tan2Θ=-2√2,π
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tan2Θ=-2√2,πtan2Θ=-2√2,πtan2Θ=-2√2,π2cos^2Θ/2-sinΘ-1/√2sin(Θ+π/4)=(cosΘ-sinΘ)/(cosΘ+sinΘ){在分子和分母上同时除
tan2Θ=-2√2,π
tan2Θ=-2√2,π
tan2Θ=-2√2,π
2cos^2Θ/2-sinΘ-1 / √2sin(Θ+π/4) =(cosΘ-sinΘ) /(cosΘ+sinΘ) {在分子和分母上同时除以cosΘ}
=(1-tanΘ) / (1+tanΘ)
tan2Θ=-2√2 =2tanΘ / (1-tan^2 Θ) 解得 tanΘ =√2或-√2/2
tanΘ =√2时 原式 = 2√2 -3
tanΘ =-√2/2时 原式 = 3 + 2√2
cosπ/12 x cos5π/12=1/2(cos(π/12+5π/12)+cos(π/12-5π/12))
=1/2(cos(π/2+cos(-π/3))
=1/2cosπ/3
=1/4
cosπ/12 x cos5π/12
=1/2(cos(π/12+5π/12)+cos(π/12-5π/12))
=1/2(cos(π/2+cos(-π/3))
=1/2cosπ/3
=1/2*1/2=1/4
tan2Θ=-2√2,π
若tan2θ=-2√2,π/2
若tan2θ=-2√2,π/2
tan2Θ=-2√2,π/4
已知tan2θ=2√2,且0
已知tan2θ=3/4(π/2
已知tan2θ=3/4(π/2
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