Example

1. Find the equation of the tangent to the curve \(y = 2\sin x + \cos 2x\) at the point \(x=\pi\).
2. Find the minimum value of \(y = 2\sin x + \cos 2x\) in the interval \(0\leq x\leq 2\pi\).

Solution

1. The gradient of the tangent is given by \(\dfrac{dy}{dx} = 2\cos x - 2\sin 2x = -2\) at \(x=\pi\). The \(y\)-value at this point is 1. Hence, the equation of the tangent is \[ y - 1 = -2(x - \pi) \quad \text{or, equivalently,} \quad y + 2x = 1 + 2\pi. \]
2. Since the function is continuous, the minimum will occur either at an end point of the interval \(0\leq x\leq 2\pi\) or at a stationary point. The \(y\)-value at each endpoint is 1. To find the stationary points, we solve \(\dfrac{dy}{dx}=0\). This gives \(\cos x= \sin 2x\). To proceed, we use a double-angle formula: \begin{align*} \cos x = \sin 2x \quad &\implies \quad \cos x = 2\,\sin x\,\cos x \\ &\implies \quad (1-2\sin x)\,\cos x = 0. \end{align*} Hence \(\cos x = 0\) or \(\sin x = \dfrac{1}{2}\). The solutions in the range \(0\leq x\leq 2\pi\) are \[ x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{\pi}{6}, \dfrac{5\pi}{6}. \] The smallest \(y\)-value, which is \(-3\), occurs at \(x = \dfrac{3\pi}{2}\). Hence the minimum value is \(-3\).