Prove $\sin x\ge x-\frac{x^3}{3!}$

Question

Prove $$\sin x\ge x-\frac{x^3}{3!}$$ for $x\ge 0$.

I know that a calculus/continuity proof exists, but I am curious if this can be proved without that. Here is a sketch I have made.;

we will use for $x\ge 0$;

$$\sin x\le x\le \tan x$$

proof of this see here

I got a weaker bound : $$\tan\left(\frac{x}{2}\right)\ge \frac{x}{2}$$ $$\sin\left(\frac{x}2\right)\cos\left(\frac{x}2\right)\ge \frac{x}{2}\cos^2\left(\frac{x}{2}\right)$$ $$\sin x= x\left[1-\sin^2\left(\frac{x}{2}\right)\right]\ge x\left(1-\frac{x^2}{4}\right)= x-\frac{x^3}{4}$$ Thus $$\sin x \ge x-\frac{x^3}{4}$$ As you can see is there any way I can take this to strengthen the inequality, to get $$\sin x\ge x-\frac{x^3}{6}.$$ Also could this possibly be extended by strengthening inequality further to get the Taylor series. I think I am going too far!

@Surb not at all my question is very different , i want to solve this differntly — Albus Dumbledore, Nov 13 '20 at 14:23
Does this answer your question [ it gives the limit as $x\to 0$ and then use the fact that $\sin(x)<x$ for $x>0$]? Are all limits solvable without L'Hôpital Rule or Series Expansion — Integrand, Nov 13 '20 at 15:29
@Integrand pls explain how this is relevant to this problem — Albus Dumbledore, Nov 13 '20 at 17:31
Use robjohn's geometric arguments to get $\lim_{x\to 0}(\sin(x) - x +x^3/6)/x^3=0^+$, i.e. the term has a strict minimum at $x=0$ — Integrand, Nov 13 '20 at 17:46

xpaul · Answer 1 · 2020-11-13T17:38:28.253

3

Using $\sin^2(\frac t2)\le (\frac{t}{2})^2$, one has $$ \sin(x)=\int_0^x\cos(t)dt=\int_0^x\bigg(1-2\sin^2(\frac t2)\bigg)dt\ge \int_0^x(1-\frac12t^2)dt=x-\frac{x^3}{3!}.$$

edited Nov 13 '20 at 17:38

answered Nov 13 '20 at 14:53

xpaul

44,000

That's nice but arent you still using calculus – Albus Dumbledore Nov 13 '20 at 15:03
Never mind ,this proof is also nice,but if you do get one without calculus please considr sharing – Albus Dumbledore Nov 13 '20 at 15:15
I will think about it. Thanks. – xpaul Nov 13 '20 at 15:16
The cubic term can be tightened to $x^3/3!=x^3/6$, as in the question. Note $2\sin^2(t/2)\sim2(t/2)^2=t^2/2$ for small $t$. – J.G. Nov 13 '20 at 15:16
@J.G., thanks, I didn't see clearly. – xpaul Nov 13 '20 at 15:53

score 2 · Accepted Answer · answered Nov 13 '20 at 18:47

Well, from the celebrated triplication formula, we get $$\sin3x=3\,\sin x-4\,\sin^3x.$$ This, of course, means that $$3^{k}\sin x/3^k-3^{k+1}\sin x/3^{k+1}=-4\cdot3^{k}\,\sin^3 x/3^{k+1}.$$ Summing the telescopic series and observing $3^{k}\sin x/3^k\to x$ as $k\to\infty$, we get $$\sin x-x= -4\,\sum^\infty_{k=0}3^{k}\,\sin^3 x/3^{k+1},$$ i.e. $$x-\sin x= 4\,\sum^\infty_{k=0}3^{k}\,\sin^3 x/3^{k+1}.$$ Since $\sin x\le x$ for non-negative $x$, this means $$x-\sin x\le4\,\sum^\infty_{k=0}3^k\,\left(x/3^{k+1}\right)^3=4\cdot\frac1{27}\cdot\frac1{1-\frac19}\,x^3=\frac16\,x^3.$$

wow! thats nice i never would have thought of that,could a similar argument be taken to prove $\sin x\ge x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^5}{7!}$ — Albus Dumbledore, Nov 14 '20 at 04:09

Prove $\sin x\ge x-\frac{x^3}{3!}$

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