Proving $ 2\arcsin\left(\sinh(x)\right) + \arccos\left( 2 - \cosh(2x) \right) = 0 $

Question

$$ 2\arcsin\big( \sinh(x) \big) + \arccos\big( 2 - \cosh(2x) \big) = 0 $$

If this is true, then for what values of $x$? (Clearly, the $\arcsin(\cdot)$ and $\arccos(\cdot)$ are not valid for every real argument.)

Answer 1

Recall a couple double-angle formulas valid for all $x$:

$$ \cos(2x)=1-2\sin^2(x), \\ \cosh(2x)=1+2\sinh^2(x). \tag{1}$$

Apply $u\mapsto 2-u$ to the latter so we may write

$$ 1-2\sinh^2(x)=2-\cosh(2x). \tag{2}$$

Use the substitution $\sinh(x)=\sin(\theta)$, where $\theta$ is in $[-\frac{\pi}{2},\frac{\pi}{2}]$, the range of $\arcsin$. This way the left side of $(2)$ is just $\cos(2\theta)$, so we may write this as

$$ \cos(2\theta)=2-\cosh(2x). \tag{3}$$

Split into two cases: either $2\theta$ is in $[0,\pi]$, the range of $\arccos$, or it's in $[-\pi,0]$.

In the first case, we can apply $\arccos$ to both sides of $(3)$ to obtain

$$ 2\theta=\arccos(2-\cosh(2x)) \tag{4a}$$

In the second case, we have $2\theta=-\arccos(\cos(2\theta))$, so from $(3)$ we get

$$ 2\theta=-\arccos(2-\cosh(2x)). \tag{4b}$$

The conditions on $\theta$ can be turned into conditions on $x$:

$$ \begin{array}{rrrrr} 0\le\theta\le\frac{\pi}{2} & \iff & 0\le \sin\theta\le 1 & \iff & 0\le x\le\sinh^{-1}(1) \\ -\frac{\pi}{2}\le\theta\le0 & \iff & -1\le\sin\theta\le0 & \iff & -\sinh^{-1}(1)\le x\le0 \end{array} \tag{5}$$

Finally, replace $\theta$ with $\arcsin(\sinh(x))$. In conclusion,

$$ \arcsin(\sinh(x))=\mathrm{sgn}(x)\arccos(2-\cosh(2x)) $$ for all $-\sinh^{-1}(1)\le x\le \sinh^{-1}(1).~$ (Note $\sinh^{-1}(1)=\ln(1+\sqrt{2})\approx 0.88$.)

Answer 2

Looks wrong to Desmos:

So...I guess that it's probably almost never true (except perhaps for $x = 0$).

Details: Let's look at $x = \frac{1}{2}$.

Now here's a matlab program to test that out:

function silly()
x = 0.5
a = sinh(x)
b = asin(a)
c = 2 * b

'part 2'
d = cosh(1)
f = 2 - d
g = acos(f)

'sum'
c + g

and the results of executing it:

>> silly
x =
    0.5000
a =
    0.5211
b =
    0.5481
c =
    1.0963
ans =
   'part 2'
d =
    1.5431
f =
    0.4569
g =
    1.0963
ans =
    'sum'
ans =
    2.1925

If $2.1925$ looks like zero to within roundoff error, then you are correct; if not, then Desmos and I are correct. It's telling that the two parts look identical, so that their difference at $x = 0.5$ would be zero, while their sum is nonzero.

(To be fair, I should say that I was half-correct --- it's almost never true for positive $x$; it appears to be identically true for negative $x$.)

Answer 3

If we use the Taylor series, $\arcsin(\sinh(x))\approx x+\frac {x^3}3+\frac {x^5}6$ and $\arccos(2-2-\cosh(2x))\approx |2x+\frac {2x^3}3+\frac {x^5}3|$ so for $x \gt 0$ the dominant term is $4x$. For $x \lt 0$ the first through fifth order terms cancel.

Answer 4

I suppose a typo; changing the $+$ to $-$ makes

$$2\arcsin\big( \sinh(x) \big) \color{red}{-} \arccos\big( 2 - \cosh(2x) \big) = 0$$ true for all $x \geq 0$.

Composing Taylor series built at $x=0$ and assuming $x >0$, we have $$2\arcsin\big( \sinh(x) \big)=2 x+\frac{2 x^3}{3}+\frac{x^5}{3}+\frac{79 x^7}{315}+\frac{493 x^9}{2268}+O\left(x^{11}\right)$$ $$\arccos\big( 2 - \cosh(2x) \big)=2 x+\frac{2 x^3}{3}+\frac{x^5}{3}+\frac{79 x^7}{315}+\frac{493 x^9}{2268}+O\left(x^{11}\right)$$

Answer 5

Let $$ L(x) = 2\arcsin(\sinh(x)) \\ R(x) = -\arccos(2-2\cosh(x)) $$ I'll show that these functions are equal for $x \ge 0$ when $|\sinh(x)| \le 1$.

Step 1: check that $L(0) = R(0)$. Yes.

Step 2: Check that $L'(x) = R'(x)$ for all $x$.

Start from the following: for $0 \le u \le 1$, we have

$$ \sqrt{4u^2 - 4u^4} = 2u \sqrt{1-u^2}, $$ because $u > 0$, so we don't need the absolute value sign on the right (although including it would probably deal with the case where $x < 0$ later, but I leave that to OP).

Then \begin{align} \sqrt{1 - (1 - 4u^2 + 4u^4)} &= 2u \sqrt{1-u^2} \\ \sqrt{1 - (1 - 2u^2)^2} &= 2u \sqrt{1-u^2}. \end{align} If we replace $u$ with $\sinh(x)$, we get \begin{align} \sqrt{1 - (1 - 2\sinh^2(x))^2} &= 2\sinh(x) \sqrt{1-\sinh^2(x)}\\ \sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\sinh(x) \sqrt{1-\sinh^2(x)}\\ \cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\cosh(x)\sinh(x) \sqrt{1-\sinh^2(x)}\\ \cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= \sinh(2x) \sqrt{1-\sinh^2(x)}\\ 2\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\sinh(2x) \sqrt{1-\sinh^2(x)}\\ \frac{2\cosh(x)}{\sqrt{1-\sinh^2(x)}} &= \frac{2\sinh(2x)}{\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2}}\\ \frac{2\cosh(x)}{\sqrt{1-\sinh^2(x)}} &= \frac{2\sinh(2x)}{\sqrt{1 - (2 - \cosh(2x))^2}}\\ \end{align}

The left hand side of this is just $L'(x)$; the right hand side is $R'(x)$. Because they're equal, and $L(0) = R(0)$, we have that $L(x) = R(x)$ within the specified domain (by the FTC, if you want to get fancy).

In short: the OP's statement, with a sign-fix, is correct for positive $x$; a similar proof, without the sign-change, presumably demonstrates the original claim for negative $x$. The whole thing is an exercise in algebra and the FTC.