Question

已知tanα=-1/2,求下列各式的值（1）4sinα-3cosα/2sinα+5cosα (2)2sin^2α-3sinαcosα-5cos^2α

Answer 1

（1）(4sinα-3cosα)/(2sinα+5cosα)
=(4tana-3)/(2tana+5) (上下同时除以cosα)
=(-2-3)/(-1+5)
=-5/4
(2)2sin^2α-3sinαcosα-5cos^2α
=(2sin^2α-3sinαcosα-5cos^2α)/(sin²α+cos²α)
=(2tan²α-3tanα-5)/(tan²α+1) (上下同时除以cos²α)
=(1/2+3/2-5)/(1/4+1)
=-12/5

Answer 2

4sinα-3cosα/2sinα+5cosα
=(4tana-3)/(2tana+5)
=[4*(-1/2)-3]/[2*(-1/2)+5]
=(-5)/4
=-5/4

2sin^2α-3sinαcosα-5cos^2α
=cos²a(2tan²a-3tana-5)
cos²a=(1+cos2a)/2
cos2a=(1-tan²a)/(1+tan²a)=(1-1/4)/(1+1/4)=3/5
cos²a=(1+3/5)/2=4/5
2sin^2α-3sinαcosα-5cos^2α
=(4/5)[1/4*2+3/2-5]
=4/5*(-3)
=-12/5