Prove the hyperbolic function corresponding to the given trigonometric function

[chwala]

...Must improve before it can be proved.

(A variant of #6 )
Since $\sin (ix) = i \sinh (x)$,
then $\sin (x) = i \sinh (x/i) = i \sinh (-ix)= -i\sinh(ix) = -i\sinh(u)$ where $u=ix$.

check: $\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right) \stackrel{u=ix}{=}-i \left(\frac{e^{u}-e^{-u}}{2}\right) =-i\sinh(u)$​

And since $\cos ix = \cosh (x)$,
then $\cos x = \cosh (x/i)= \cosh(-ix) =\cosh(ix) = \cosh(u)$ where $u=ix$.

check: $\displaystyle \cos(x)=\frac{e^{ix}+e^{-ix}}{2} \stackrel{u=ix}{=} \left(\frac{e^{u}+e^{-u}}{2}\right) =\cosh(u)$​

my revisions in redSo,
\begin{align*}
8 \sin^4x &= 3-4\cos 2x+\cos 4x\\
8 \color{red}{(-i \sinh u)}^4 &= 3-4(\color{red}{\cosh 2u})+(\color{red}{\cosh 4u})\\
8 (-i)^4 \sinh^4 u &= \\
8 \phantom{XX} \sinh^4 u &= \\
8 \sinh^4 u &= 3-4\cosh 2u+\cosh 4u\\
&= \ldots\\
&\ \ \vdots\\
\end{align*}
Because $(-i)^4=1$ (ref: #17 ), you were lucky that you never saw the problem.

In general, one might not be so lucky. (Ref: #2 )
\begin{align*}
1 &= \cos^2 x +\sin^2x \\
&= \color{red}{(\cosh u)}^2 + \color{red}{(-i \sinh u)}^2 \\
&=\cosh^2 u +(-i)^2 \sinh^2 u \\
&=\cosh^2 u +(-1) \sinh^2 u \\
&=\cosh^2 u - \sinh^2 u \\
\end{align*}
which is not what one gets by
replacing [incorrectly] "$\sin x$ with $\sinh x$ and $\cos x$ with $\cosh x$".

check: $\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right)$

just to be absolutely clear on the referenced...you multiplied the left hand side by $-i$ to arrive at the right hand side. Correct?

 

[robphy]

check: $\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right)$

just to be absolutely clear on the referenced...you multiplied the left hand side by $-i$ to arrive at the right hand side. Correct?

I multiplied by 1 = $\frac{-i}{-i}$.

 

[vanhees71]

To me, the question sounds odd.

How would you answer the question if the given equation were
$$1=\cos^2 u +\sin^2 u$$
instead?

I guess the idea is to use
$$\cosh u=\cos(\mathrm{i} u), \quad \sinh u=-\mathrm{i} \sin(\mathrm{i} u).$$
This gives
$$1=\cos^2(\mathrm{i} u) + \sin^2(\mathrm{i} u)=\cosh^2 u-\sinh^2 u.$$
;-)).