Question

$ \sin 2A = 2 \sin A $ is true when $ A $ is equal to:

• A. $ 0^\circ $
• B. $ 30^\circ $
• C. $ 45^\circ $
• D. $ 60^\circ $

Hint:

When solving trigonometric equations involving double-angle identities, always simplify using basic trigonometric formulas and consider all possible solutions within the given range. Factorization can help identify key angles that satisfy the equation.

Answer 1

The Correct Option is A

Step 1: Use the Double-Angle Identity for Sine. 
The double-angle identity for sine states:
$$ \sin 2A = 2 \sin A \cos A. $$ Thus, the given equation becomes: $$ 2 \sin A \cos A = 2 \sin A. $$ Step 2: Simplify the Equation. 
Rearrange the equation:
$$ 2 \sin A \cos A - 2 \sin A = 0. $$ Factor out $ 2 \sin A $: $$ 2 \sin A (\cos A - 1) = 0. $$ Step 3: Solve for $ A $. 
Set each factor equal to zero:
1. $ 2 \sin A = 0 $: $$ \sin A = 0 \implies A = 0^\circ, 180^\circ, 360^\circ, \dots $$ The simplest solution within the standard range is $ A = 0^\circ $. 2. $ \cos A - 1 = 0 $: $$ \cos A = 1 \implies A = 0^\circ, 360^\circ, \dots $$ Again, the simplest solution is $ A = 0^\circ $. Step 4: Analyze the Options. 
Option (1): $ 0^\circ $ — Correct, as it satisfies the equation. 
Option (2): $ 30^\circ $ — Incorrect, as substituting $ A = 30^\circ $ does not satisfy the equation. 
Option (3): $ 45^\circ $ — Incorrect, as substituting $ A = 45^\circ $ does not satisfy the equation. 
Option (4): $ 60^\circ $ — Incorrect, as substituting $ A = 60^\circ $ does not satisfy the equation.
Step 5: Final Answer. \((1) \mathbf{0^\circ}\)