已知tan=2,那么1/2sinαcosα+cosα²α和4sin²α+2sinαcosα-cos²α?
人气:254 ℃ 时间:2020-05-23 03:42:19
解答
1/2sinαcosα+cosα²α
=(1/2sinαcosα+cos²α)/1
=(1/2sinαcosα+cos²α)/(sin²a+cos²a)
分子分母同除cos²a得
=(1/2tana+1)/(tan²a+1)
=(1+1)/(4+1)
=2/5
(2)
4sin²α+2sinαcosα-cos²α
=(4sin²α+2sinαcosα-cos²α)/1
=(4sin²α+2sinαcosα-cos²α)/(sin²a+cos²a)
分子分母同除cos²a得
=(4tan²a+2tana-1)/(tan²a+1)
=(16+4-1)/(4+1)
=19/5
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