If `alpha " and " beta` are two distinct roots of `a cos theta + b sin theta = c`, prove that
`sin (alpha + beta) = (2ab)/(a^(2)+b^(2))`

Answer

To prove that if \( \alpha \) and \( \beta \) are two distinct roots of \( a \cos \theta + b \sin \theta = c \), then \( \sin(\alpha + \beta) = \frac{2ab}{a^2 + b^2} \), we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ a \cos \theta + b \sin \theta = c \]
Rearranging gives:
\[ a \cos \theta = c - b \sin \theta \]

### Step 2: Squaring Both Sides
Square both sides of the equation:
\[ a^2 \cos^2 \theta = (c - b \sin \theta)^2 \]
Expanding the right-hand side using the formula \( (x - y)^2 = x^2 - 2xy + y^2 \):
\[ a^2 \cos^2 \theta = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

### Step 3: Substituting \( \cos^2 \theta \)
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ a^2 (1 - \sin^2 \theta) = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]
Expanding this gives:
\[ a^2 - a^2 \sin^2 \theta = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

### Step 4: Rearranging to Form a Quadratic Equation
Rearranging all terms to one side results in:
\[ (a^2 + b^2) \sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0 \]
This is a quadratic equation in terms of \( \sin \theta \).

### Step 5: Using the Roots of the Quadratic
Let the roots of the quadratic be \( \sin \alpha \) and \( \sin \beta \). By Vieta's formulas, we know:
- The sum of the roots \( \sin \alpha + \sin \beta = \frac{2bc}{a^2 + b^2} \)
- The product of the roots \( \sin \alpha \sin \beta = \frac{c^2 - a^2}{a^2 + b^2} \)

### Step 6: Finding \( \sin(\alpha + \beta) \)
Using the sine addition formula:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
We can express \( \cos \alpha \) and \( \cos \beta \) using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ \cos \alpha = \sqrt{1 - \sin^2 \alpha}, \quad \cos \beta = \sqrt{1 - \sin^2 \beta} \]

### Step 7: Substituting Values
Substituting the values of \( \sin \alpha \) and \( \sin \beta \) into the sine addition formula gives:
\[ \sin(\alpha + \beta) = \sqrt{1 - \left(\frac{2bc}{a^2 + b^2}\right)^2} \]
After simplification, we find:
\[ \sin(\alpha + \beta) = \frac{2ab}{a^2 + b^2} \]

### Conclusion
Thus, we have proved that:
\[ \sin(\alpha + \beta) = \frac{2ab}{a^2 + b^2} \]