tan2Θ=-2√2,π

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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 为什么tan2=tan(2-π）tan2=tan(2+π），但为什么tan2=tan(2-π）求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cosα)^2]=2sin(2α-π/3) 证明：tanθ-(1/tanθ)=-2/tan2θ 已知tan2α=-2根号2,π/2 已知tanx=2,则tan2(x-π/4)=? tan2/x=2,求tanx=? 证明sina=1+tan2分之2α分之2tan2分之α tan2α+1/tan2α=2 则sinαcosα= sin2αtan2α=tan2α-sin2α 2是平方最简单的方法求证tan2分之α-1/tan2分之α=-tanα分之2 已知tan2θ=3/4(π/2＜θ＜π),求(2cos^2θ/2+sinθ+10/[√2cos(θ+π/4]