If \[ \cos\alpha+\cos\beta=\frac{1}{3} \] and \[ \sin\alpha+\sin\beta=\frac{1}{4}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \pm\frac{5}{24} \]
Question
If \[ \cos\alpha+\cos\beta=\frac{1}{3} \] and \[ \sin\alpha+\sin\beta=\frac{1}{4}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \pm\frac{5}{24}. \]
Solution
Using the sum-to-product identities,
\[ \cos\alpha+\cos\beta = 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Therefore,
\[ 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \frac{1}{3} \]
\[ \cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \frac{1}{6} \]
Also,
\[ \sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Hence,
\[ 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \frac{1}{4} \]
\[ \sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \frac{1}{8} \]
Now square and add the two equations:
\[ \left( \cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \right)^2 + \left( \sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \right)^2 \]
\[ = \left(\frac{1}{6}\right)^2 + \left(\frac{1}{8}\right)^2 \]
Taking \[ \cos^2\frac{\alpha-\beta}{2} \] common,
\[ \cos^2\frac{\alpha-\beta}{2} \left( \cos^2\frac{\alpha+\beta}{2} + \sin^2\frac{\alpha+\beta}{2} \right) = \frac{1}{36} + \frac{1}{64} \]
Using
\[ \sin^2\theta+\cos^2\theta=1, \]
we get
\[ \cos^2\frac{\alpha-\beta}{2} = \frac{1}{36} + \frac{1}{64} \]
\[ = \frac{16+9}{576} \]
\[ = \frac{25}{576} \]
Taking square root,
\[ \cos\frac{\alpha-\beta}{2} = \pm\frac{5}{24} \]
Hence proved.
Final Answer
\[ \boxed{ \cos\frac{\alpha-\beta}{2} = \pm\frac{5}{24} } \]