已知tanα=2,则3sin²α-4cos²α=,sinαcosα=
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![已知tanα=2,则3sin²α-4cos²α=,sinαcosα=](/uploads/image/z/2738576-56-6.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B1%EF%BC%9D2%2C%E5%88%993sin%26%23178%3B%CE%B1%EF%BC%8D4cos%26%23178%3B%CE%B1%EF%BC%9D%2Csin%CE%B1cos%CE%B1%EF%BC%9D)
已知tanα=2,则3sin²α-4cos²α=,sinαcosα=
已知tanα=2,则3sin²α-4cos²α=,sinαcosα=
已知tanα=2,则3sin²α-4cos²α=,sinαcosα=
3sin²α-4cos²α
=cos²α(3tan²α-4)
=8cos²α
=4(cos2α+1)
=4cos2α +4
=4(1-tan²α)/(1+tan²α) +4
=4(1-4)/(1+4) +4
=4- (12/5)
=8/5
sinαcosα
=(sin2α)/2
=tanα/(1+tan²α)
=2/(1+4)
=2/5