求证:(1)2sin(π+θ)•cosθ−11−2sin
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求证:
(1)
(1)
2sin(π+θ)•cosθ−1 |
1−2sin
证明:(1)左边=
−2sinθcosθ−1 cos2θ−sin2θ= −(sinθ+cosθ)2 (sinθ+cosθ)cosθ−sinθ)= (sinθ+cosθ) (sinθ−cosθ)= tanθ+1 tanθ−1= −sinθ−cosθ cosθ−sinθ= −tanθ−1 1−tanθ= tanθ+1 tanθ−1; 右边= tan(8π+π+θ)+1 tanθ−1= tanθ+1 tanθ−1, ∴左=右,得证; (2)左边= sinθ cosθ•sinθ sinθ cosθ−sinθ= sin2θ sinθ(1−cosθ)= sinθ 1−cosθ, 右边= cosθ•( sinθ cosθ+sinθ) sin2θ= sinθ(1+cosθ) 1−cos2θ= sinθ 1−cosθ, ∴左=右,得证. 再问: =(sinα+cosα)/(sinα-cosα)怎么变成 =(tanα+1)/(tanα-1)
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化简:(1)sin(α+β)−2sinαcosβ2sinαsinβ+cos(α+β)
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