已知tanα=-1/2,则5cos^2α+3sinαcosα-2sin^2α
问题描述:
已知tanα=-1/2,则5cos^2α+3sinαcosα-2sin^2α
答
5cos^2α+3sinαcosα-2sin^2α
=(5cos^2α+3sinαcosα-2sin^2α)/(sin²α+cos²α)
=(5+3tanα-2tan²α)/(tan²α+1)
=(5-3/2-2/4)/((-1/2)²+1)
=3/(5/4)
=12/5
答
5cos^2α+3sinαcosα-2sin^2α
=(5cos^2α+3sinαcosα-2sin^2α)/1
=(5cos^2α+3sinαcosα-2sin^2α)/(cos^2α+sin^2α)分子分母同时除以cos^2α
=(5cos^2α/cos^2α+3sinαcosα/cos^2α-2sin^2α/cos^2α)/(cos^2α/cos^2α+sin^2α/cos^2α)
=(5+3tanα-2tan^2α)/(1+tan^2α)
=(5-3*1/2-2*1/4)/(1+1/4)
=3/(5/4)
=3*4/5
=12/5