Question

If $ A + B + C + D = 2\pi $, then $$ \sin A + \sin B + \sin C + \sin D = ? $$

• A. \( 4\sin\left( \frac{A + B}{4} \right)\sin\left( \frac{A + C}{4} \right)\sin\left( \frac{A + D}{4} \right) \)
• B. \( 4\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A + C}{4} \right)\cos\left( \frac{A + D}{4} \right) \)
• C. \( 4\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A + C}{2} \right)\sin\left( \frac{A + D}{2} \right) \)
• D. \( 4\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A + C}{4} \right)\sin\left( \frac{A + D}{4} \right) \)

Hint:

Remember: if sum of four angles is \( 2\pi \), then their sine sum has a known product form identity.

Answer 1

The Correct Option is C

Use sum-to-product identities. Also, known identity for sum of four sines when angles add up to \( 2\pi \): \[ \sin A + \sin B + \sin C + \sin D = 4\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A + C}{2} \right)\sin\left( \frac{A + D}{2} \right) \]