Question

高一数学。求过程谢谢~

已知tanα=2,求值 (1)（cosα+sinα)/(cosα+sinα) ; (2) 2×(sinα)^2－sinαcosα＋(cosα)^2

Answer 1

已知tanα=2,求值 (1)（cosα+sinα)/(cosα-sinα) ; (2) 2sin²α－sinαcosα＋cos²α
解：(1)(cosα+sinα)/(cosα-sinα)=(1+tanα)/(1-tanα)=(1+2)/(1-2)=-3
(2) 2sin²α-sinαcosα+cos²α=1-cos2α-(1/2)sin2α+(1+cos2α)/2=(3/2)-(1/2)(sin2α+cos2α)
=(3/2)-(1/2)[2tanα/(1+tan²α)+(1-tan²α)/(1+tan²α)]
=(3/2)-(1/2)[(2tanα+1-tan²α)/(1+tan²α)]=(3/2)-(1/2)[(4+1-4)/(1+4)]=7/5

Answer 2

(1)（cosα+sinα)/(cosα+sinα) =1
(2) 2×(sinα)^2－sinαcosα＋(cosα)^2
=[2×(sinα)^2－sinαcosα＋(cosα)^2]/((sinα)^2＋(cosα)^2)
=[ 2×(tanα)^2－tanα＋1]/((tanα)^2＋1)
=(2*2^2-2+1)/(2^2+1)
=7/5