# A calculus of the absurd

#### 10.9 A bunch of trigonometric limits

There are some useful properties about what happens to the values of trig functions when the angles we input into them tend to zero^{75}^{75} They’re also in the formula sheet, where they are called the "small angle
approximations". These are that

\[\lim _{x \to 0} \frac {\sin (x)}{x} = 1 \text { and } \lim _{x \to 0} \frac {1-\cos (x)}{x}=0\]

and it isn’t immediately clear why this is true. I hope you like geometry, because if (like me) you don’t, then proving this is just a bit painful (note: the proof is definitely not in the A Level, so you can just skip to the next section where the actual differentiation of trig functions - which is in the A Level - is).

First, we can draw a diagram of the unit circle, with a bunch of additional triangles

from which we can work out the areas of some of the shapes. The area of the small triangle (ADO) is just half the base times the height. The height is AB, which (by applying trigonometry to ABO) is just \(\sin (\theta )\) and thus the area of ADO is \(\frac {1}{2} 1 \sin (\theta ) = \frac {\sin (\theta )}{2}\). The area of the section of the circle, with angle \(\theta \) is \(\pi 1^2 \frac {\theta }{2\pi }\), i.e. \(\frac {\theta }{2}\). The area of OCD is \(\frac {\tan (\theta )}{2}\).

From here, we can write down an inequality, and use a principle called various things (including the "squeeze principle" and the "sandwich principle") but meaning one thing; if for every value of \(x\) it is true that \(a(x) < b(x) < c(x)\), then if as \(x \to p\) both \(a(x)\) and \(c(x)\) both tend towards \(L\), then \(b(x)\) will also tend towards \(l\).

Our inequality states that the area of ABO is smaller than the section of the circle, which in turn is smaller than the area of triangle OCD. Thus we have that

\(\seteqnumber{0}{10.}{85}\)\begin{equation} \frac {\sin (\theta )}{2} < \frac {\theta }{2} < \frac {\tan (\theta )}{2} \end{equation}

which we can then start to rearrange. Firstly, all the \(2\)s can go

\(\seteqnumber{0}{10.}{86}\)\begin{equation} \sin (\theta )< \theta < \tan (\theta ) \end{equation}

Then, we can divide through by \(\sin (\theta )\), which gives that ^{76}^{76} Remember that

\[\tan (\theta ) = \frac {\sin (\theta )}{\cos (\theta )}\]

\begin{equation} 1 < \frac {\theta }{\sin (\theta )} < \frac {1}{\cos (\theta )} \end{equation}

The middle bit looks pretty close to what we’re trying to prove! If we apply \(f(x) = \frac {1}{x}\) to each part of the inequality, we get that

\(\seteqnumber{0}{10.}{88}\)\begin{equation} 1 > \frac {\sin (\theta )}{\theta } > \cos (\theta ) \end{equation}

^{77}^{77} Why did we flip the inequality? If you look at the graph of \(\frac {1}{x}\) above, and pick any two values of \(x\) you like (e.g. \(2\) and \(3\)), and then take the reciprocal bigger values become
*smaller*! This means that while we previously had \(3 > 2\), we now have \(\frac {1}{3} < \frac {1}{2}\). This is because \(f(x) = \frac {1}{x}\) is a *decreasing* function.

From here, we can now take a limit as \(\theta \to 0\).

\(\seteqnumber{0}{10.}{89}\)\begin{equation} \lim _{\theta \to 0} \cos (\theta ) < \lim _{\theta \to 0} \frac {\sin (\theta )}{\theta } < \lim _{\theta \to 0} 1 \end{equation}

From the graph of \(\cos (\theta )\), as \(\theta \to 0\) we can see that \(\cos (\theta ) \to 1\), and so by the squeeze principle (mentioned above) because limits on either side of the middle are equal to \(1\), then

\(\seteqnumber{0}{10.}{90}\)\begin{equation} \label {lim sinx/x} \lim _{\theta \to 0} \frac {\sin (\theta )}{\theta } = 1 \end{equation}

To then prove that

\(\seteqnumber{0}{10.}{91}\)\begin{equation*} \lim _{\theta \to 0} \frac {1-\cos (\theta )}{\theta } = 0 \end{equation*}

we can just rewrite in terms of Equation 10.91, by multiplying by \(1+\cos (\theta )\).

\(\seteqnumber{0}{10.}{91}\)\begin{align*} \lim _{\theta \to 0} \frac {1-\cos (\theta )}{\theta } &= \lim _{\theta \to 0} \frac {1-\cos (\theta )}{\theta } \frac {1+\cos (\theta )}{1+\cos (\theta )} \\ &= \lim _{\theta \to 0} \frac {1-\cos ^2(\theta )}{\theta } \frac {1}{1+\cos (\theta )} \\ &= \lim _{\theta \to 0} \frac {\sin ^2(\theta )}{\theta } \frac {1}{1+\cos (\theta )} \end{align*}

But we want to apply \(\lim _{\theta \to 0} \frac {\sin (\theta )}{\theta } = 1\) somewhere! If we split \(\sin ^2(\theta )\) into \(\sin (\theta )\sin (\theta )\) we then have that

\(\seteqnumber{0}{10.}{91}\)\begin{align*} \lim _{\theta \to 0} \frac {\sin ^2(\theta )}{\theta } \frac {1}{1+\cos (\theta )} &= \lim _{\theta \to 0} \sin (\theta ) \frac {\sin (\theta )} {\theta } \frac {1} {1+\cos (\theta )} \end{align*}

Which is equal to \(0\), because as \(\theta \) approaches \(0\), so does \(\sin (\theta )\).