Form a quadratic equations defined over rational coefficients whose one root is sin ` 18^(@) `

Answer

To form a quadratic equation with rational coefficients where one root is \(\sin 18^\circ\), we will follow these steps:

### Step 1: Determine the value of \(\sin 18^\circ\)
We know from trigonometric identities that:
\[ \sin 18^\circ = \frac{\sqrt{5} - 1}{4} \]

### Step 2: Identify the conjugate root
Since \(\sin 18^\circ\) is an irrational number, its conjugate must also be a root of the quadratic equation. The conjugate of \(\frac{\sqrt{5} - 1}{4}\) is:
\[ \sin 18^\circ' = \frac{-\sqrt{5} - 1}{4} \]

### Step 3: Calculate the sum of the roots
Let the roots be \(r_1 = \sin 18^\circ = \frac{\sqrt{5} - 1}{4}\) and \(r_2 = \sin 18^\circ' = \frac{-\sqrt{5} - 1}{4}\).

The sum of the roots \(S\) is given by:
\[ S = r_1 + r_2 = \frac{\sqrt{5} - 1}{4} + \frac{-\sqrt{5} - 1}{4} \]
Combining the fractions:
\[ S = \frac{(\sqrt{5} - 1) + (-\sqrt{5} - 1)}{4} = \frac{-2}{4} = -\frac{1}{2} \]

### Step 4: Calculate the product of the roots
The product of the roots \(P\) is given by:
\[ P = r_1 \cdot r_2 = \left(\frac{\sqrt{5} - 1}{4}\right) \cdot \left(\frac{-\sqrt{5} - 1}{4}\right) \]
Using the difference of squares:
\[ P = \frac{(-1)(\sqrt{5} - 1)(\sqrt{5} + 1)}{16} = \frac{-((\sqrt{5})^2 - 1^2)}{16} = \frac{-(5 - 1)}{16} = \frac{-4}{16} = -\frac{1}{4} \]

### Step 5: Form the quadratic equation
Using the standard form of a quadratic equation:
\[ x^2 - Sx + P = 0 \]
Substituting the values of \(S\) and \(P\):
\[ x^2 - \left(-\frac{1}{2}\right)x - \frac{1}{4} = 0 \]
This simplifies to:
\[ x^2 + \frac{1}{2}x - \frac{1}{4} = 0 \]

### Step 6: Clear the fractions
To eliminate the fractions, multiply the entire equation by 4:
\[ 4x^2 + 2x - 1 = 0 \]

### Final Answer
The quadratic equation with rational coefficients whose one root is \(\sin 18^\circ\) is:
\[ \boxed{4x^2 + 2x - 1 = 0} \]