# Derive the inequality $x<\arcsin \left(x\right)<\frac{x}{\sqrt{1-x^2}}$

I found that I could do this with the Mean Value theorem and also found a similar question here in the stackexchange and Socratic but couldn't solve the right hand side of the inequality.

I figured that by specifically choosing the function and the interval (combinations of the given question) we could prove that. But I am lost tin selecting the appropriate function.

Can you guys please suggest some ideas and also some example sums where I could check these kind of sums.

Thanks

• You can't prove it, because it is not true. Take $x=1$. Jan 6 '21 at 5:39
• @Hanul Jeon Oops sorry I wrote the question wrong Jan 6 '21 at 5:40
• Substitute $x = \sin u$ Jan 6 '21 at 5:43
• Is this only for x>0? Because the inequality breaks down if x<0. Jan 6 '21 at 6:00

Let $$f(x)=x-\sin^{-1} x \implies f'(x)=1-\frac{1}{\sqrt{1-x^2}}=\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}}<0$$ So $$f(x)$$ is a decreasing function. $$x\ge 0 \implies f(x)\le f(0) \implies f(x)\le 0$$ Next, we take $$g(x)=\sin^{-1} x-\frac{x}{\sqrt{1-x^2}} \implies g'(x)=\frac{-x^2}{\sqrt{1-x^2}}<0.$$ So $$g(x)$$ is decreasing function, then $$x \ge 0 \implies g(x)\le g(0)=0 \implies \sin^{-1}x \le \frac{x}{\sqrt{1-x^2}}.$$

• @Z Ahmed I understand the second part but the first part doesn't prove $x< sin^{-1}x$ Jan 6 '21 at 7:43
• @Aghilan Oh! thanks now please see I have corrected it. Jan 6 '21 at 8:12

For $$|x|<1$$,

$$1<\frac1{\sqrt{1-x^2}}<\frac1{(1-x^2)^{3/2}}$$ seems obvious. Then by integration between $$0$$ and $$x$$ you get the claimed inequalities.

• thanks for the interesting answer Jan 6 '21 at 9:27
• @Aghilan: any analytical proof will be equivalent to this. For geometric proofs, you need to show equivalence of the geometric axioms and analytic ones.
– user65203
Jan 6 '21 at 9:56

This is pretty standard property of $$\arcsin$$ which is equivalent to the more famous inequality $$\sin x for $$x\in(0, \pi/2)$$.

A direct proof can be given by noting that if $$x\in(0,1)$$ and $$t\in(0,x)$$ then $$1<\frac{1}{\sqrt{1-t^2}}<\frac{1}{\sqrt{1-x^2}}$$ integrating the above with respect to $$t$$ on interval $$[0,x]$$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$ for all $$x\in(0,1)$$.

You can avoid integrals and achieve the same result via mean value theorem. If $$x\in(0,1)$$ then by mean value theorem we have $$\arcsin x=\arcsin x - \arcsin 0=\frac {x}{\sqrt{1-c^2}}$$ for some $$c\in(0,x)$$. But then we can obviously see that $$1<\frac{1}{\sqrt{1-c^2}}<\frac{1}{\sqrt{1-x^2}}$$ Multiplying by positive $$x$$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$

• I just have a small doubt. The c taken must satisfy for any value in the interval of (0,1) right? So we are taking it for some c in the interval (0,x) Jan 6 '21 at 12:27
• @Aghilan: yes $c\in(0,x)$ ie $0<c<x<1$. Jan 6 '21 at 12:45