθcosθ)=-10/3
[(sinθ)^2+(cosθ)^2]/(sinθcosθ)
=(sinθ)^2/(sinθcosθ)+(cosθ)^2/(sinθcosθ)
=sinθ/cosθ+cosθ/sinθ
=tanθ+1/tanθ=-10/3
3(tanθ)^2+10tanθ+3=0
(3tanθ+1)(tanθ+3)=0
tanθ=-1/3或tanθ=-3
sinθ+cosθ=－（根号10）/5,两边平方得,
1+2sinθcosθ=2/5,
sin2θ=-3/5,cos2θ=-√[1+sin^2(2θ)]=-4/5.
tan2θ=sin2θ/cos2θ=3/4=2tanθ/(1-tan^2θ),
3tan^2θ+8tanθ-3=0,
tanθ=1/3(不合,舍去),或
tanθ=-3.
(sinθ+cosθ)/2^(1/2) = -1/5^(1/2)
sin(θ+PI/4) = -1/5^(1/2)
2kPI + PI < θ+PI/4 < 2kPI + 2PI,
2kPI + 3PI/4 < θ < 2kPI + PI + 3PI/4
当2kPI + 3PI/4 < θ < 2kPI + PI/2 + 3PI/4时，
2kPI + PI ...
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(sinθ+cosθ)/2^(1/2) = -1/5^(1/2)
sin(θ+PI/4) = -1/5^(1/2)
2kPI + PI < θ+PI/4 < 2kPI + 2PI,
2kPI + 3PI/4 < θ < 2kPI + PI + 3PI/4
当2kPI + 3PI/4 < θ < 2kPI + PI/2 + 3PI/4时，
2kPI + PI < θ+PI/4 < 2kPI + 3PI/2,
cos(θ+PI/4) = -2/5^(1/2)
tan(θ+PI/4) = 1/2
tan(θ) = tan(θ+PI/4 - PI/4)
= [tan(θ+PI/4) - tan(PI/4)]/[1+tan(θ+PI/4)tan(PI/4)]
= [1/2 - 1]/[1 + 1/2]
= -1/3
当2kPI + PI/2 + 3PI/4 <= θ < 2kPI + PI + 3PI/4时，
2kPI + 3PI/2 <= θ+PI/4 < 2kPI + 2PI,
cos(θ+PI/4) = 2/5^(1/2)
tan(θ+PI/4) = -1/2
tan(θ) = tan(θ+PI/4 - PI/4)
= [tan(θ+PI/4) - tan(PI/4)]/[1+tan(θ+PI/4)tan(PI/4)]
= [-1/2 - 1]/[1 - 1/2]
= -3
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