Here

there's the proof of a very nice trigonometric identity. is there a way to generate other identities that are similar to this-one by using also the complex numbers ?

thanks, p.

Pascal,

There are plenty of ways to generate identities. Here's a simple one:

To express powers of cosq in terms of cosq, cos2q etc.:

Consider cos³ qas an example:

cos³ q = [1/2(e^(iq)+ e^-iq)]³

=1/4[1/2(e^3iq+ e^-3iq) +3/2(e^iq+ e^-iq)]

=1/4[cos3q+3cosq]

The method works for other powers of cosq, and can similarly be applied to powers of sinq.

It can also be applied in reverse to express cosnq in terms of powers of cosq.
Andre