Science:Math Exam Resources/Courses/MATH100/December 2010/Question 08/Solution 1

Notice that the function $\displaystyle T(\theta )$ is $\displaystyle 2\pi$ periodic since the equator is shaped like a circle. Mathematically this in particular means

$\displaystyle T(2\pi )=T(0).$

We exploit this fact by looking at

$\displaystyle f(0)=T(\pi )-T(0)$

and

${\begin{aligned}\displaystyle f(\pi )&=T(\pi +\pi )-T(\pi )\\&=T(2\pi )-T(\pi )\\&=T(0)-T(\pi )\\&=-f(0).\end{aligned}}$

At this point we want to distinguish two cases:

Case 1. f(0) = 0. Then

$\displaystyle 0=f(0)=T(\pi )-T(0)$

and hence $\displaystyle T(\pi )=T(0)$, which is what we wanted to show.

Case 2. f(0) ≠ 0. Then we have that one of $\displaystyle f(0)$ and $\displaystyle f(\pi )$ is positive and one is negative. As T is continuous, we have that f is continuous and hence we may invoke the intermediate value theorem to see that there exists a point $\displaystyle c\in (0,\pi )$ such that $\displaystyle 0=f(c)=T(c+\pi )-T(c)$. Thus, we have $\displaystyle T(c+\pi )=T(c)$ and this completes the proof.