Question

If \( P + Q + R = \frac{\pi}{4} \), then \[ \cos \left( \frac{\pi}{8} - P \right) + \cos \left( \frac{\pi}{8} - Q \right) + \cos \left( \frac{\pi}{8} - R \right) = P + Q + R = \frac{\pi}{4}. \]

• A. \( 4 \cos \frac{P}{2} \cos \frac{Q}{2} - \cos \frac{\pi}{8} \)
• B. \( 4 \cos \frac{P}{2} \sin \frac{R}{2} + \cos \frac{\pi}{8} \)
• C. \( 4 \sin \frac{P}{2} \sin \frac{R}{2} - \cos \frac{\pi}{8} \)
• D. \( 4 \sin \frac{P}{2} \cos \frac{R}{2} + \cos \frac{\pi}{8} \)

Hint:

Using the sum-to-product identities for trigonometric functions can simplify the expression of multiple trigonometric terms into one.

Answer 1

The Correct Option is A

We are given the equation: \[ \cos \left( \frac{\pi}{8} - P \right) + \cos \left( \frac{\pi}{8} - Q \right) + \cos \left( \frac{\pi}{8} - R \right) = P + Q + R = \frac{\pi}{4}. \] Using the sum-to-product identity for cosines: \[ \cos A - \cos B = -2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right), \] we can express the sum of cosines as a sum of sines, simplifying to: \[ 4 \cos \frac{P}{2} \cos \frac{Q}{2} - \cos \frac{\pi}{8}. \] Thus, the correct answer is: \[ \boxed{4 \cos \frac{P}{2} \cos \frac{Q}{2} - \cos \frac{\pi}{8}}. \]