1. Stated problem: Find the equation of the tangent line to the curve $y = \log_e \sqrt{1} = \sin 2x$ at the point where $x = \frac{\pi}{2}$. 2. Simplify the function: $\log_e \sqrt{1}$ is $\log_e 1$ since $\sqrt{1} = 1$. 3. Since $\log_e 1 = 0$, the function reduces to $y = 0 = \sin 2x$, meaning the curve is $y = \sin 2x$. 4. Find the value of the function at $x = \frac{\pi}{2}$: $$y\left(\frac{\pi}{2}\right) = \sin\left(2 \times \frac{\pi}{2}\right) = \sin \pi = 0$$ 5. Compute the derivative $y' = \frac{dy}{dx}$ to find the slope of the tangent line: $$y = \sin 2x$$ Using the chain rule, $$y' = 2 \cos 2x$$ 6. Evaluate the slope at $x = \frac{\pi}{2}$: $$y'\left(\frac{\pi}{2}\right) = 2 \cos \pi = 2 \times (-1) = -2$$ 7. Using point-slope form of a line $y - y_1 = m(x - x_1)$ with point $(x_1, y_1) = \left(\frac{\pi}{2}, 0\right)$ and slope $m = -2$, the equation is: $$y - 0 = -2 \left(x - \frac{\pi}{2}\right)$$ Simplified: $$y = -2x + \pi$$ 8. Final answer: The equation of the tangent line is $$y = -2x + \pi$$