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Step-by-Step Solution
Step 1: Write down the given equations
We have:
$$\sin \alpha + \sin \beta = -\frac{21}{65} \quad ...(1)$$
$$\cos \alpha + \cos \beta = -\frac{27}{65} \quad ...(2)$$
Additionally, we are told that
$$\pi < \alpha - \beta < 3\pi.$$
Step 2: Square and add the two given equations
Consider
$$(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2.$$
Expand and simplify:
$$(\sin \alpha + \sin \beta)^2 = \sin^2 \alpha + 2\sin \alpha \sin \beta + \sin^2 \beta,$$
$$(\cos \alpha + \cos \beta)^2 = \cos^2 \alpha + 2\cos \alpha \cos \beta + \cos^2 \beta.$$
Summing these,
$$
(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2
= (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2(\sin \alpha \sin \beta + \cos \alpha \cos \beta).
$$
Since $\sin^2 \theta + \cos^2 \theta = 1$, the above becomes
$$
2 + 2(\sin \alpha \sin \beta + \cos \alpha \cos \beta).
$$
But $\sin \alpha \sin \beta + \cos \alpha \cos \beta = \cos(\alpha - \beta).$
Thus,
$$
(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2
= 2 + 2\cos(\alpha - \beta).
$$
Step 3: Evaluate the left-hand side using the given numerical values
From (1) and (2):
$$
(\sin \alpha + \sin \beta)^2 = \left(-\frac{21}{65}\right)^2 = \frac{441}{4225},
$$
$$
(\cos \alpha + \cos \beta)^2 = \left(-\frac{27}{65}\right)^2 = \frac{729}{4225}.
$$
Adding these,
$$
\frac{441}{4225} + \frac{729}{4225} = \frac{1170}{4225}.
$$
So,
$$
(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2
= \frac{1170}{4225}.
$$
Step 4: Relate the numerical sum to the identity
We have found:
$$
\frac{1170}{4225} = 2 + 2\cos(\alpha - \beta).
$$
Hence,
$$
2\cos(\alpha - \beta) = \frac{1170}{4225} - 2.
$$
But $2 = \frac{8450}{4225}$, so
$$
\frac{1170}{4225} - 2 = \frac{1170}{4225} - \frac{8450}{4225} = -\frac{7280}{4225}.
$$
Thus,
$$
2\cos(\alpha - \beta) = -\frac{7280}{4225}
\quad \Longrightarrow \quad
\cos(\alpha - \beta) = -\frac{3640}{4225}.
$$
However, a more straightforward approach (also as often used) is to note the step in the provided solution that mentions the standard identity:
$$
(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = 2\bigl(1 + \cos(\alpha - \beta)\bigr).
$$
Then we equate
$$
2\bigl(1 + \cos(\alpha - \beta)\bigr) = \frac{1170}{4225}.
$$
Step 5: Solve for $\cos(\alpha - \beta)$
From
$$
2\bigl(1 + \cos(\alpha - \beta)\bigr) = \frac{1170}{4225},
$$
we get
$$
1 + \cos(\alpha - \beta) = \frac{1170}{2 \cdot 4225} = \frac{1170}{8450}.
$$
Hence,
$$
\cos(\alpha - \beta) = \frac{1170}{8450} - 1 = \frac{1170 - 8450}{8450} = -\frac{7280}{8450}.
$$
We can proceed to find $\cos^2\frac{\alpha - \beta}{2}$:
$$
\cos(\alpha - \beta) = 1 - 2\sin^2\left(\frac{\alpha - \beta}{2}\right)
$$
or
$$
\cos(\alpha - \beta) = 2\cos^2\left(\frac{\alpha - \beta}{2}\right) - 1.
$$
Either approach leads to
$$
\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{1 + \cos(\alpha - \beta)}{2}.
$$
Substituting the value found, we simplify to get
$$
\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{9}{130}.
$$
Thus,
$$
\cos\left(\frac{\alpha - \beta}{2}\right) = \pm \frac{3}{\sqrt{130}}.
$$
Step 6: Determine the correct sign of $\cos\frac{\alpha - \beta}{2}$
We are given that
$$\pi < \alpha - \beta < 3\pi.$$
Dividing by 2 implies
$$
\frac{\pi}{2} < \frac{\alpha - \beta}{2} < \frac{3\pi}{2}.
$$
In the interval $\bigl(\frac{\pi}{2}, \frac{3\pi}{2}\bigr),$ the cosine function is negative. Therefore,
$$
\cos\left(\frac{\alpha - \beta}{2}\right) = -\frac{3}{\sqrt{130}}.
$$
Final Answer
$$
\cos\left(\frac{\alpha - \beta}{2}\right) = -\frac{3}{\sqrt{130}}.
$$
