1. **State the problem:** We have the curve $y = \sin\left(x + \frac{3}{5}\pi\right) \cos x$ and need to find its derivative $\frac{dy}{dx}$ and show a trigonometric identity at stationary points. 2. **Find $\frac{dy}{dx}$:** Use the product rule: if $y = u v$, then $\frac{dy}{dx} = u' v + u v'$. Here, $u = \sin\left(x + \frac{3}{5}\pi\right)$ and $v = \cos x$. Calculate derivatives: $$u' = \cos\left(x + \frac{3}{5}\pi\right), \quad v' = -\sin x$$ Apply product rule: $$\frac{dy}{dx} = \cos\left(x + \frac{3}{5}\pi\right) \cos x + \sin\left(x + \frac{3}{5}\pi\right)(-\sin x)$$ Simplify: $$\frac{dy}{dx} = \cos\left(x + \frac{3}{5}\pi\right) \cos x - \sin\left(x + \frac{3}{5}\pi\right) \sin x$$ 3. **Show the formula for cos(A + B):** Recall the identity: $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ 4. **Apply the identity:** Let $A = x + \frac{3}{5}\pi$ and $B = x$, then $$\cos\left(x + \frac{3}{5}\pi + x\right) = \cos\left(x + \frac{3}{5}\pi\right) \cos x - \sin\left(x + \frac{3}{5}\pi\right) \sin x$$ This matches $\frac{dy}{dx}$, so $$\frac{dy}{dx} = \cos\left(2x + \frac{3}{5}\pi\right)$$ 5. **At stationary points,** $\frac{dy}{dx} = 0$, so $$\cos\left(2x + \frac{3}{5}\pi\right) = 0$$ **Final answers:** (i) $$\frac{dy}{dx} = \cos\left(x + \frac{3}{5}\pi\right) \cos x - \sin\left(x + \frac{3}{5}\pi\right) \sin x$$ (ii) $$\cos\left(2x + \frac{3}{5}\pi\right) = 0$$ at stationary points.