In one of the problems in 8.04 problem set, the wave function given (29) has both growing exponential and a decaying exponential which is an even function by itself. But they are splitting it into odd and even function again using $\cosh(x)$ and $\sinh(x)$? They say $\cosh(x)$ and $\sinh(x)$ are the odd and even combination for real exponentials (which is true). but how is it possible to split combination real exponentials as $\cosh(x)$ and $\sinh(x)$?
No. Arbitrary combinations of the two functions is not even or odd. $$f(x)=Ae^{\kappa x}+B e^{-\kappa x} $$ $$f(-x)= Ae^{-\kappa x}+B e^{\kappa x}$$ Now if we impose $f(x)=\pm f(-x)$, we get the condition: $$A=\pm B$$ This is nothing but the combinations for the hyperbolic trigonometric functions. $$\cosh(x)=\frac{e^x+e^{-x}}{2}$$ $$\sinh(x)=\frac{e^x-e^{-x}}{2}$$
Take your (29) and replace $x\to -x$ to get \begin{align} \psi(x)=\left\{\begin{array}{l} Ae^{-\kappa x} \\ Ce^{-\kappa x} +De^{\kappa x}\\ F e^{\kappa x}\end{array}\right. \end{align} and compare with your initial (29).
Clearly to get $+\psi(x)$ back you need $A=F$ and $C=D$: this produces an even function since by construction $\psi(x)=\psi(-x)$.
To get $-\psi(x)$ you need $A=-F$ and $C=-D$, which produces an odd function since $\psi(x)=-\psi(-x)$.
