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

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 $1-\dfrac{\theta^2}{2}<\cos \theta<1-\dfrac{\theta^2}{2}+\dfrac{\theta^4}{24}?$
• Did you mean $\theta^4/24$? – J.G. May 16 '18 at 6:18
• This is true for all $\theta\ne0$ and the second inequality can be improved to $\cos\theta<1-\theta^2/2+\theta^4/24$. It can be proved by integration, or the MVT. – Lord Shark the Unknown May 16 '18 at 6:18
• @J.G.: Oh yes! Sorry for the typo! – Omojola Micheal May 16 '18 at 6:19
• This answer may be of help: math.stackexchange.com/questions/2268559/… – rae306 May 16 '18 at 6:24
• Here is another helpful post: math.stackexchange.com/questions/2140643 In there, it is shown that IN GENERAL, truncations of the Taylor series are valid inequalities if $f^{(n)}(x)$ has alternating signs independent of $x$. See MartinR's answer. In particular, see the last comment in the discussion where $\cos(x)$ is explicitly mentioned as a valid example. – Andreas May 16 '18 at 6:37

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.