2(sinα)^2-3sinαcosα-1
=[2(sinα)^2-3sinαcosα-1]/1
=[2(sinα)^2-3sinαcosα-(sinα)^2-(cosα)^2]/[(sinα)^2+(cosα)^2]
=[(sinα)^2-3sinαcosα-(cosα)^2]/[(sinα)^2+(cosα)^2]
上下同除以(cosα)^2
=[(tanα)^2-3tanα-1]/[(tanα)^2+1]
=(1/4+3/2-1)/(1/4+1)
=3/5