已知tan平方θ=4/9θ,属于(π/2,π) 求tan(θ-π/4) 求2sin(π-θ)cos
已知tan平方θ=4/9θ,属于(π/2,π)求tan(θ-π/4)求2sin(π-θ)cos(-2π-θ)/sin平方(5π/2-θ)-3sin平方(-θ)...
已知tan平方θ=4/9θ,属于(π/2,π) 求tan(θ-π/4) 求2sin(π-θ)cos(-2π-θ)/sin平方(5π/2-θ)-3sin平方(-θ)
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tan^2θ=4/9
∵θ∈(π/2,π)
∴tanθ=-2/3
sinθ/cosθ=-2/3
sin^2θ/cos^2θ=4/9
sin^2θ=4/9cos^2θ
sin^2θ=4/9(1-sin^2θ)
5/9sin^2θ=4/9
sin^2θ=4/5
tan(θ-π/4)
=(tanθ-tanπ/4)/(1+tanθtanπ/4)
=(-2/3-1)/[1+(-2/3)×1]
=(-5/3)/(1/3)
=-5
2sin(π-θ)cos(-2π-θ)/sin^2(5π/2-θ)-3sin^2(-θ)
=2sinθcos(-θ)/sin^2[2π+(π/2-θ)]-3(-sinθ)^2
=2sinθcosθ/cos^2θ-3sin^2θ
=2tanθ-3sin^2θ
=2(-2/3)-3(4/5)
=-4/3-12/5
=-20/15-36/15
=-56/15
∵θ∈(π/2,π)
∴tanθ=-2/3
sinθ/cosθ=-2/3
sin^2θ/cos^2θ=4/9
sin^2θ=4/9cos^2θ
sin^2θ=4/9(1-sin^2θ)
5/9sin^2θ=4/9
sin^2θ=4/5
tan(θ-π/4)
=(tanθ-tanπ/4)/(1+tanθtanπ/4)
=(-2/3-1)/[1+(-2/3)×1]
=(-5/3)/(1/3)
=-5
2sin(π-θ)cos(-2π-θ)/sin^2(5π/2-θ)-3sin^2(-θ)
=2sinθcos(-θ)/sin^2[2π+(π/2-θ)]-3(-sinθ)^2
=2sinθcosθ/cos^2θ-3sin^2θ
=2tanθ-3sin^2θ
=2(-2/3)-3(4/5)
=-4/3-12/5
=-20/15-36/15
=-56/15
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谢谢大侠!虽然不知道对不对但已解燃眉之急!
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