...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##".