Solve the given trigonometry equation

[chwala]

Homework Statement

Solve for $x$ in exact form given;

$\sinh^{-1}x = 2\cosh^{-1} 2$

Relevant Equations

hyperbolic trig equation

I was able to solve with a rather longer way; there could be a more straightforward approach;

My steps are along these lines;

$\sinh^{-1} x = 2 \ln (2+ \sqrt{3})$

$\sinh^{-1} x = \ln (7+ 4\sqrt{3})$

$x = \sinh[ \ln (7+ 4\sqrt{3})]$

$x = \dfrac {e^{\ln (7+ 4 \sqrt{3})} - e^{-[\ln 7+ 4 \sqrt{3}]}} {2}$


$x = \dfrac{(7+ 4 \sqrt{3})^2 -1}{\ 7+ 4 \sqrt{3}} \div 2$

$x = \dfrac{49+28\sqrt{3} + 28\sqrt{3} +48-1}{2(7+4\sqrt{3})}$

$x = \dfrac{672-384\sqrt{3}+392\sqrt{3} -672}{2(7+4\sqrt{3})}=\dfrac{8\sqrt{3}}{2}$

$x=4\sqrt{3}$

 

[Orodruin]

$$
x = \sinh(2\cosh^{-1}(2)) = 2\cdot 2 \sinh(\cosh^{-1}(2))
= 4 \sqrt{4-1} = 4 \sqrt 3
$$

 

[chwala]

$$
x = \sinh(2\cosh^{-1}(2)) = 2\cdot 2 \sinh(\cosh^{-1}(2))
= 4 \sqrt{4-1} = 4 \sqrt 3
$$

Informative- I was not aware of the identities as indicated on wikipedia;

https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions

I am still trying to understand how you factored out the extra $2$ in your substitution for;

$\sin[\cosh^{-1} x ]= \sqrt{x^2-1}$

Otherwise, from my further study i was able to realize,

$2 \cosh^{-1} 2 = \cosh^{-1} (2⋅2^2-1)=\cosh^{-1} 7$

and

$\sin[\cosh^{-1} 7 ]= \sqrt{7^2-1}=\sqrt{48}=4\sqrt{3}$

Cheers.

 

[Orodruin]

Informative- I was not aware of the identities as indicated on wikipedia;

https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions

I am still trying to understand how you factored out the extra $2$ in your substitution for;

$\sin[\cosh^{-1} x ]= \sqrt{x^2-1}$

Not sure which 2 you think is the ”extra” …

 

[chwala]

$$
x = \sinh(2\cosh^{-1}(2)) = 2\cdot 2 \sinh(\cosh^{-1}(2))
= 4 \sqrt{4-1} = 4 \sqrt 3
$$

highlighted...

 

[Orodruin]

highlighted...

That doesn’t really help. One of the twos is from the 2 in $\sinh(2x) = 2\sinh(x)\cosh(x)$. The other from the cosh(x) and $x = \cosh^{-1}(2)$.

 

[chwala]

Aaaaaah boss I can see what you did on that part now; a bit tricky will post the response later.

One has to take ... $\cosh[\cosh^{-1} (2)]= 2$. Wah.


...

From

$x=\sinh[2\cosh^{-1}(2)]$
$= 2\sinh[\cosh^{-1} (2)]⋅\cosh[\cosh^{-1} (2)]$
$= 2\sinh[\cosh^{-1} (2)]⋅2$
$= 4 \sinh[\cosh^{-1} (2)]=4\sqrt{4-1}=4\sqrt {3}$

 

[SammyS]

Aaaaaah boss I can see what you did on that part now; a bit tricky will post the response later.

One has to take ... $\cosh[\cosh^{-1} (2)]= 2$. Wah.

A couple of identities for hyperbolic functions are also useful. @Orodruin used the hyperbolic function double argument identity:

$\displaystyle \quad \quad \quad \quad \sinh(2u)=2\sinh u \cosh u \$ .

In addition to this hyperbolic identity, one may also make use of $\displaystyle \quad \sinh^2 u = \cosh^2 u -1 \ ,\quad$ to find that

$\displaystyle \quad \quad \sinh[\cosh^{-1} (2)] = \sqrt{ \cosh^2 [ \cosh^{-1} (2) ]-1 \ }=\sqrt{3}$

So, its not so tricky after all.
Moreover, there's no need to resort to tricky identities for Inverse hyperbolic functions.