How can I know the values of: $\sin(i)$ and $\cos(i)$?
I am studying complex variables for the first time, so I asked myself if these functions exist (considering that the complex plane can be seen as $\mathbb{R}^2$); and, if so, how can I calculate their values?
Thanks :)
$\endgroup$ 24 Answers
$\begingroup$In Complex Analysis, we define$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots\text{ and }\cos z=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots$$In particular\begin{align}\sin i&=i-\frac{i^3}{3!}+\frac{i^5}{5!}-\cdots\\&=i\times\left(1+\frac1{3!}+\frac1{5!}+\cdots\right)\\&=\sinh(1)i.\end{align}By a similar argument, $\cos i=\cosh1$.
$\endgroup$ $\begingroup$HINT
Recall that
$$\cos z = \frac{e^{iz}+e^{-iz}}{2} \implies \cos i = \frac{e^{-1}+e^{1}}{2}$$
and similarly for $\sin z$.
$\endgroup$ 0 $\begingroup$Just take the definitions:$$ \sin z=\frac{\mathrm e^{iz}-\mathrm e^{-iz}}{2i},\quad \cos z=\frac{\mathrm e^{iz}+\mathrm e^{-iz}}{2}.$$Applying these formulæ with $z=i$, we obtain $$\sin(i)=\frac{\mathrm e^{-1}-\mathrm e^{1}}{2i}=i\sinh(1),\qquad \cos(i)=\cosh(1).$$
$\endgroup$ 2 $\begingroup$$\sin(i)=iG(\frac{i}{\pi})G(-\frac{i}{\pi})$ and
$\cos(i)=\sqrt{1-\sin^2(i)},$
where
$\displaystyle G(z)=\prod_{n=1}^\infty\big(1+\frac{z}{n}\big)e^{\frac{z}{n}},G$ is an entire function.
$\endgroup$
