化简,求值
1)化简:(1+sinθ-cosθ/1+sinθ+cosθ)+(1+cosθ+sinθ/1-cosθ+sinθ) 2)已知tg(π/4+α)=-1/2,求sin2α-2cos^2(α)/1+tgα的值. 3)已知tgθ=√2,求2cos^2(θ/2)-sinθ-1/√2sin(π/4+θ)的值.
1 (1+sinθ-cosθ)/(1+sinθ+cosθ)+(1+sinθ+cosθ)/(1+sinθ-cosθ) =[(1+sinθ-cosθ)^2+(1+sinθ+cosθ)^2]/[(1+sinθ+cosθ)*(1+sinθ-cosθ)] =(1+sinθ^2+cosθ^2-2sinθcosθ+2sinθ-2cosθ+1+sinθ^2+cosθ^2+2sinθcosθ+2sinθ+2cosθ)/(1+sinθ^2+2sinθ-cosθ^2) =(1+1+4sinθ+1+1)/(sinθ^2+cosθ^2+sinθ^2-cosθ^2+2sinθ) =(4+4sinθ)/(2sinθ^2+2sinθ) =4(1+sinθ)/[2sinθ(sinθ+1)] =2/sinθ 2 tg(π/4+α)=(tgπ/4+tgα)/(1-tgπ/4tgα)=(1+tgα)/(1-tgα)=-1/2 解得tgα=-3 (tgα-1)/(1+tgα)=2 tgα=sinα/cosα=-3——sinα=-3cosα sin^2(α)+cos^2(α)=1 9cos^2(α)+cos^2(α)=1 cos^2(α)=1/10 (sin2α-2cos^2(α))/(1+tgα) =(2sinαcosα-2cos^2(α))/(1+tgα) =2cos^2(α)*(tgα-1)/(1+tgα) =2*(1/10)*2 =2/5 3 (2cos^2(θ/2)-sinθ-1)/√2sin(π/4+θ) =(cos^2(θ/2)-sinθ-sin^2(θ/2))/√2(sinπ/4cosθ+cosπ/4sinθ) =(cosθ-sinθ)/√2(cosθ/√2+sinθ/√2) =(cosθ-sinθ)/(cosθ+sinθ) =(1-tgθ)/(1+tgθ) =(1-√2)/(1+√2) =(1-√2)*(√2-1)/((1+√2)*(√2-1)) =-(3-2√2) =2√2-3。
1)化简:(1+sinθ-cosθ/1+sinθ+cosθ)+(1+cosθ+sinθ/1-cosθ+sinθ)
=[(1+sinθ-cosθ)^2+(1+cosθ+sinθ)^2]/[(1+sinθ+cosθ)(1-cosθ+sinθ)]
=(2+2sinθ)^2/[(1+sinθ+cosθ)(1-cosθ+sinθ)]-2
=2cscθ
2)已知tg(π/4+α)=-1/2,求sin2α-2cos^2(α)/1+tgα的值。
tg2(π/4+α)=2tg(π/4+α)/[1-tg^(π/4+α)]
=-1/(3/4)
tg(π/2+2α)=-4/3
ctg2α=4/3
sin2α=3/5,cos^2α=16/25,tgα=sin2α/(1+cos2α)=1/3
sin2α-2cos^2(α)/1+tgα=3/5-(32/25)/(1+1/3)
=-0。
36
3)已知tgθ=√2,求2cos^2(θ/2)-sinθ-1/√2sin(π/4+θ)的值。
2cos^2(θ/2)-sinθ-1/√2sin(π/4+θ)
=cosθ+1-sinθ-1/(cosθ+sinθ)
=√3/3+1-√6/3-1/(√3/3+√2/3)=3√2-8√3/3-√6/3
。
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