已知tan=2,求下列各式的值4sinα-3sinαcosα-5cosα

问题描述:

已知tan=2,求下列各式的值4sinα-3sinαcosα-5cosα

4sin^2α-3sinαcosα-5cos^2α
=(4sin^2α-3sinαcosα-5cos^2α)/1
=(4sin^2α-3sinαcosα-5cos^2α)/(sin^2a+cos^2a) (上下同除以cos^2a得)
=(4tan^2α-3tanα-5)/(tan^2a+1)
=(16-6-5)/(5)
=1