Question

若tan(π-α)=1/3,则cos2α/2sinαcosα+cos2α的值为

Answer 1

∵tan(π-α)=-tanα=1/3
tanα=-1/3
∴cos2α/(2sinαcosα+cos2α)
=(cos²α-sin²α)/(2sinαcosα+cos²α-sin²α)
=(1-sin²α/cos²α)/(2sinα/cosα+1-sin²α/cos²α) 【分子分母同时除以cos²α】
=(1-tan²α)/(2tanα+1-tan²α) 【sinα/cosα=tanα】
=(1-1/9)/(-2/3+1-1/9)
=8/(-6+9-1)
=8/2
=

Answer 2

tan(π-α)=-tana=1/3,则tana=-1/3
cos2α/[2sinαcosα+cos2α]
=[cos²α-sin²a]/[2sinαcosα+cos²α-sin²]
=(1-tan²a)/[2tana+1-tan²a]
=(1-1/9)/[-2/3+1-1/9]
=(8/9)/(2/9)
=4