Using the definitions of the corresponding functions, prove…

Using the definitions of the corresponding functions, prove: $\ \cosh^2\left(\dfrac z2\right)=\dfrac12(\cosh z+1)$.

Step 1 · Left-hand side

$$ \cosh^2\left(\frac z2\right)=\left(\frac{e^{z/2}+e^{-z/2}}2\right)^2=\frac{e^z+2e^{z/2}e^{-z/2}+e^{-z}}4=\frac{e^z+2+e^{-z}}4 $$

Step 2 · Right-hand side

$$ \frac12(\cosh z+1)=\frac12\left(\frac{e^z+e^{-z}}2+1\right)=\frac{e^z+e^{-z}+2}4=\frac{e^z+2+e^{-z}}4 $$

Step 3 · Compare

Both sides equal $\dfrac{e^z+2+e^{-z}}4$, so the identity holds.

Alternative route · via the double-angle formula

From $\cosh^2w-\sinh^2w=1$ (previous problem), $\sinh^2w=\cosh^2w-1$, and the addition formula gives $\cosh(2w)=\cosh^2w+\sinh^2w=2\cosh^2w-1$, so $\cosh^2w=\dfrac{\cosh(2w)+1}2$. Setting $w=z/2$ reproduces the same identity.