1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程
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1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程
1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程
1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程
1-sin^2θ-cos^2θ/1-sinθ-cosθ
=1-1/1-sinθ-cosθ
=0
1-sin^6θ-cos^6θ/1-sin^4θ-cos^4θ
=1-(sin^6θ+cos^6θ)/1-sin^4θ-cos^4θ
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)/1-sin^4θ-cos^4θ
=1-1×(sin^4θ-sin^2θcos^2θ+cos^4θ)/1-sin^4θ-cos^4θ
=1-sin^4θ+sin^2θcos^2θ-cos^4θ]/1-sin^4θ-cos^4θ
=1 - (sin^2θcos^2θ/1-sin^4θ-cos^4θ)
=1 - (sin^2θcos^2θ/sin^2θ+cos^2θ-sin^4θ-cos^4θ)
=1 - (sin^2θcos^2θ/sin^2θ(1-sin^2θ)+cos^2θ(1-cos^2θ))
=1 - (sin^2θcos^2θ/sin^2θcos^2θ+cos^2θsin^2θ)
=1 - (sin^2θcos^2θ/2sin^2θcos^2θ)
=1-1/2
=1/2
sin^2θ+cos^2θ=1
1-sin^2θ-cos^2θ=1-(sin^2θ+cos^2θ)=0
所以1-sin^2θ-cos^2θ/1-sinθ-cosθ =0
0
(sinθ)^2+(cosθ)^2=1