已知tanα=-2,求下列各式值
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/05 14:46:25
已知tanα=-2,求下列各式值
(1) cosα+3sinα/3cosα-sinα
(2)sin^2α+2sinαcosα-10cos^2α+1
(1) cosα+3sinα/3cosα-sinα
(2)sin^2α+2sinαcosα-10cos^2α+1
∵tanα=-2
∴cosα≠0
(1)原式=(cosα+3sinα)/(3cosα-sinα)
=[(cosα+3sinα)×1/cosα]/[(3cosα-sinα)×1/cosα]
=(1+3sinα/cosα)/(3-sinα/cosα)
=(1+3tanα)/(3-tanα)
=[1+3×(-2)]/(3+2)
=-1
(2) ∵sin²α+cos²α=1
∴原式=sin²α+2sinαcosα-10cos²α+sin²α+cos²α
=2sin²α+2sinαcosα-9cos²α
=(2sin²α+2sinαcosα-9cos²α)/(sin²α+cos²α)
=[(2sin²α+2sinαcosα-9cos²α)×1/cos²α]/[(sin²α+cos²α)×1/cos²α]
=(2sin²α/cos²α+2sinα/cosα-9)/(sin²α/cos²α+1)
=(2tan²α+2tanα-9)/(tan²α+1)
=[2×(-2)²+2×(-2)-9]/[(-2)²+1]
=(8-4-9)/(4+1)
=-1
∴cosα≠0
(1)原式=(cosα+3sinα)/(3cosα-sinα)
=[(cosα+3sinα)×1/cosα]/[(3cosα-sinα)×1/cosα]
=(1+3sinα/cosα)/(3-sinα/cosα)
=(1+3tanα)/(3-tanα)
=[1+3×(-2)]/(3+2)
=-1
(2) ∵sin²α+cos²α=1
∴原式=sin²α+2sinαcosα-10cos²α+sin²α+cos²α
=2sin²α+2sinαcosα-9cos²α
=(2sin²α+2sinαcosα-9cos²α)/(sin²α+cos²α)
=[(2sin²α+2sinαcosα-9cos²α)×1/cos²α]/[(sin²α+cos²α)×1/cos²α]
=(2sin²α/cos²α+2sinα/cosα-9)/(sin²α/cos²α+1)
=(2tan²α+2tanα-9)/(tan²α+1)
=[2×(-2)²+2×(-2)-9]/[(-2)²+1]
=(8-4-9)/(4+1)
=-1