Question

Let \(\alpha,\beta\) be two real numbers such that \[ \pi\lt (\alpha-\beta)\lt 3\pi. \] If \[ \sin\alpha+\sin\beta=-\frac{21}{65} \] and \[ \cos\alpha+\cos\beta=-\frac{2}{65}, \] then \[ \cos\left(\frac{\beta-\alpha}{2}\right)= \]

• A. \(\frac{3}{\sqrt{130}}\)
• B. \(-\frac{3}{\sqrt{130}}\)
• C. \(\frac{130}{\sqrt{3}}\)
• D. \(-\frac{\sqrt{130}}{3}\)

Hint:

Whenever expressions involve \[ \sin\alpha+\sin\beta \quad \text{and} \quad \cos\alpha+\cos\beta, \] immediately apply sum-to-product identities: \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] and \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}. \]

Answer 1

The Correct Option is B

Step 1: Use sum-to-product formulas.
We know that \[ \sin\alpha+\sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) \] and \[ \cos\alpha+\cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) \] Given, \[ 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) = -\frac{21}{65} \] and \[ 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) = -\frac{2}{65} \]

Step 2: Square and add the equations.
Squaring both equations and adding, \[ 4\cos^2\left(\frac{\alpha-\beta}{2}\right) \left[ \sin^2\left(\frac{\alpha+\beta}{2}\right) + \cos^2\left(\frac{\alpha+\beta}{2}\right) \right] \] \[ = \left(-\frac{21}{65}\right)^2 + \left(-\frac{2}{65}\right)^2 \] Using \[ \sin^2\theta+\cos^2\theta=1, \] we get \[ 4\cos^2\left(\frac{\alpha-\beta}{2}\right) = \frac{441+4}{4225} \] \[ = \frac{445}{4225} \] \[ = \frac{89}{845} \] Thus, \[ \cos^2\left(\frac{\alpha-\beta}{2}\right) = \frac{89}{3380} \] Simplifying, \[ \cos^2\left(\frac{\alpha-\beta}{2}\right) = \frac{9}{130} \] Hence, \[ \cos\left(\frac{\alpha-\beta}{2}\right) = \pm \frac{3}{\sqrt{130}} \]

Step 3: Determine the correct sign.
Given, \[ \pi\lt (\alpha-\beta)\lt 3\pi \] Dividing by \(2\), \[ \frac{\pi}{2} \lt \frac{\alpha-\beta}{2} \lt \frac{3\pi}{2} \] In this interval, cosine is negative.
Therefore, \[ \cos\left(\frac{\alpha-\beta}{2}\right) = -\frac{3}{\sqrt{130}} \] Now, \[ \cos\left(\frac{\beta-\alpha}{2}\right) = \cos\left(-\frac{\alpha-\beta}{2}\right) \] Since cosine is an even function, \[ \cos(-\theta)=\cos\theta \] Hence, \[ \cos\left(\frac{\beta-\alpha}{2}\right) = -\frac{3}{\sqrt{130}} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{-\frac{3}{\sqrt{130}}} \]