You can use FullSimplify and play with the ComplexityFunction Option until you obtain a satisfactory result. For example: Let's define our function in terms of LeafCount
c[n_][e_] := n Count[e, _Sin | _ArcTan, Infinity] + LeafCount[e]
Then:
FullSimplify[Sin[1/2 ArcTan[(2 Log[5])/(Log[5]^2 - 2)]],
ComplexityFunction -> c[#]] & /@ Range[40, 60, 4]
Which gives:
{I Sinh[1/4 (Log[1 - (I Log[25])/(-2 + Log[5]^2)] -
Log[1 + (I Log[25])/(-2 + Log[5]^2)])],
I Sinh[1/4 (Log[1 - (I Log[25])/(-2 + Log[5]^2)] -
Log[1 + (I Log[25])/(-2 + Log[5]^2)])],
I Sinh[1/4 (Log[1 - (I Log[25])/(-2 + Log[5]^2)] -
Log[1 + (I Log[25])/(-2 + Log[5]^2)])],
I Sinh[1/4 (Log[1 - (I Log[25])/(-2 + Log[5]^2)] -
Log[1 + (I Log[25])/(-2 + Log[5]^2)])],
Log[5] Sqrt[2/( 4 + Log[5]^4 - 2 Sqrt[4 + Log[5]^4] + Log[5]^2 Sqrt[4
+ Log[5]^4])],
Log[5] Sqrt[2/( 4 + Log[5]^4 - 2 Sqrt[4 + Log[5]^4] + Log[5]^2
Sqrt[4 + Log[5]^4])]}
Where the last 2 answers are the same as that obtained from FunctionExpand. Only this one is more flexible.