The equation `(a+b)^(2)=4ab sin^(2)theta` is true if and only if

Options

A. `2a=b`
B. `a=b`
C. `a=2b`
D. `agtb`

Answer

To solve the equation \((a+b)^{2} = 4ab \sin^{2} \theta\) and determine under what conditions it holds true, we can follow these steps:

### Step 1: Expand the Left Side
Start by expanding the left side of the equation:
\[ (a+b)^{2} = a^{2} + 2ab + b^{2} \]
So, we rewrite the equation as:
\[ a^{2} + 2ab + b^{2} = 4ab \sin^{2} \theta \]

### Step 2: Rearrange the Equation
Rearranging the equation gives:
\[ a^{2} + b^{2} + 2ab - 4ab \sin^{2} \theta = 0 \]
This simplifies to:
\[ a^{2} + b^{2} + 2ab(1 - 2\sin^{2} \theta) = 0 \]

### Step 3: Use the Identity for Sine
Recall the identity \(1 - 2\sin^{2} \theta = \cos(2\theta)\). Thus, we can rewrite the equation as:
\[ a^{2} + b^{2} + 2ab \cos(2\theta) = 0 \]

### Step 4: Analyze the Equation
The expression \(a^{2} + b^{2} + 2ab \cos(2\theta) = 0\) can be interpreted. Since \(a^{2} + b^{2} \geq 0\) for all real \(a\) and \(b\), we need:
\[ 2ab \cos(2\theta) \leq 0 \]

### Step 5: Determine Conditions
- If \(ab > 0\) (both \(a\) and \(b\) are positive), then \(\cos(2\theta)\) must be less than or equal to 0, which occurs when \(2\theta\) is in the second or third quadrant.
- If \(ab < 0\) (one of \(a\) or \(b\) is negative), then \(\cos(2\theta)\) must be greater than or equal to 0, which occurs when \(2\theta\) is in the first or fourth quadrant.
- If \(ab = 0\) (either \(a\) or \(b\) is zero), then the equation holds true for any \(\theta\).

### Step 6: Special Case
The equation is particularly satisfied when \(a = b\). In this case, substituting \(b\) with \(a\) gives:
\[ (2a)^{2} = 4a^{2} \sin^{2} \theta \]
This simplifies to:
\[ 4a^{2} = 4a^{2} \sin^{2} \theta \]
This holds true if \(\sin^{2} \theta = 1\), which occurs when \(\theta = 90^\circ\).

### Conclusion
Thus, the equation \((a+b)^{2} = 4ab \sin^{2} \theta\) is true if and only if:
\[ a = b \]