Prove the following identity. ?

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1 Answer
Feb 21, 2018

Answer:

The identity is:
#cosh(a) + cosh(b) = 2cosh( (a+b)/2) cosh( (a-b)/2)#
See explanation for its derivation.

Explanation:

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!