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Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$

I am getting:

$$\sin(x) = x - \frac{x^3}{3!} + R_4(x)$$

where $R_4(x) = \frac{\cos(c)x^5}{5!}$ for some $c$ between $0$ and $x$

I want to prove $R_4(x)\geq 0$ to arrive at the result $x-x^3/3 \leq \sin(x)$.

for:$$0 \leq x \leq \pi/2 \Rightarrow 0<c<\pi/2 \Rightarrow R_4(x)\geq 0$$ but for: $$-\pi/2 \leq x < 0 \Rightarrow -\pi/2 < c < 0 \Rightarrow R_4(x) < 0$$

How can I proceed with this ? There are many cases that I need to check

Somos
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So Lo
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2 Answers2

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The aim is to show that:

$$ x - \frac{x^3}{3!} \leq \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + ...\leq x $$

For $sin x\leq x$ it suffices to use MVT:

$$\cos c =\frac{\sin x- \sin 0}{x-0}=\frac{\sin x}{x}\implies -1\leq \frac{\sin x}{x}\leq \implies \frac{\sin x}{x}\leq 1 \implies sin x\leq x$$

For $x - \frac{x^3}{3!}\leq \sin(x) $ see here

Proof for $\sin(x) > x - \frac{x^3}{3!}$

user
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This is an easy consequence of Taylor's theorem. We have via Taylor's theorem with Langrange's form of remainder $$\sin x=x-\frac{x^{3}}{3!}\cos\theta x$$ where $0<\theta<1$. Since $\cos\theta x\leq 1$ it follows that $$\sin x\geq x-\frac{x^{3}}{6},\,\forall x\geq 0$$ The inequality $\sin x\leq x$ for $x\geq 0$ is a fundamental property of $\sin x$ which is an immediate consequence of any suitable definition of $\sin x$. Thus for example the geometrical definition based on arc length of a circle implies $$x=\int_{0}^{\sin x} \frac{dt} {\sqrt{1-t^{2}}} $$ for $x\in(-\pi/2,\pi/2)$ and gives the inequality directly as the integrand is greater than or equal to $1$.

More generally if $x\geq 0$ then Taylor's theorem can be used to show that $$x-\frac{x^{3}}{3!}+\dots+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}\leq \sin x\leq x-\frac{x^{3}}{3!}+\dots+(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$$ whenever $n$ is an even positive integer.

Paramanand Singh
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