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