Question:

Step 1: Determine $\sin \alpha \cos \alpha$ Given the equation $\sin \alpha + \cos \alpha = \frac{\sqrt{3}}{2}$. Square both sides of the equation: $$(\sin \alpha + \cos \alpha)^2 = \left(\frac{\sqrt{3}}{2}\right)^2$$ $$\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha = \frac{3}{4}$$ Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$: $$1 + 2\sin \alpha \cos \alpha = \frac{3}{4}$$ $$2\sin \alpha \cos \alpha = \frac{3}{4} - 1$$ $$2\sin \alpha \cos \alpha = -\frac{1}{4}$$ $$\sin \alpha \cos \alpha = -\frac{1}{8}$$ Step 2: Determine $(\sin \alpha - \cos \alpha)^2$ The expression $(\sin \alpha - \cos \alpha)^2$ can be expanded as: $$(\sin \alpha - \cos \alpha)^2 = \sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha$$ Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$: $$(\sin \alpha - \cos \alpha)^2 = 1 - 2\sin \alpha \cos \alpha$$ Substitute the value of $\sin \alpha \cos \alpha$ from Step 1: $$(\sin \alpha - \cos \alpha)^2 = 1 - 2\left(-\frac{1}{8}\right)$$ $$(\sin \alpha - \cos \alpha)^2 = 1 + \frac{1}{4}$$ $$(\sin \alpha - \cos \alpha)^2 = \frac{5}{4}$$ Taking the square root of both sides: $$\sin \alpha - \cos \alpha = \pm\sqrt{\frac{5}{4}} = \pm\frac{\sqrt{5}}{2}$$ Step 3: Determine the sign of $\sin \alpha - \cos \alpha$ We have $\sin \alpha + \cos \alpha = \frac{\sqrt{3}}{2}$ and $\sin \alpha \cos \alpha = -\frac{1}{8}$. Since $\sin \alpha \cos \alpha < 0$, $\sin \alpha$ and $\cos \alpha$ must have opposite signs. Since $\sin \alpha + \cos \alpha > 0$, the positive term must be greater in magnitude than the negative term. Consider the quadratic equation $x^2 - (\sin \alpha + \cos \alpha)x + \sin \alpha \cos \alpha = 0$, whose roots are $\sin \alpha$ and $\cos \alpha$: $$x^2 - \frac{\sqrt{3}}{2}x - \frac{1}{8} = 0$$ $$8x^2 - 4\sqrt{3}x - 1 = 0$$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$x = \frac{4\sqrt{3} \pm \sqrt{(-4\sqrt{3})^2 - 4(8)(-1)}}{2(8)}$$ $$x = \frac{4\sqrt{3} \pm \sqrt{48 + 32}}{16}$$ $$x = \frac{4\sqrt{3} \pm \sqrt{80}}{16}$$ $$x = \frac{4\sqrt{3} \pm 4\sqrt{5}}{16}$$ $$x = \frac{\sqrt{3} \pm \sqrt{5}}{4}$$ Thus, $\{\sin \alpha, \cos \alpha\} = \left\{\frac{\sqrt{3} + \sqrt{5}}{4}, \frac{\sqrt{3} - \sqrt{5}}{4}\right\}$. Since $\sqrt{5} > \sqrt{3}$, $\frac{\sqrt{3} + \sqrt{5}}{4}$ is positive and $\frac{\sqrt{3} - \sqrt{5}}{4}$ is negative. For $\sin \alpha + \cos \alpha > 0$, we must have $\sin \alpha = \frac{\sqrt{3} + \sqrt{5}}{4}$ and $\cos \alpha = \frac{\sqrt{3} - \sqrt{5}}{4}$. Therefore, $\sin \alpha - \cos \alpha = \frac{\sqrt{3} + \sqrt{5}}{4} - \frac{\sqrt{3} - \sqrt{5}}{4} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2}$. Step 4: Solve for $\sin \alpha$ and $\cos \alpha$ We have a system of two linear equations: 1) $\sin \alpha + \cos \alpha = \frac{\sqrt{3}}{2}$ 2) $\sin \alpha - \cos \alpha = \frac{\sqrt{5}}{2}$ Adding (1) and (2): $$2\sin \alpha = \frac{\sqrt{3}}{2} + \frac{\sqrt{5}}{2}$$ $$2\sin \alpha = \frac{\sqrt{3} + \sqrt{5}}{2}$$ $$\sin \alpha = \frac{\sqrt{3} + \sqrt{5}}{4}$$ Subtracting (2) from (1): $$2\cos \alpha = \frac{\sqrt{3}}{2} - \frac{\sqrt{5}}{2}$$ $$2\cos \alpha = \frac{\sqrt{3} - \sqrt{5}}{2}$$ $$\cos \alpha = \frac{\sqrt{3} - \sqrt{5}}{4}$$ Step 5: Find $\tan \alpha$ $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{(\sqrt{3} + \sqrt{5})/4}{(\sqrt{3} - \sqrt{5})/4}$$ $$\tan \alpha = \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} - \sqrt{5}}$$ Rationalize the denominator: $$\tan \alpha = \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} - \sqrt{5}} \times \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}$$ $$\tan \alpha = \frac{(\sqrt{3} + \sqrt{5})^2}{(\sqrt{3})^2 - (\sqrt{5})^2}$$ $$\tan \alpha = \frac{3 + 2\sqrt{15} + 5}{3 - 5}$$ $$\tan \alpha = \frac{8 + 2\sqrt{15}}{-2}$$ $$\tan \alpha = -(4 + \sqrt{15})$$ Step 6: Find $\tan(\alpha/2)$ Using the half-angle formula $\tan(\alpha/2) = \frac{1 - \cos \alpha}{\sin \alpha}$: $$\tan(\alpha/2) = \frac{1 - \left(\frac{\sqrt{3} - \sqrt{5}}{4}\right)}{\frac{\sqrt{3} + \sqrt{5}}{4}}$$ $$\tan(\alpha/2) = \frac{\frac{4 - (\sqrt{3} - \sqrt{5})}{4}}{\frac{\sqrt{3} + \sqrt{5}}{4}}$$ $$\tan(\alpha/2) = \frac{4 - \sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}$$ Rationalize the denominator: $$\tan(\alpha/2) = \frac{4 - \sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}}$$ $$\tan(\alpha/2) = \frac{(4 - \sqrt{3} + \sqrt{5})(\sqrt{5} - \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2}$$ $$\tan(\alpha/2) = \frac{4\sqrt{5} - 4\sqrt{3} - \sqrt{15} + 3 + 5 - \sqrt{15}}{5 - 3}$$ $$\tan(\alpha/2) = \frac{8 - 4\sqrt{3} + 4\sqrt{5} - 2\sqrt{15}}{2}$$ $$\tan(\alpha/2) = 4 - 2\sqrt{3} + 2\sqrt{5} - \sqrt{15}$$