Calculator test

[Commie]

Hi Guys,

Although not new to me, when I first saw Carey's post, the penny dropped, he stated that there is a workaround for complex trig. for the ti 84+ this being....



\[\cos(z) = \frac{e^{iz} + e^{-iz}}{2}\]
\[\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}\]

These two equations are the true equations which describe sin(x) and cos(x), i.e., if you want it to output only real, then simply leave the real part of z and set coefficient of i to zero, and hey presto,
it gives the same answer as sin(x) real. And, since the hp 15c(e) can do this for any function(more or less), means everything is complex, real is just a subset of complex.

I've checked my Derive 6.2 and it works for most functions.

Cheers Guys