There is a very nice relation between trigonometric functions and exponential function:
$$e^{i \theta} = \cos(\theta) + i \sin(\theta)$$
$$\cos(\theta) = \dfrac{e^{i\theta}+e^{-i\theta}}{2} $$
$$\sin(\theta) = \dfrac{e^{i\theta}-e^{-i\theta}}{2i} $$
If we use the expansion for $\cos$ function then it becomes trivial for:
$$ H(e^{i\omega T}) = (1+e^{-i\omega T})$$
- Expand your equation in order to get $-i\omega T / 2$ factor in exponent. This will give you a sum of two exponents, multiplied by the other one:
$$ \begin{align}
H(e^{i\omega T}) = \left(e^{i\omega T/2}+e^{-i\omega T/2} \right)e^{-i\omega T/2} &= \\e^{i\omega T/2} e^{-i\omega T/2} + e^{-i\omega T/2} e^{-i\omega T/2} &=\\ e^0 + e^{-i\omega T}
\end{align}$$
- Use the identity mentioned above in order to obtain $\cos$ term, where $\theta = \omega T/2$. Multiply both sides by $2$:
$$2\cos(\omega T/2) = 2\dfrac{e^{i\omega T/2}+e^{-i\omega T/2}}{2} $$
$$2\cos(\omega T/2) =e^{i\omega T/2}+e^{-i\omega T/2} $$
- Substitute the result:
$$ H(e^{i\omega T}) = \left(e^{i\omega T/2}+e^{-i\omega T/2} \right)e^{-i\omega T/2} = 2\cos(\omega T/2) e^{-i\omega T/2} $$
QED.