$$\alpha = \sin 36°$$ is a root of which of the following equation

We need to find which equation has $$\alpha = \sin 36°$$ as a root.

Use the key identity $$5 \times 36° = 180°$$

Since $$5 \times 36° = 180°$$, we have $$\sin(5 \times 36°) = \sin 180° = 0$$.

Let $$\theta = 36°$$. Then $$\sin 5\theta = 0$$.

Expand $$\sin 5\theta$$ using the multiple angle formula

The formula for $$\sin 5\theta$$ in terms of $$\sin\theta$$ is:

$$\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$$

To derive this, we use:

$$\sin 5\theta = \sin(3\theta + 2\theta) = \sin 3\theta \cos 2\theta + \cos 3\theta \sin 2\theta$$

Where:

$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$

$$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$

$$\sin 2\theta = 2\sin\theta\cos\theta$$

$$\cos 2\theta = 1 - 2\sin^2\theta$$

Substituting:

$$\sin 5\theta = (3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta) + (4\cos^3\theta - 3\cos\theta)(2\sin\theta\cos\theta)$$

First part: $$(3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta)$$

$$= 3\sin\theta - 6\sin^3\theta - 4\sin^3\theta + 8\sin^5\theta$$

$$= 3\sin\theta - 10\sin^3\theta + 8\sin^5\theta$$

Second part: $$(4\cos^3\theta - 3\cos\theta)(2\sin\theta\cos\theta)$$

$$= 2\sin\theta\cos\theta(4\cos^3\theta - 3\cos\theta)$$

$$= 2\sin\theta(4\cos^4\theta - 3\cos^2\theta)$$

Using $$\cos^2\theta = 1 - \sin^2\theta$$:

$$\cos^4\theta = (1-\sin^2\theta)^2 = 1 - 2\sin^2\theta + \sin^4\theta$$

$$= 2\sin\theta[4(1-2\sin^2\theta+\sin^4\theta) - 3(1-\sin^2\theta)]$$

$$= 2\sin\theta[4 - 8\sin^2\theta + 4\sin^4\theta - 3 + 3\sin^2\theta]$$

$$= 2\sin\theta[1 - 5\sin^2\theta + 4\sin^4\theta]$$

$$= 2\sin\theta - 10\sin^3\theta + 8\sin^5\theta$$

Adding both parts:

$$\sin 5\theta = (3\sin\theta - 10\sin^3\theta + 8\sin^5\theta) + (2\sin\theta - 10\sin^3\theta + 8\sin^5\theta)$$

$$= 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta$$

$$= \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5)$$

Set $$\sin 5\theta = 0$$ and simplify

$$\sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) = 0$$

Since $$\theta = 36°$$, $$\sin\theta \neq 0$$. So we divide by $$\sin\theta$$:

$$16\sin^4\theta - 20\sin^2\theta + 5 = 0$$

Write in terms of $$x = \sin 36°$$

$$16x^4 - 20x^2 + 5 = 0$$

Verify

We know $$\sin 36° = \frac{\sqrt{10-2\sqrt{5}}}{4}$$, so $$\sin^2 36° = \frac{10-2\sqrt{5}}{16}$$.

$$16x^4 = 16 \times \frac{(10-2\sqrt{5})^2}{256} = \frac{100 - 40\sqrt{5} + 20}{16} = \frac{120-40\sqrt{5}}{16}$$

$$20x^2 = \frac{200-40\sqrt{5}}{16}$$

$$16x^4 - 20x^2 + 5 = \frac{120-40\sqrt{5} - 200+40\sqrt{5}}{16} + 5 = \frac{-80}{16} + 5 = -5 + 5 = 0$$ ✓

Therefore, $$\sin 36°$$ is a root of $$16x^4 - 20x^2 + 5 = 0$$.

The correct answer is Option A: $$16x^4 - 20x^2 + 5 = 0$$.