If `sin 2A=(1)/(2) and sin 2B=-(1)/(2)`, then which of the following is false?

Options

A. `sin(A+B)" may be "0`
B. `cos (A-B)" may be "0`
C. `sin(A+B) or cos (A-B)` is zero
D. `sin(A+B)=0`

Answer

To solve the problem, we are given two equations involving sine functions:

1. \( \sin 2A = \frac{1}{2} \)
2. \( \sin 2B = -\frac{1}{2} \)

We need to determine which of the given statements is false based on these equations.

### Step 1: Analyze \( \sin 2A = \frac{1}{2} \)

The sine function equals \( \frac{1}{2} \) at specific angles. The general solutions for \( 2A \) can be found using the sine inverse function:

\[ 2A = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]

This gives us two sets of solutions:

1. \( 2A = \frac{\pi}{6} + 2k\pi \)
2. \( 2A = \frac{5\pi}{6} + 2k\pi \)

From these, we can find:

\[ A = \frac{\pi}{12} + k\pi \quad \text{and} \quad A = \frac{5\pi}{12} + k\pi \]

### Step 2: Analyze \( \sin 2B = -\frac{1}{2} \)

The sine function equals \( -\frac{1}{2} \) at specific angles as well. The general solutions for \( 2B \) are:

\[ 2B = n\pi + (-1)^n \frac{7\pi}{6}, \quad n \in \mathbb{Z} \]

This gives us two sets of solutions:

1. \( 2B = \frac{7\pi}{6} + 2k\pi \)
2. \( 2B = \frac{11\pi}{6} + 2k\pi \)

From these, we can find:

\[ B = \frac{7\pi}{12} + k\pi \quad \text{and} \quad B = \frac{11\pi}{12} + k\pi \]

### Step 3: Evaluate \( \sin 2A + \sin 2B \)

Now, we can add \( \sin 2A \) and \( \sin 2B \):

\[ \sin 2A + \sin 2B = \frac{1}{2} - \frac{1}{2} = 0 \]

### Step 4: Use the Sine Addition Formula

Using the sine addition formula, we have:

\[ \sin(2A + 2B) = \sin 2A \cos 2B + \cos 2A \sin 2B \]

Since \( \sin 2A + \sin 2B = 0 \), we can conclude:

\[ \sin(2A + 2B) = 0 \]

This implies:

\[ 2A + 2B = n\pi \quad \text{for some integer } n \]

### Step 5: Determine Possible Conditions

From \( \sin(2A + 2B) = 0 \), we can derive:

1. \( \sin(A + B) = 0 \)
2. \( \cos(A - B) = 0 \)

Thus, we have two conditions that can be true:

- \( A + B = m\pi \) for some integer \( m \)
- \( A - B = \frac{\pi}{2} + k\pi \) for some integer \( k \)

### Conclusion

Now, we need to determine which of the statements provided in the options is false. Based on our analysis, we can conclude that:

- The conditions derived from \( \sin 2A + \sin 2B = 0 \) are valid.
- However, we cannot guarantee that \( \sin(A + B) = 0 \) or \( \cos(A - B) = 0 \) will always hold true without specific values for \( A \) and \( B \).

Thus, the false statement is likely the one that asserts one of these conditions must hold true without exception.