已知tanα=2求下列各值 .求4sin平方α-3sinαcosα-5cos平方α 的值

【α应为锐角吧。】
∵tanα = sinα / cosα = 2
∴ sinα = 2cosα cosα = √5 / 5
∴ 4sin²α - 3sinαcosα - 5cos²α
= 16cos²α - 6cos²α - 5cos²α
= 5cos²α
= 1

4sin^2α-3sinαcosα-5cos^2α
依题意的：原式=(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

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