已知tanα=2,则cos(π÷2-2α)+cos2α=

问题描述:

已知tanα=2,则cos(π÷2-2α)+cos2α=

cos(π÷2-2α)+cos2α
=sin2α+cos2α
=2tanα/(1+tan²α)+(1-tan²α)/(1+tan²α)
=(2tanα+1-tan²α)/(1+tan²α)
=(2*2+1-2²)/(1+2²)
=(4+1-4)/5
=1/5

2sinacosa+cosa^2-sina^2=1/5

cos(π÷2-2α)+cos2α=sin2a+cos2a
=2sinacosa+(cosa)^2-(sina)^2
=[2sinacosa+(cosa)^2-(sina)^2]/[(cosa)^2+(sina)^2]
=[2tana+1-(tana)^2]/[1+(tana)^2]
=1/5