You must know the definition of sinh and cosh to find the identity.
By definition:
sinh x = (e^x - e^-x)/2 and
cosh x = (e^x + e^-x)/2
Hint: see how is the identity of sinh a + sinh b.
We start off by rewriting:
cosh a + cosh b = (e^a + e^-a)/2 + (e^b + e^-b)/2
factoring out 1/2 gives:
=1/2 [e^a + e^-a + e^b + e^-b]
factoring out e^(1/2) gives:
=e^(1/2)/2 [e^(2a) + e^(-2a) + e^(2b) + e^(-2b)]
Trick! Add "zeroes" in the form of a-a or b-b:
=e^(1/2)/2 [e^(a+a+b-b) + e^(-a-a+b-b) + e^(b+b+a-a) + e^(-b-b+a-a)]
Separate a from a and b from b (and rearrange) to get :
=e^(1/2)/2 [e^(a+b)e^(a-b) + e^-(a+b)e^-(a-b) + e^(a+b)e^-(a-b) + e^-(a+b)e^(a-b)]
This becomes:
=e^(1/2)/2 [(e^(a+b) +e^-(a+b))(e^(a-b)+e^-(a-b))]
Re-introduce e^(1/2)
=1/2 [(e^((a+b)/2) +e^(-(a+b)/2))(e^((a-b)/2)+e^(-(a-b)/2))]
Trick! Multiply by 1=2/2:
=2 [(e^((a+b)/2) +e^(-(a+b)/2))]/2 [(e^((a-b)/2)+e^(-(a-b)/2))]/2
which is really just
=2cosh( (a+b)/2) cosh( (a-b)/2)
The identity is:
cosh(a) + cosh(b) = 2cosh( (a+b)/2) cosh( (a-b)/2)
Quite tricky!
