Question

Let $\alpha, \beta$ be such that $\pi < \alpha - \beta < 3\pi$. If $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$, then the value of $\cos \frac{\alpha - \beta}{2}$ is:

A) $-\frac{3}{\sqrt{130}}$ B) $\frac{3}{\sqrt{130}}$ C) $\frac{6}{65}$ D) $-\frac{6}{65}$

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Answer


We are given that:

$$ \sin \alpha + \sin \beta = -\frac{21}{65} $$

and

$$ \cos \alpha + \cos \beta = -\frac{27}{65} $$

We need to find the value of $ \cos \frac{\alpha - \beta}{2} $.

Step 1: Square and sum the given equations.

[ (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 ]

Substituting the given values:

[ \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2 ]

Step 2: Simplify the expressions under the squares.

[ \left(\frac{21}{65}\right)^2 + \left(\frac{27}{65}\right)^2 = \frac{441}{4225} + \frac{729}{4225} ]

[ = \frac{1170}{4225} ]

Step 3: Apply the Pythagorean identity.

[ (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2 (\sin \alpha \sin \beta + \cos \alpha \cos \beta) = \frac{1170}{4225} ]

Since $ \sin^2 \alpha + \cos^2 \alpha = 1 $ and $ \sin^2 \beta + \cos^2 \beta = 1 $,

Step 4: Simplify the above equation.

[ 2 + 2(\sin \alpha \sin \beta + \cos \alpha \cos \beta) = \frac{1170}{4225} ]

Divide both sides by 2:

[ 1 + \cos (\alpha - \beta) = \frac{585}{4225} ]

[ \cos (\alpha - \beta) = \frac{585}{4225} - 1 = \frac{585 - 4225}{4225} = \frac{-3640}{4225} ]

[ \cos (\alpha - \beta) = -\frac{3640}{4225} ]

Step 5: Use the double angle formula to find $ \cos \frac{\alpha - \beta}{2} $.

[ \cos (\alpha - \beta) = 2 \cos^2 \left( \frac{\alpha - \beta}{2} \right) - 1 ]

[ -\frac{3640}{4225} = 2 \cos^2 \left( \frac{\alpha - \beta}{2} \right) - 1 ]

[ 2 \cos^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \frac{3640}{4225} = \frac{4225 - 3640}{4225} = \frac{585}{4225} ]

[ \cos^2 \left( \frac{\alpha - \beta}{2} \right) = \frac{585}{2 \times 4225} = \frac{585}{8450} = \frac{1170}{16900} = \frac{9}{130} ]

Step 6: Solve for $ \cos \left( \frac{\alpha - \beta}{2} \right) $.

Since $ \pi < \alpha - \beta < 3 \pi $

The value is negative:

[ \cos \left( \frac{\alpha - \beta}{2} \right) = -\frac{3}{\sqrt{130}} ]

Thus, the correct option is:

A) $-\frac{3}{\sqrt{130}}$


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