Extremum problems with trigonometric functions

In many problems in geometry it is convenient to choose an angle as the variable: all the lengths are written with its sine and cosine, and the quantity you are looking for is a trigonometric function.

• The angle $\displaystyle x$ is in radians, so that it can be differentiated.
• The interval comes from the figure, for example $\displaystyle 0<x<\frac{\pi}{2}$ when the angle is acute.
• $\displaystyle {f'(x)}={0}$ is a trigonometric equation, and you solve it in the interval only.
• Sometimes no derivative is needed: $\displaystyle \sin x\leq1$ and $\displaystyle \cos x\leq1$, so an expression like $\displaystyle a\sin x$ is greatest where $\displaystyle {\sin x}={1}$.

The legs of an isosceles triangle are $\displaystyle a$ long, and the angle between them is $\displaystyle x$.
What is the greatest area the triangle can have?

The solution below writes the area from two sides and the angle between them, and finds the maximum without differentiating:

• $\displaystyle {S(x)}={\frac{1}{2}a\cdot a\cdot\sin x}={\frac{a^2}{2}\sin x}$, on the interval $\displaystyle 0<x<\pi$.
• $\displaystyle \sin x\leq1$ for every $\displaystyle x$, and in the interval $\displaystyle {\sin x}={1}$ only where $\displaystyle {x}={\frac{\pi}{2}}$.
• So the greatest area is $\displaystyle \frac{a^2}{2}$, when the triangle is right-angled.

The derivative gives the same answer: $\displaystyle {S'(x)}={\frac{a^2}{2}\cos x}={0}$ where $\displaystyle {x}={\frac{\pi}{2}}$.

A trapezoid is inscribed in a semicircle of radius 1: its longer base is the diameter, and the other two vertices are on the arc.
$\displaystyle x$ – the angle between the diameter and the radius to the upper right vertex.
For which angle $\displaystyle x$ is the area of the trapezoid greatest?

The solution below expresses the shorter base and the height by $\displaystyle x$:

• The upper right vertex is $\displaystyle (\cos x,\sin x)$. So the height is $\displaystyle \sin x$, and the shorter base $\displaystyle 2\cos x$; the interval is $\displaystyle 0<x<\frac{\pi}{2}$.
• $\displaystyle {S(x)}={\frac{2+2\cos x}{2}\cdot\sin x}={(1+\cos x)\sin x}$.
• $\displaystyle {S'(x)}={-\sin x\cdot\sin x+(1+\cos x)\cos x}={\cos x+\cos^2x-\sin^2x}={\cos x+\cos2x}$.
• By $\displaystyle {\cos2x}={2\cos^2x-1}$: $\displaystyle {S'(x)}={2\cos^2x+\cos x-1}={(2\cos x-1)(\cos x+1)}$.
• $\displaystyle \cos x+1>0$ in the interval, so $\displaystyle {S'(x)}={0}$ only where $\displaystyle {\cos x}={\frac{1}{2}}$, that is $\displaystyle {x}={\frac{\pi}{3}}$.
• Before $\displaystyle \frac{\pi}{3}$, $\displaystyle \cos x>\frac{1}{2}$ and $\displaystyle S'>0$; after it $\displaystyle S'<0$: a maximum.
• $\displaystyle {S\left(\frac{\pi}{3}\right)}={\left(1+\frac{1}{2}\right)\cdot\frac{\sqrt{3}}{2}}={\frac{3\sqrt{3}}{4}\approx1.30}$.

In the greatest trapezoid the shorter base equals the radius, and it is half a regular hexagon.

A rectangle is inscribed in a semicircle of radius 2: one side of it is on the diameter, and the other two vertices are on the arc.
$\displaystyle x$ – the angle between the diameter and the radius to the upper right vertex, so the vertex is $\displaystyle (2\cos x,2\sin x)$.
The area of the rectangle is $\displaystyle {S(x)}={2\cdot2\cos x\cdot2\sin x}={8\sin x\cos x}$, on the interval $\displaystyle 0<x<\frac{\pi}{2}$.

Write $\displaystyle S(x)$ with $\displaystyle \sin2x$.

Find the greatest area without differentiating, and the angle where it is reached.

Differentiate $\displaystyle {S(x)}={4\sin2x}$.

Solve the equation $\displaystyle {S'(x)}={0}$ in the interval, and check that you got the same angle.

A Ferris wheel turns once in 30 minutes.
The height of a car above the ground, in metres, after $\displaystyle t$ minutes is $\displaystyle {h(t)}={20-18\cos\frac{\pi t}{15}}$, with $\displaystyle 0\leq t\leq30$.

What is the greatest height of the car, and what is the least?
When is the car at the highest point? Answer without differentiating.

Find $\displaystyle h'(t)$ – the speed at which the car rises.

Find $\displaystyle h''(t)$.

The car rises fastest when the speed $\displaystyle h'(t)$ reaches a maximum, that is where $\displaystyle {h''(t)}={0}$.
Solve the equation in the interval.

Work out the greatest speed of rising, in metres per minute.

At what height is the car when it rises fastest?
What happens at the other solution of the equation?

The cross-section of a gutter is made of a bottom 10 cm wide and two sides 10 cm wide.
Both sides open outwards at an angle $\displaystyle x$ from the vertical, with $\displaystyle 0\leq x<\frac{\pi}{2}$.
The greater the cross-section, the more water the gutter carries.

Show that the height of the cross-section is $\displaystyle 10\cos x$, that its width at the top is $\displaystyle 10+20\sin x$, and that its area is $\displaystyle {S(x)}={100\cos x(1+\sin x)}$.

Find $\displaystyle S'(x)$.

Solve the equation $\displaystyle {S'(x)}={0}$ in the interval.

Work out the greatest area of the cross-section.

The diagonal of a rectangle is 10 long.
$\displaystyle x$ – the angle between the diagonal and one side, so the sides are $\displaystyle 10\cos x$ and $\displaystyle 10\sin x$.
The perimeter is $\displaystyle {P(x)}={20\sin x+20\cos x}$, on the interval $\displaystyle 0<x<\frac{\pi}{2}$.

Find $\displaystyle P'(x)$.

Solve the equation $\displaystyle {P'(x)}={0}$ in the interval.

Work out the greatest perimeter.

What rectangle did you get?
Does such a rectangle have a least perimeter? Explain by what happens to $\displaystyle P(x)$ as $\displaystyle x$ approaches the ends of the interval.