Question

In \( \triangle ABC \), prove the identity: $$ a^2 \sin 2B + b^2 \sin 2A = $$ 

• A. \( 2ab \cos A \)
• B. \( 2ab \sin A \)
• C. \( 2ab \sin C \)
• D. \( 2ab \cos C \)

Hint:

For triangle identities, apply the sine rule and double-angle formulas for simplification.

Answer 1

The Correct Option is C

Step 1: Using the sine rule We use the sine rule in a triangle: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R. \] Step 2: Expressing the given terms We rewrite: \[ a^2 \sin 2B + b^2 \sin 2A. \] Using \( \sin 2\theta = 2 \sin \theta \cos \theta \), we get: \[ a^2 (2 \sin B \cos B) + b^2 (2 \sin A \cos A). \] Step 3: Simplifying Using trigonometric identities and law of sines, we derive: \[ 2ab \sin C. \]