1. **State the problem:** Find the equation of the tangent line to the curve $y = \sin x \cdot \tan\left(\frac{x}{2}\right)$ at $x = \frac{\pi}{3}$. 2. **Formula for tangent line:** The equation of the tangent line at $x = a$ is given by: $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of $f(x)$ evaluated at $x = a$. 3. **Find $f(a)$:** Calculate $y$ at $x = \frac{\pi}{3}$: $$f\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) \cdot \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2}$$ 4. **Find $f'(x)$:** Use the product rule: $$f'(x) = \frac{d}{dx}[\sin x] \cdot \tan\left(\frac{x}{2}\right) + \sin x \cdot \frac{d}{dx}\left[\tan\left(\frac{x}{2}\right)\right]$$ Calculate derivatives: $$\frac{d}{dx}[\sin x] = \cos x$$ $$\frac{d}{dx}\left[\tan\left(\frac{x}{2}\right)\right] = \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2}$$ So, $$f'(x) = \cos x \cdot \tan\left(\frac{x}{2}\right) + \sin x \cdot \frac{1}{2} \sec^2\left(\frac{x}{2}\right)$$ 5. **Evaluate $f'(x)$ at $x = \frac{\pi}{3}$:** Calculate each term: $$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$ $$\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$$ $$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$\sec^2\left(\frac{\pi}{6}\right) = \frac{1}{\cos^2\left(\frac{\pi}{6}\right)} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$ Substitute: $$f'\left(\frac{\pi}{3}\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{4}{3} = \frac{1}{2\sqrt{3}} + \frac{\sqrt{3}}{2} \cdot \frac{2}{3} = \frac{1}{2\sqrt{3}} + \frac{\sqrt{3}}{3}$$ Find common denominator $6\sqrt{3}$: $$\frac{1}{2\sqrt{3}} = \frac{3}{6\sqrt{3}}, \quad \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{6\sqrt{3}} = \frac{2}{6}$$ Rewrite: $$f'\left(\frac{\pi}{3}\right) = \frac{3}{6\sqrt{3}} + \frac{2}{6} = \frac{3}{6\sqrt{3}} + \frac{2}{6}$$ Convert $\frac{3}{6\sqrt{3}}$ to decimal approx $0.2887$ and $\frac{2}{6} = 0.3333$, sum approx $0.622$. 6. **Write the tangent line equation:** $$y = f\left(\frac{\pi}{3}\right) + f'\left(\frac{\pi}{3}\right) \left(x - \frac{\pi}{3}\right) = \frac{1}{2} + 0.622 \left(x - \frac{\pi}{3}\right)$$ **Final answer:** $$y = \frac{1}{2} + \left(\frac{1}{2\sqrt{3}} + \frac{\sqrt{3}}{3}\right) \left(x - \frac{\pi}{3}\right)$$