Question

If \(A + B + C = \dfrac{\pi}{4}\), then \(\sin 4A + \sin 4B + \sin 4C =\)

• A. \(4 \cos 2A \cos 2B \cos 2C\)
• B. \(4 \sin 2A \sin 2B \sin 2C\)
• C. \(1 + 4 \sin 2A \sin 2B \sin 2C\)
• D. \(1 + 4 \cos 2A \cos 2B \cos 2C\)

Hint:

To solve trigonometric identities involving multiple angles, consider converting to product forms and apply sum-to-product identities.

Answer 1

The Correct Option is A

We are given \(A + B + C = \dfrac{\pi}{4}\). Multiply both sides by 4 to get: \[ 4A + 4B + 4C = \pi. \] Using the identity for sum of sines: \[ \sin 4A + \sin 4B = 2 \sin(2A + 2B) \cos(2A - 2B), \] and noting that \(2A + 2B = 2(\dfrac{\pi}{4} - C) = \dfrac{\pi}{2} - 2C\), so: \[ \sin 4A + \sin 4B = 2 \cos 2C \cos(2A - 2B). \] Now, \[ \sin 4A + \sin 4B + \sin 4C = 2 \cos 2C \cos(2A - 2B) + \sin 4C. \] But also, \[ \sin 4C = \sin(\pi - 4A - 4B) = \sin(4A + 4B), \] which can be expanded as: \[ \sin(4A + 4B) = 2 \sin(2A + 2B) \cos(2A + 2B) = 2 \cos 2C \cos(2A + 2B). \] Putting it all together gives: \[ \sin 4A + \sin 4B + \sin 4C = 4 \cos 2A \cos 2B \cos 2C. \]