Question

已知sina=5/13,a属于(π/2,π),求sin2a,cos2a,tan2a的值.

Answer 1

二倍角公式:

sin2α = 2cosαsinα

cos2a=2cosa^2-1=1-2sina^2=cosa^2-sina^2

tan2α=2tanα/[1-(tanα)^2]

sina=5/13,a∈(π/2,π)

cosa=-12/13

tana=-5/12

sin2a=2sanacosa=2*(5/13)*(-12/13)=-120/169

cos2a=2cos^2(a)-1=119/169

tan2a=2tana/(1-tan^2a)=-120/119