Prove inequalities (Calculus 1) $\ x - \frac{x^{3}}{6}< \sin x < x-\frac{x^{3}}{6} +\frac{x^{5}}{120}$

Question

Prove the following inequality:

$$x - \frac{x^{3}}{6}< \sin x < x-\frac{x^{3}}{6} +\frac{x^{5}}{120} ,\; \forall x > 0$$

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I'm not sure my proof is correct. I separated the problem into proving the inequality on the left and on the right side. The left side wasn't really difficult,the right one,however,gave me a lot of work.I named $g(x)=x-\frac{x^{3}}{6} +\frac{x^{5}}{120}$ and $f(x) = \sin(x)$ and derived both equations multiple times until i reached the $5$th and analysed the rate of growth of each one until I came into the conlusion that $f(x) < g(x)$.I used the same method on the left side and it worked.Could anyone tell me if I'm on the right path here?Any help would be kindly appreciated. Thank you in advance.

Answer 1

Start with $\sin'(x)=\cos(x) $ and $\cos'(x)=-\sin(x) $ so

$(1) \sin(x) =\int_0^x \cos(t) dt $

and

$(2) \cos(x) =1-\int_0^x \sin(t) dt $.

For $x > 0$, starting with $\cos(x) \le 1$, (1) gives $\sin(x) \le x$.

(2) then gives $\cos(x) \ge 1-\dfrac{x^2}{2} $.

(1) then gives $\sin(x) \ge x-\dfrac{x^3}{6} $.

(2) then gives $\cos(x) \le 1-\dfrac{x^2}{2}+\dfrac{x^4}{24} $.

(1) then gives $\sin(x) \le x-\dfrac{x^3}{6}+\dfrac{x^5}{120} $.

And so on.

By induction, you can show that the power series for $\sin$ and $\cos$ are enveloping.

Note: this is not original. I found it in One Hundred Great Problems of Elementary Mathematics: Their History and Solution (Dover classics of science & mathematics) by Heinrich Dorrie.

I highly recommend this book. It is available for $10 as a Kindle book.

Answer 2

Note that $1-\cos u\ge 0$. Integrate this positive function recursively to obtain \begin{align} \int_0^x\int_0^t\int_0^s (1-\cos u)du ds dt= \frac{x^3}6-x+\sin x>0 \end{align} which is the left inequality $x - \frac{x^{3}}{6}< \sin x$. Integrate two more times to obtain the right inequality $\sin x < x-\frac{x^{3}}{6} +\frac{x^{5}}{120}$.

Answer 3

If you proved that $x-\frac{x^3}{6}<\sin{x}$, so $$g''(x)=-x+\frac{x^3}{6}+\sin{x}>0,$$ which says $g'$ increases for $x>0$ and we obtain $$g'(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\cos{x}>g'(0)=0,$$ which says $$g(x)>g(0)=0.$$ By the way, $x-\frac{x^3}{6}<\sin{x}$ because $$\left(\sin{x}-x+\frac{x^3}{6}\right)'=\cos{x}-1+\frac{x^2}{2}=2\left(\frac{x}{2}\right)^2-2\sin^2\frac{x}{2}=$$ $$=2\left(\frac{x}{2}-\sin\frac{x}{2}\right)\left(\frac{x}{2}+\sin\frac{x}{2}\right)>0.$$

Answer 4

Say $f$ on some interval $I =[0, M]$, $f(0) = 0$, and $f'>0$ on $I$ except finitely many points. Then $f>0$ on $I$. This should be clear, as $f$ is strictly increasing. Using this we can prove by induction the following: if $f(0) = f'(0) = \ldots f^{(n-1)} (0) = 0$, and $f^{(n)} > 0$ on $I$ except finitely many points then $f>0$ on $I$.

So consider \begin{eqnarray} f(x)&=& \sin x - (x - \frac{x^3}{6})&,&\ \ f(0) &=&0\\ f'(x) &=& \cos x- (1- \frac{x^2}{2})&,& \ \ f'(0) &=& 0 \\ f''(x) &=& - \sin x + x &,& \ f''(0)&=& 0 \\ f^{(3)}(x) &=& 1 - \cos x = 2 \sin^2 \frac{x}{2} &> &0 \ \ \ \ \ \ \ \text{ (except a discrete subset)} \end{eqnarray} .