How should one proceed in this trigonometric simplification involving non integer angles? The problem is as follows:

Find the value of this function 
  $$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$
  when $\omega=33^{\circ}{20}'$ and $\phi=56^{\circ}{40}'$.

Thus,
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right)^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right)^{2}$$
By solving the power I obtained the following:
$$A= \cos^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +\sin^{2}\frac{\omega}{2} +\sin^{2}\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}$$
I noticed some familiar terms and using pitagoric identities then I rearranged the equation as follows:
\begin{align}
A &= \cos^{2}\frac{\omega}{2} +\sin^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +\sin^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}   \\
  &= 1+1+2\cos\frac{\omega}{2}cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}
\end{align}
Since the latter terms are another way to write prosthapharesis formulas I did the following:
\begin{align}
\cos\alpha +\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}   \\
\cos\alpha -\cos\beta &= -2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}   \\ \\
\alpha+\beta &= \omega   \\
\alpha-\beta &= \phi
\end{align}
By solving the system I found: $\alpha=\frac{\omega+\phi}{2}$ and $\beta=\frac{\omega-\phi}{2}$.
Therefore by inserting these into the problem:
\begin{align}
A &= 1+1 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}   \\
  &= 2 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} -\left(-2\sin\frac{\omega}{2}\cos\frac{\phi}{2}\right)   \\
  &= 2 +\cos\frac{\omega+\phi}{2} +\cos\frac{\omega-\phi}{2} -\left(\cos\frac{\omega+\phi}{2} -\cos\frac{\omega-\phi}{2}\right)
\end{align}
By cancelling elements,
$$A=2 +2\cos\frac{\omega-\phi}{2}$$
However I'm stuck at trying to evaluate these values:
\begin{align}
\omega &= 33^{\circ}{20}'\; \phi=56^{\circ}{40}'   \\
\omega-\phi &= \left(33+\frac{20}{60}\right) -\left(56+\frac{40}{60}\right) = -23-\frac{20}{60}
\end{align}
Therefore,
$$A=2+2\cos\left(\frac{-23-\frac{20}{60}}{2}\right).$$
However the latter answer does not appear in the alternative neither seems to be right. Is there something wrong on what I did?
 A: There is a mistake in your expansion of the second bracket.
We should have 
$A=\cos^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2\frac{\phi}{2}$
$=2+2(\cos\frac{\omega}{2} \cos\frac{\phi}{2} -\sin\frac{\omega}{2}\sin\frac{\phi}{2})$ .
In the bracket we have the expansion for $\cos(\frac{\omega}{2}+\frac{\phi}{2})$
and fortunately $\omega+\phi=90^\circ$.
Thus $A=2+2\cos(45^\circ)$
$=2+\sqrt{2}$
A: Hint: You did a miscalculation:
$$A=\left (\cos\frac{\omega}{2}+\cos\frac{\phi}{2}  \right )^{2}+\left (\sin\frac{\omega}{2}-\sin\frac{\phi}{2}  \right )^{2}$$
$$=\cos ^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin ^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2 \frac{\phi}{2}$$
$$=2+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}$$
$$=2+2\left(\cos\frac{\omega}{2}\cos\frac{\phi}{2}-\sin\frac{\omega}{2}\sin\frac{\phi}{2}\right)$$
$$=2+2\cos\left(\frac{\omega}{2}+\frac{\phi}{2}\right)$$
In the last step I used $\cos(x+y)=\cos x\cos y-\sin x \sin y$.
