1,已知sinθ-cosθ=-1/5求(1)sinθcosθ(2)sin^4θ+cos^4θ2,证明下列恒等式(1)2cos²θ+sin^4θ=cos^θ+1(2)sin^4θ+sin²θcos²θ+cos²θ=1

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1,已知sinθ-cosθ=-1/5求(1)sinθcosθ(2)sin^4θ+cos^4θ2,证明下列恒等式(1)2cos²θ+sin^4θ=cos^θ+1(2)sin^4θ+sin²θcos²θ+cos²θ=1

1,已知sinθ-cosθ=-1/5求(1)sinθcosθ(2)sin^4θ+cos^4θ2,证明下列恒等式(1)2cos²θ+sin^4θ=cos^θ+1(2)sin^4θ+sin²θcos²θ+cos²θ=1
1,已知sinθ-cosθ=-1/5
求(1)sinθcosθ
(2)sin^4θ+cos^4θ
2,证明下列恒等式
(1)2cos²θ+sin^4θ=cos^θ+1
(2)sin^4θ+sin²θcos²θ+cos²θ=1

1,已知sinθ-cosθ=-1/5求(1)sinθcosθ(2)sin^4θ+cos^4θ2,证明下列恒等式(1)2cos²θ+sin^4θ=cos^θ+1(2)sin^4θ+sin²θcos²θ+cos²θ=1
1、(1)sinθ-cosθ=-1/5两边平方得:1-2cosθsinθ=1/25,所以sinθcosθ=12/25
(2)sin^4θ+cos^4θ=(sin^2θ+cos^2θ)^2-2(sinθcosθ)^2=1-2*(144/625)=
337/625
2、(1)2cos²θ+sin^4θ=cos²θ+cos²θ+sin^4θ=cos²θ+cos²θ(sin^2θ+
cos^2θ)+sin^4θ=cos^4θ+cos²θ+(cos²θsin^2θ+sin^4θ)=cos^4θ+cos²θ
+sin^2θ=cos^4θ+1
(2)sin^4θ+sin²θcos²θ+cos²θ=sin²θ(cos²θ+sin²θ)+cos²θ=sin²θ
+cos²θ=1