tanα=2,则sin^2α+sinαcosα+3cos^2α 谁会帮忙做下 3Q
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tanα=2,则sin^2α+sinαcosα+3cos^2α 谁会帮忙做下 3Q
(sinα)^2+sinαcosα+3(cosα)^2
= {[(sinα)^2+sinαcosα+3(cosα)^2]/ (cosα)^2}*(cosα)^2
= [(sinα)^2/(cosα)^2+sinαcosα/(cosα)^2+3(cosα)^2/(cosα)^2]*(cosα)^2
= [(tanα)^2+tanα +1]*(cosα)^2
=[(tanα)^2+tanα +1]*[1/(secα)^2]
=[(tanα)^2+tanα +1]*{1/[(tanα)^2+1]}
=(4+2+1)*(1/5)
=7/5
= {[(sinα)^2+sinαcosα+3(cosα)^2]/ (cosα)^2}*(cosα)^2
= [(sinα)^2/(cosα)^2+sinαcosα/(cosα)^2+3(cosα)^2/(cosα)^2]*(cosα)^2
= [(tanα)^2+tanα +1]*(cosα)^2
=[(tanα)^2+tanα +1]*[1/(secα)^2]
=[(tanα)^2+tanα +1]*{1/[(tanα)^2+1]}
=(4+2+1)*(1/5)
=7/5
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