If `sin A + cos A =1,` then `sin 2A ` is equal to

Options

A. 1
B. 2
C. 0
D. `1/2`

Answer

To solve the problem where \( \sin A + \cos A = 1 \) and we need to find \( \sin 2A \), we can follow these steps:

### Step 1: Square both sides of the equation
Given:
\[ \sin A + \cos A = 1 \]
Square both sides:
\[ (\sin A + \cos A)^2 = 1^2 \]
This expands to:
\[ \sin^2 A + 2\sin A \cos A + \cos^2 A = 1 \]

### Step 2: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 A + \cos^2 A = 1 \]
Substituting this into our equation gives:
\[ 1 + 2\sin A \cos A = 1 \]

### Step 3: Simplify the equation
Now, we can simplify:
\[ 2\sin A \cos A = 1 - 1 \]
This simplifies to:
\[ 2\sin A \cos A = 0 \]

### Step 4: Solve for \( \sin A \) or \( \cos A \)
From the equation \( 2\sin A \cos A = 0 \), we have:
\[ \sin A \cos A = 0 \]
This implies that either:
\[ \sin A = 0 \quad \text{or} \quad \cos A = 0 \]

### Step 5: Find \( \sin 2A \)
Using the double angle identity for sine:
\[ \sin 2A = 2\sin A \cos A \]
Since we established that \( \sin A \cos A = 0 \), we can conclude:
\[ \sin 2A = 2 \cdot 0 = 0 \]

### Final Answer
Thus, the value of \( \sin 2A \) is:
\[ \sin 2A = 0 \]

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