Choose the correct option and justify your choice.
`sin2A = 2sinA ` is true when A =

Options

A. `0^@`
B. `30^@`
C. `45^@`
D. `60^@`

Answer

To solve the equation \( \sin 2A = 2 \sin A \) and determine the values of \( A \) for which this is true, we can follow these steps:

### Step 1: Use the double angle identity for sine
The double angle identity states that:
\[ \sin 2A = 2 \sin A \cos A \]
Thus, we can rewrite the equation as:
\[ 2 \sin A \cos A = 2 \sin A \]

### Step 2: Simplify the equation
We can divide both sides of the equation by 2 (assuming \( \sin A \neq 0 \)):
\[ \sin A \cos A = \sin A \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ \sin A \cos A - \sin A = 0 \]
Factoring out \( \sin A \):
\[ \sin A (\cos A - 1) = 0 \]

### Step 4: Solve for \( A \)
This equation gives us two cases to consider:
1. \( \sin A = 0 \)
2. \( \cos A - 1 = 0 \)

**Case 1:** \( \sin A = 0 \)
This occurs when:
\[ A = n\pi \quad (n \in \mathbb{Z}) \]
In degrees, this corresponds to:
\[ A = 0^\circ, 180^\circ, 360^\circ, \ldots \]

**Case 2:** \( \cos A - 1 = 0 \)
This occurs when:
\[ \cos A = 1 \]
This happens at:
\[ A = 2n\pi \quad (n \in \mathbb{Z}) \]
In degrees, this corresponds to:
\[ A = 0^\circ, 360^\circ, 720^\circ, \ldots \]

### Conclusion
Thus, the values of \( A \) for which \( \sin 2A = 2 \sin A \) is true include \( A = 0^\circ \) and any integer multiples of \( 180^\circ \) (or \( n\pi \) in radians).

### Justification
The correct option is \( A = 0^\circ \) because substituting \( A = 0^\circ \) into the original equation gives:
\[ \sin 2(0) = 2 \sin(0) \implies 0 = 0 \]
This confirms that the equation holds true.