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

∵sina=5/13,a属于(π/2,π)
∴cosa=-12/13
∴sin2a=2sinacosa=-120/169
cos2a=1-2sin²a=1-2*(5/13)²=119/169
tan2a=sin2a/cos2a=-120/119

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

cosa=-12/13
sin2a=2sinacosa=120/169
cos2a=cos^2a-sin^2a=119/169
tan2a=sin2a/cos2a=120/119