10.10 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 zero1515 15 They’re also in the formula sheet, where they are called the ”small angle approximations”. These are that

limx0sin(x)x=1 and limx01cos(x)x=0\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(θ)\sin(\theta) and thus the area of ADO is 121sin(θ)=sin(θ)2\frac{1}{2}1\sin(\theta)=\frac{\sin(\theta)}{2}. The area of the section of the circle, with angle θ\theta is π12θ2π\pi 1^{2}\frac{\theta}{2\pi}, i.e. θ2\frac{\theta}{2}. The area of OCD is tan(θ)2\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 xx it is true that a(x)<b(x)<c(x)a(x)<b(x)<c(x), then if as xpx\to p both a(x)a(x) and c(x)c(x) both tend towards LL, then b(x)b(x) will also tend towards ll.

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

sin(θ)2<θ2<tan(θ)2\frac{\sin(\theta)}{2}<\frac{\theta}{2}<\frac{\tan(\theta)}{2} (10.87)

which we can then start to rearrange. Firstly, all the 22s can go

sin(θ)<θ<tan(θ)\sin(\theta)<\theta<\tan(\theta) (10.88)

Then, we can divide through by sin(θ)\sin(\theta), which gives that 1616 16 Remember that tan(θ)=sin(θ)cos(θ)\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}

1<θsin(θ)<1cos(θ)1<\frac{\theta}{\sin(\theta)}<\frac{1}{\cos(\theta)} (10.89)

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

1>sin(θ)θ>cos(θ)1>\frac{\sin(\theta)}{\theta}>\cos(\theta) (10.90)
1717 17 Why did we flip the inequality? If you look at the graph of 1x\frac{1}{x} above, and pick any two values of xx you like (e.g. 22 and 33), and then take the reciprocal bigger values become smaller! This means that while we previously had 3>23>2, we now have 13<12\frac{1}{3}<\frac{1}{2}. This is because f(x)=1xf(x)=\frac{1}{x} is a decreasing function.

From here, we can now take a limit as θ0\theta\to 0.

limθ0cos(θ)<limθ0sin(θ)θ<limθ01\lim_{\theta\to 0}\cos(\theta)<\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta}<% \lim_{\theta\to 0}1 (10.91)

From the graph of cos(θ)\cos(\theta), as θ0\theta\to 0 we can see that cos(θ)1\cos(\theta)\to 1, and so by the squeeze principle (mentioned above) because limits on either side of the middle are equal to 11, then

limθ0sin(θ)θ=1\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta}=1 (10.92)

To then prove that

limθ01cos(θ)θ=0\lim_{\theta\to 0}\frac{1-\cos(\theta)}{\theta}=0

we can just rewrite in terms of Equation 10.92, by multiplying by 1+cos(θ)1+\cos(\theta).

limθ01cos(θ)θ\displaystyle\lim_{\theta\to 0}\frac{1-\cos(\theta)}{\theta} =limθ01cos(θ)θ1+cos(θ)1+cos(θ)\displaystyle=\lim_{\theta\to 0}\frac{1-\cos(\theta)}{\theta}\frac{1+\cos(% \theta)}{1+\cos(\theta)}
=limθ01cos2(θ)θ11+cos(θ)\displaystyle=\lim_{\theta\to 0}\frac{1-\cos^{2}(\theta)}{\theta}\frac{1}{1+% \cos(\theta)}
=limθ0sin2(θ)θ11+cos(θ)\displaystyle=\lim_{\theta\to 0}\frac{\sin^{2}(\theta)}{\theta}\frac{1}{1+\cos% (\theta)}

But we want to apply limθ0sin(θ)θ=1\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta}=1 somewhere! If we split sin2(θ)\sin^{2}(\theta) into sin(θ)sin(θ)\sin(\theta)\sin(\theta) we then have that

limθ0sin2(θ)θ11+cos(θ)\displaystyle\lim_{\theta\to 0}\frac{\sin^{2}(\theta)}{\theta}\frac{1}{1+\cos(% \theta)} =limθ0sin(θ)sin(θ)θ11+cos(θ)\displaystyle=\lim_{\theta\to 0}\sin(\theta)\frac{\sin(\theta)}{\theta}\frac{1% }{1+\cos(\theta)}

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