If cos x= 5/13 and sin x < 0 find cos (x/2) and sin (2x)

We know that
if cos x is positive
and
sin x is negative
so
the angle x belong to the IV quadrant
cos x=5/13
we know that
sin²x+cos²x=1> sin²x=1-cos²x> 1-(5/13)²---> 144/169
sin x=√(144/169)> sin x=12/13
but remember that x is on the IV quadrant
so
sin x=-12/13

Part A) cos (x/2)
cos (x/2)=(+/-)√[(1+cos x)/2]
cos (x/2)=(+/-)√[(1+5/13)/2]
cos (x/2)=(+/-)√[(18/13)/2]
cos (x/2)=(+/-)√[36/13]
cos (x/2)=(+/-)6/√13> cos (x/2)=(+/-)6√13/13
the angle (x/2) belong to the II quadrant
so
cos (x/2)=-6√13/13

the answer Part A) is
 cos (x/2)=-6√13/13

Part B) sin (2x)
sin (2x)=2*sin x* cos x> 2*[-12/13]*[5/13]> -120/169

the answer Part B)  is
 sin(2x)=-120/169

A: 18/15b : 7 - -8 = 15 , 12 - -6 = 18 the y variable always goes over the x variables, c : same thing a