When input is \(x\left [ n\right ] =e^{j\Omega _{0}n}\) and system is given by \(H\left ( \Omega \right ) \) then the output is \(y\left [ n\right ] =e^{j\Omega _{0}n}H\left ( \Omega _{0}\right ) \) which is the same as \(y\left [ n\right ] =e^{j\Omega _{0}n}\left \vert H\left ( \Omega _{0}\right ) \right \vert e^{j\arg H\left ( \Omega _{0}\right ) }\).
Hence when the input is linear combination of complex exponentials \begin{align*} A\cos \left ( \Omega _{0}n+\theta \right ) & =\frac{A}{2}\left ( e^{j\left ( \Omega _{0}n+\theta \right ) }+e^{-j\left ( \Omega _{0}n+\theta \right ) }\right ) \\ & =\left ( \frac{A}{2}e^{j\theta }\right ) e^{j\Omega _{0}n}+\left ( \frac{A}{2}e^{-j\theta }\right ) e^{-j\Omega _{0}n} \end{align*}
Then, and since the system is linear, then the output will be scaled and linear sum of each output corresponding to each term above. In other words, when the input is \(\left ( \frac{A}{2}e^{j\theta }\right ) e^{j\Omega _{0}n}\) then the output is\begin{align} y_{1}\left [ n\right ] & =\left ( \frac{A}{2}e^{j\theta }\right ) e^{j\Omega _{0}n}\left \vert H\left ( \Omega _{0}\right ) \right \vert e^{j\arg H\left ( \Omega _{0}\right ) }\nonumber \\ & =\left \vert H\left ( \Omega _{0}\right ) \right \vert \frac{A}{2}e^{j\left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) }\tag{1} \end{align}
And when the input is \(\left ( \frac{A}{2}e^{-j\theta }\right ) e^{-j\Omega _{0}n}\) then the output is\begin{align*} y_{2}\left [ n\right ] & =\left ( \frac{A}{2}e^{-j\theta }\right ) e^{-j\Omega _{0}n}\left \vert H\left ( -\Omega _{0}\right ) \right \vert e^{j\arg H\left ( -\Omega _{0}\right ) }\\ & =\left \vert H\left ( -\Omega _{0}\right ) \right \vert \frac{A}{2}e^{-j\left ( \Omega _{0}n+\theta -\arg H\left ( -\Omega _{0}\right ) \right ) } \end{align*}
But for real input, which is the case here, \(\left \vert H\left ( \Omega _{0}\right ) \right \vert \) is symmetrical. Hence \(\left \vert H\left ( \Omega _{0}\right ) \right \vert =\left \vert H\left ( -\Omega _{0}\right ) \right \vert \) and \(\arg H\left ( -\Omega _{0}\right ) =-\arg H\left ( \Omega _{0}\right ) \) (see table 4.6 for these properties). Hence \begin{equation} y_{2}[n]=\left \vert H\left ( \Omega _{0}\right ) \right \vert \frac{A}{2}e^{-j\left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) }\tag{2} \end{equation} Therefore, by linearity, \(y\left [ n\right ] =y_{1}\left [ n\right ] +y_{2}\left [ n\right ] \) or by adding (1) and (2)\begin{align*} y\left [ n\right ] & =\left \vert H\left ( \Omega _{0}\right ) \right \vert \frac{A}{2}e^{j\left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) }+\left \vert H\left ( \Omega _{0}\right ) \right \vert \frac{A}{2}e^{-j\left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) }\\ & =\left \vert H\left ( \Omega _{0}\right ) \right \vert A\left ( \frac{e^{j\left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) }+e^{-j\left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) }}{2}\right ) \\ & =\left \vert H\left ( \Omega _{0}\right ) \right \vert A\cos \left ( \Omega _{0}n+\theta +\arg H\left ( \Omega _{0}\right ) \right ) \end{align*}