## Content

### 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.

#### 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 $$-3$$.

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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