\begin{aligned} &\text{Let } x = 36^\circ. \text{ Then, } 5x = 180^\circ. \\ \\ &\text{Rewriting: } 2x + 3x = 180^\circ, \text{ so } 2x = 180^\circ - 3x. \\ \\ &\text{Taking the sine of both sides:} \\ &\sin(2x) = \sin(180^\circ - 3x). \\ \\ &\text{Using trigonometric identities:} \\ &\sin(2x) = 2\sin(x)\cos(x), \\ &\sin(180^\circ - 3x) = \sin(3x) = 3\sin(x) - 4\sin^3(x). \\ \\ &\text{Substituting:} \quad 2\sin(x)\cos(x) = 3\sin(x) - 4\sin^3(x). \\ \\ &\text{Since } \sin(x) \neq 0 \text{ (as } x = 36^\circ \text{), divide by } \sin(x): \\ &2\cos(x) = 3 - 4\sin^2(x). \\ \\ &\text{Using the identity } \cos^2(x) + \sin^2(x) = 1, \text{ we get } \cos(x) = \sqrt{1 - \sin^2(x)}. \\ &\text{Also, replacing } \sin^2(x) \text{ with } 1 - \cos^2(x), \text{ we rewrite:} \\ &2\cos(x) = 3 - 4(1 - \cos^2(x)). \\ \\ &\text{Simplifying:} \\ &2\cos(x) = 3 - 4 + 4\cos^2(x). \\ &4\cos^2(x) - 2\cos(x) - 1 = 0. \\ \\ &\text{Solving the quadratic equation for } \cos(x): \\ &\cos(x) = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{1 \pm \sqrt{5}}{4}. \\ \\ &\text{Since } x = 36^\circ \text{ is in the first quadrant, we take the positive value:} \\ &\cos(36^\circ) = \frac{1 + \sqrt{5}}{4}. \\ \\ &\text{Using the identity } \sin^2(x) = 1 - \cos^2(x), \text{ we find:} \\ &\sin^2(36^\circ) = 1 - \left(\frac{1 + \sqrt{5}}{4}\right)^2. \\ \\ &= 1 - \frac{1 + 2\sqrt{5} + 5}{16} \\ &= 1 - \frac{6 + 2\sqrt{5}}{16} \\ &= \frac{16 - 6 - 2\sqrt{5}}{16} \\ &= \frac{10 - 2\sqrt{5}}{16} \\ &= \frac{5 - \sqrt{5}}{8}. \\ \\ &\text{Taking the square root:} \\ &\sin(36^\circ) = \sqrt{\frac{10 - 2\sqrt{5}}{4}}. \\ \\ &\text{Since } 36^\circ \text{ is in the first quadrant, } \sin(36^\circ) \text{ is positive.} \\ \end{aligned}