Calculator test

[Idnarn]

(12-10-2024 05:44 PM)Commie Wrote:  \[\cos(z) = \frac{e^{iz} + e^{-iz}}{2}\]
\[\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}\]

\[\sin(a+b\times i) = \sin(a)\times\cosh(b)+\cos(a)\times \sinh(b)\times i\]
\[\cos(a+b\times i) = \cos(a)\times\cosh(b)-\sin(a)\times \sinh(b)\times i\]

If you have a calculator capable of sinh and cosh on real numbers, you can compute it.