已知tanα/(tanα-1)=-1 求sin²α+sinαcosα+2的值
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tanα/(tanα-1)=-1
tanα=-(tanα-1)
tanα=1-tanα
tanα=1/2
sin²α+sinαcosα+2
=(sin²α+sinαcosα+2*1)/1
=[sin²α+sinαcosα+2*(sin²α+cos²α)]/(sin²α+cos²α)
=(sin²α+sinαcosα+2sin²α+2cos²α)/(sin²α+cos²α)
=(tan²α+tanα+2tanα+2)/(tan²α+1)
将tanα=1/2代入上式得
(1/4+1/2+1/2+2)/(1/4+1)=13/5
tanα=-(tanα-1)
tanα=1-tanα
tanα=1/2
sin²α+sinαcosα+2
=(sin²α+sinαcosα+2*1)/1
=[sin²α+sinαcosα+2*(sin²α+cos²α)]/(sin²α+cos²α)
=(sin²α+sinαcosα+2sin²α+2cos²α)/(sin²α+cos²α)
=(tan²α+tanα+2tanα+2)/(tan²α+1)
将tanα=1/2代入上式得
(1/4+1/2+1/2+2)/(1/4+1)=13/5
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