设α∈(π,2π)若tan(α+π/6)=2,则cos(π/6-2α)=如题

Question

设α∈(π,2π) 若tan(α+π/6)=2,则cos(π/6-2α)=
如题

Answer 1

tan(α+π/6)=(tanα+tan(π/6))/(1-tanαtan(π/6))
=(tanα+1/√3)/(1-tanα/√3)=(1+√3tanα)/(√3-tanα)=2
解得tanα=(2√3-1)/(√3+2)=5√3-8
tan²α=64+75-80√3=139-80√3
sin2α=2tanα/(1+tan²α)=2(5√3-8)/(140-80√3)
cos2α=(1-tan²α)/(1+tan²α)=(-138+80√3)/(140-80√3)
cos(π/6-2α)=cos(π/6)cos(2α)+sin(π/6)sin(2α)
=√3/2*(-138+80√3)/(140-80√3)+1/2*2(5√3-8)/(140-80√3)
=[(-69√3+40*3)+(5√3-8)]/(140-80√3)
=(112-64√3)/(140-80√3)
=(28-16√3)/(35-20√3)
=4/5*(7-4√3)/(7-4√3)
=4/5