If `cos^(- 1)x-cos^(- 1)(y/2)=alpha` then `4x^2-4xycosalpha+y^2=`

Options

A. 4
B. `2 sin^(2) alpha`
C. `-4 sin^(2) alpha`
D. `4 sin^(2) alpha`

Explanation

To solve the problem, we start with the equation given:

\[ \cos^{-1} x - \cos^{-1} \left( \frac{y}{2} \right) = \alpha \]

### Step 1: Rewrite the equation using cosine
Using the property of inverse cosine, we can rewrite the equation as:

\[ \cos^{-1} x = \alpha + \cos^{-1} \left( \frac{y}{2} \right) \]

### Step 2: Apply cosine to both sides
Now, we apply cosine to both sides:

\[ \cos(\cos^{-1} x) = \cos\left(\alpha + \cos^{-1} \left( \frac{y}{2} \right)\right) \]

This simplifies to:

\[ x = \cos\left(\alpha + \cos^{-1} \left( \frac{y}{2} \right)\right) \]

### Step 3: Use the cosine addition formula
Using the cosine addition formula, we have:

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

Letting \( A = \alpha \) and \( B = \cos^{-1} \left( \frac{y}{2} \right) \), we get:

\[ x = \cos \alpha \cdot \cos\left(\cos^{-1} \left( \frac{y}{2} \right)\right) - \sin \alpha \cdot \sin\left(\cos^{-1} \left( \frac{y}{2} \right)\right) \]

### Step 4: Simplify using properties of cosine
Since \( \cos\left(\cos^{-1} \left( \frac{y}{2} \right)\right) = \frac{y}{2} \) and \( \sin\left(\cos^{-1} \left( \frac{y}{2} \right)\right) = \sqrt{1 - \left(\frac{y}{2}\right)^2} = \sqrt{\frac{4 - y^2}{4}} = \frac{\sqrt{4 - y^2}}{2} \), we can substitute these into our equation:

\[ x = \cos \alpha \cdot \frac{y}{2} - \sin \alpha \cdot \frac{\sqrt{4 - y^2}}{2} \]

### Step 5: Multiply through by 2
Multiplying both sides by 2 gives:

\[ 2x = y \cos \alpha - \sin \alpha \sqrt{4 - y^2} \]

### Step 6: Rearranging the equation
Rearranging gives:

\[ y \cos \alpha - 2x = \sin \alpha \sqrt{4 - y^2} \]

### Step 7: Square both sides
Squaring both sides results in:

\[ (y \cos \alpha - 2x)^2 = \sin^2 \alpha (4 - y^2) \]

### Step 8: Expand both sides
Expanding both sides gives:

\[ y^2 \cos^2 \alpha - 4xy \cos \alpha + 4x^2 = 4\sin^2 \alpha - y^2 \sin^2 \alpha \]

### Step 9: Combine like terms
Rearranging terms leads to:

\[ 4x^2 - 4xy \cos \alpha + y^2 (\cos^2 \alpha + \sin^2 \alpha) = 4\sin^2 \alpha \]

Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):

\[ 4x^2 - 4xy \cos \alpha + y^2 = 4\sin^2 \alpha \]

### Final Result
Thus, we find that:

\[ 4x^2 - 4xy \cos \alpha + y^2 = 4\sin^2 \alpha \]