Prove that if $0<\theta<\frac\pi2$, then $1-\frac{\theta^2}2<\cos\theta<1-\frac{\theta^2}2+\frac{\theta^4}{24}$

Question

Prove that if $0<\theta<\dfrac{\pi}{2}$, then $1-\dfrac{\theta^2}{2}<\cos \theta<1-\dfrac{\theta^2}{2}+\dfrac{\theta^4}{24}$.

Here is what I have done:

If $0<\theta<\dfrac{\pi}{2}$, then by Mean Value Theorem, $\exists\; c\in(0,\theta)$ such that

$$f'(c)=\frac{\cos \theta-\cos 0}{\theta- 0}=\frac{\cos \theta}{\theta}=-\sin c.\tag{$\ast$}$$

I know that $$\cos \theta=\sum_{n=0}^{\infty}(-1)^{n}\frac{\theta\,^{2n}}{(2n)!}>1-\dfrac{\theta^2}{2}.\tag{$\ast\ast$}$$

Questions:

1. How do I relate $(**)$ with $(*)?$
2. How do I bring $1-\dfrac{\theta^2}{2}+\dfrac{\theta^4}{24}$ in, so that $1-\dfrac{\theta^2}{2}<\cos \theta<1-\dfrac{\theta^2}{2}+\dfrac{\theta^4}{24}?$

Answer 1

The simplest proof of these inequalities is using the power series. For example $\cos \theta =1-\frac {\theta ^{2}} {2!}+(\frac {\theta ^{4}} {4!} -\frac {\theta ^{6}} {6!} )+...$ (group the terms two by two). Note that $(\frac {\theta ^{2n}} {(2n)!} -\frac {\theta ^{2n+2}} {(2n+2)!} ) \geq 0$ for $n \geq 2$ because $\theta ^{2} \leq (\frac {\pi} 2)^{2} \leq (2n+1)(2n+2)$. This gives the left hand inequality and the right hand inequality also follows by the same method. Note this method is useful in many other situations: we can get similar inequalities for $\sin (\theta)$, for example.