已知tanθ=3,则sin^2θ+sinθcosθ-2cos^2θ=
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sin^2θ+sinθcosθ-2cos^2θ
=1/cos^2θ (sin^2θ/cos^2θ+sinθcosθ/cos^2θ-2cos^2θ/cos^2θ)
=1/cos^2θ (tan^2θ+tanθ-2)
= (tan^2θ+1) (3^2+3-2)
=(3^2+1)*(9+1)
=100
=1/cos^2θ (sin^2θ/cos^2θ+sinθcosθ/cos^2θ-2cos^2θ/cos^2θ)
=1/cos^2θ (tan^2θ+tanθ-2)
= (tan^2θ+1) (3^2+3-2)
=(3^2+1)*(9+1)
=100
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2011-12-07 · 知道合伙人教育行家
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tanθ=3
sin^2θ+sinθcosθ-2cos^2θ
= 1/cos^2θ * (tan^2θ+tanθ-2)
= (tan^2θ+1) * (tan^2θ+tanθ-2)
= (3^2+1) * (3^2+3-2)
= 10 * 10
= 100
sin^2θ+sinθcosθ-2cos^2θ
= 1/cos^2θ * (tan^2θ+tanθ-2)
= (tan^2θ+1) * (tan^2θ+tanθ-2)
= (3^2+1) * (3^2+3-2)
= 10 * 10
= 100
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sin^2θ+sinθcosθ-2cos^2θ=(sin^2θ/cos^2θ+sinθ/cosθ-2)cos^2θ
=(tan^2θ+tanθ-2)/sec^2θ
=(tan^2θ+tanθ-2)/(tan^2θ+1)
=(3^2+3-2)/(3^2+1)
=1
=(tan^2θ+tanθ-2)/sec^2θ
=(tan^2θ+tanθ-2)/(tan^2θ+1)
=(3^2+3-2)/(3^2+1)
=1
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