Here's the question you clicked on:
virtus
express sin 3theta in terms of sin theta
http://en.wikipedia.org/wiki/List_of_trigonometric_identities
It's located under Double-, triple-, and half-angle formulae
\[\sin(3 \phi)=\frac{e^{3i \phi}-e^{-3i \phi}}{2i}\] Then use: \[e^{3i \phi}=(e^{i \phi})^3; e^{i \phi}=\cos(\phi)+i \sin(\phi)\] Then set the real part equal.
thanks but i do not really understand the working out
If you know the formula for \[\sin (A+B)\] then you can derive the formula for \[\sin 2A = \sin (A + A)\] and \[\sin 3A = \sin (A + 2A) \]