已知θ是第三象限角,且sin^4θ+cos^4θ=5/9,则sinθcosθ=
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已知θ是第三象限角,且sin^4θ+cos^4θ=5/9,则sinθcosθ=
已知θ是第三象限角,且sin^4θ+cos^4θ=5/9,则sinθcosθ=
已知θ是第三象限角,且sin^4θ+cos^4θ=5/9,则sinθcosθ=
5/9=sin^4θ+cos^4θ=(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=1-2sin^2θcos^2θ,
所以,sin^2θcos^2θ=2/9,
因为θ是第三象限角,所以sinθ
将条件平方相加减,求出余弦,就可以了
谢谢(sin^2θ+cos^2θ)^2=sin^4θ+cos^4θ+2(sinθcosθ)^2=1 so sin2θ=2sinθcosθ=2根号((1-5/9)/2)=2根号2/3
sin^4θ+cos^4θ=5/9
sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ=5/9
(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9
1^2-2sin^2θcos^2θ=5/9
1-2sin^2θcos^2θ=5/9
2sin^2θcos^2θ=1-5/9
2sin^2θcos^2θ=...
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sin^4θ+cos^4θ=5/9
sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ=5/9
(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ=5/9
1^2-2sin^2θcos^2θ=5/9
1-2sin^2θcos^2θ=5/9
2sin^2θcos^2θ=1-5/9
2sin^2θcos^2θ=4/9
sin^2θcos^2θ=2/9
sinθcosθ=(±√2)/3
θ是第三象限角
sinθ<0,cosθ<0
所以sinθcosθ=(√2)/3
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