Applications of the derivatives

Armed with the ability to differentiate trigonometric functions, we can now find the equations of tangents to trigonometric functions and find local maxima and minima.


  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 \(0\leq x\leq 2\pi\).


  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 \(-3\).

Exercise 8

Show that the function \(y = \dfrac{\sin x}{3+4\cos x}\) is increasing in each interval in which the denominator is not zero.

Exercise 9

Suppose an isosceles triangle has two equal sides of length \(a\) and equal base angles \(\theta\). Show that the perimeter of the triangle is \(2a(1+\cos\theta)\). Deduce that, of all isosceles triangles with fixed perimeter, the triangle of largest area is equilateral.

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