若tan2θ=-2√2,π/2
问题描述：
若tan2θ=-2√2,π/2
数学人气：224 ℃时间：2020-03-29 02:27:01
优质解答
∵π/2∴sin2θ＜0,∴sin2θ= -(2/3)√2,cos2θ=1/3
2cos²(θ/2)-1=cosθ
√2sin(θ+π/4)
=√2sinθcosπ/4+√2cosθsinπ/4
=sinθ+cosθ
∴
[2cos²(θ/2)-sinθ-1]/[√2sin(θ+π/4)]
=(cosθ-sinθ)/(cosθ+sinθ)
=(cosθ-sinθ)·(cosθ+sinθ)/(cosθ+sinθ)²
=(cos²θ-sin²θ)/(cos²θ+sin²θ+2sinθcosθ)
=cos2θ/(1+sin2θ)
=1/[1/(cos2θ)+tan2θ]
=1/(3-2√2)
=3+2√2
2cos²(θ/2)-1=cosθ
√2sin(θ+π/4)
=√2sinθcosπ/4+√2cosθsinπ/4
=sinθ+cosθ
∴
[2cos²(θ/2)-sinθ-1]/[√2sin(θ+π/4)]
=(cosθ-sinθ)/(cosθ+sinθ)
=(cosθ-sinθ)·(cosθ+sinθ)/(cosθ+sinθ)²
=(cos²θ-sin²θ)/(cos²θ+sin²θ+2sinθcosθ)
=cos2θ/(1+sin2θ)
=1/[1/(cos2θ)+tan2θ]
=1/(3-2√2)
=3+2√2
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答
∵π/2∴sin2θ＜0,∴sin2θ= -(2/3)√2,cos2θ=1/3
2cos²(θ/2)-1=cosθ
√2sin(θ+π/4)
=√2sinθcosπ/4+√2cosθsinπ/4
=sinθ+cosθ
∴
[2cos²(θ/2)-sinθ-1]/[√2sin(θ+π/4)]
=(cosθ-sinθ)/(cosθ+sinθ)
=(cosθ-sinθ)·(cosθ+sinθ)/(cosθ+sinθ)²
=(cos²θ-sin²θ)/(cos²θ+sin²θ+2sinθcosθ)
=cos2θ/(1+sin2θ)
=1/[1/(cos2θ)+tan2θ]
=1/(3-2√2)
=3+2√2