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数学
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已知θ为第三象限角,1-sinθcosθ-3cos^2=0则5sin^2θ+3sinθcosθ=?
人气:428 ℃ 时间:2019-08-31 01:30:39
解答
sinθcosθ+3cos^2θ=1
θ在第三象限,令tanθ=t,t>0
则,sinθ=-t/√(1+t^2),cosθ=-1/√(1+t^2)
代入上式得:
t+3=1+t^2
(t-2)(t+1)=0
t=2
同样,用tanθ表示5sin^2θ+3sinθcosθ
5sin^2θ+3sinθcosθ
=(5t^2+5t)/(1+t^2)
=(20+10)/(1+4)=6
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