化简,求值1)化简:(1+sin
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^...全部
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。
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