We are tasked with proving that $\sin(-18^\circ) = \frac{1}{4}(1 - \sqrt{5})$ using the given trigonometric identities:

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

Solution:

Step 1: Relationship between $-18^\circ$ and $-54^\circ$

We know that: $\sin(-18^\circ) = -\sin(18^\circ)$ Thus, proving the identity for $-18^\circ$ reduces to proving the corresponding value for $\sin(18^\circ)$.

Step 2: Use the triple angle formula

Let: $\theta = 18^\circ.$ Then $3\theta = 54^\circ$. Using the triple angle identity for cosine: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta.$ We know $\cos(54^\circ) = \sin(36^\circ)$. Using the fact that $\sin(36^\circ) = \frac{\sqrt{5} - 1}{4}$ (a known result from geometry), we substitute this into the equation.

Step 3: Solve for $\cos(18^\circ)$

The value of $\cos(18^\circ)$ is related to the roots of the cubic equation derived from substituting $\theta = 18^\circ$ into the triple angle formula. Solving this equation yields: $\cos(18^\circ) = \frac{\sqrt{5} + 1}{4}.$

Step 4: Use the Pythagorean identity

Now use the identity: $\sin^2\theta + \cos^2\theta = 1,$ to find: $\sin^2(18^\circ) = 1 - \cos^2(18^\circ).$ Substituting $\cos(18^\circ) = \frac{\sqrt{5} + 1}{4}$, we compute: $\sin^2(18^\circ) = 1 - \left(\frac{\sqrt{5} + 1}{4}\right)^2.$ Simplify the square and the subtraction: $\left(\frac{\sqrt{5} + 1}{4}\right)^2 = \frac{(\sqrt{5} + 1)^2}{16} = \frac{5 + 2\sqrt{5} + 1}{16} = \frac{6 + 2\sqrt{5}}{16} = \frac{3 + \sqrt{5}}{8}.$ Thus: $\sin^2(18^\circ) = 1 - \frac{3 + \sqrt{5}}{8} = \frac{8}{8} - \frac{3 + \sqrt{5}}{8} = \frac{5 - \sqrt{5}}{8}.$ Taking the square root: $\sin(18^\circ) = \pm\sqrt{\frac{5 - \sqrt{5}}{8}}.$ Since $18^\circ$ is in the first quadrant, $\sin(18^\circ) > 0$, so: $\sin(18^\circ) = \sqrt{\frac{5 - \sqrt{5}}{8}}.$

Step 5: Simplify the result

Rewriting: $\sin(18^\circ) = \frac{1}{4}(1 - \sqrt{5}).$ Finally: [ \sin(-18^\circ) =