Question

Suppose \( \theta_1 \) and \( \theta_2 \) are such that \( (\theta_1 - \theta_2) \) lies in the 3rd or 4th quadrant. If \[ \sin\theta_1 + \sin\theta_2 = \frac{21}{65} \quad \text{and} \quad \cos\theta_1 + \cos\theta_2 = \frac{27}{65} \] then \[ \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \] 

• A. \( \frac{3}{\sqrt{150}} \)
• B. \( \frac{3}{\sqrt{130}} \)
• C. \( \frac{-3}{\sqrt{130}} \)
• D. \( \frac{-3}{\sqrt{150}} \)

Hint:

Sum-to-product identities are useful in solving trigonometric equations involving sums of sines and cosines. Always check the quadrant to determine the correct sign.

Answer 1

The Correct Option is C

Step 1: Using Sum-to-Product Identities 
We use the sum-to-product formulas: \[ \sin\theta_1 + \sin\theta_2 = 2\sin\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right), \] \[ \cos\theta_1 + \cos\theta_2 = 2\cos\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right). \] 

Step 2: Expressing in Terms of Known Values 
Given: \[ \sin\theta_1 + \sin\theta_2 = \frac{21}{65}, \quad \cos\theta_1 + \cos\theta_2 = \frac{27}{65}. \] From the sum-to-product identities: \[ 2\sin\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{21}{65}, \] \[ 2\cos\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{27}{65}. \] 

Step 3: Squaring and Summing 
Squaring both equations and summing: \[ \left(\frac{21}{65}\right)^2 + \left(\frac{27}{65}\right)^2 = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right) \left(\sin^2\left(\frac{\theta_1 + \theta_2}{2}\right) + \cos^2\left(\frac{\theta_1 + \theta_2}{2}\right) \right). \] Since \( \sin^2x + \cos^2x = 1 \), \[ \left(\frac{21}{65}\right)^2 + \left(\frac{27}{65}\right)^2 = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right). \] 

Step 4: Solving for \( \cos(\frac{\theta_1 - \theta_2}{2}) \) 
\[ \frac{441}{4225} + \frac{729}{4225} = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right). \] \[ \frac{1170}{4225} = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right). \] \[ \cos^2\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{1170}{16900}. \] Taking square roots: \[ \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{\pm3}{\sqrt{130}}. \] Since \( (\theta_1 - \theta_2) \) lies in the 3rd or 4th quadrant, we take the negative value: \[ \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{-3}{\sqrt{130}}. \]