You must know the definition of sinh and cosh to show this.
By definition:
sinh x = (e^x - e^-x)/2 and
cosh x = (e^x + e^-x)/2
So we rewrite the left-hand-side of the equation as follows:
sinh(a) + sinh(b) = (e^a - e^-a)/2 + (e^b - e^-b)/2
This becomes:
= (e^a - e^-a + e^b - e^-b)/2
and rearranging gives:
= (e^a - e^-a + e^b - e^-b)/2 (Eq.A)
Now for the right-hand-side of the equation, we have:
2 sinh((a+b)/2) cosh((a-b)/2)= 2 [(e^((a+b)/2) - e^(-(a+b)/2))/2] [(e^((a-b)/2) + e^(-(a-b)/2))/2]
We can simplify this by cancelling out the 2 and 1/2 and keep only one 1/2:
= 1/2 [e^((a+b)/2) - e^(-(a+b)/2)] [e^((a-b)/2) + e^(-(a-b)/2)]
we can also factor out e^(1/2) as it appears everywhere:
= e^(1/2)/2 [e^(a+b) - e^(-a-b)] [e^(a-b) + e^(-a+b)]
which becomes:
= e^(1/2)/2 [e^(a+b+a-b) + e^(a+b-a+b)-e^(-a-b+a-b) - e^(-a-b -a+b)]
which simplifies to:
= e^(1/2)/2 [e^(2a) + e^(2b)-e^(-2b) - e^(-2a)]
multiply back by e^(1/2) and rearrange:
= 1/2 [e^(a)- e^(-a) + e^(b)-e^(-b) ]
and you readily see that this is the same as our (Eq.A):
= (e^(a)- e^(-a))/2 + (e^(b)-e^(-b))/2 = sinh(a) + sinh(b)
So the left and right parts of the equation are the same and the identity is correct.
Q.E.D.
