What trig. identity is used here?

[Peon666]

4/t [cos(wt/2)-1] = -8/t sin(wt/4)

?

[nicksauce]

This equation simply is not true. For t >0, the LHS is always <= 0, while the RHS becomes both positive and negative.

See: http://www.wolframalpha.com/input/?i=plot+-8%2Ft*sin%28t%2F4%29+and+%284%2Ft%29*%28cos%28t%2F 2%29-1%29

[g_edgar]

double-angle formula, \cos(2\theta) = 1-2 \sin^2(\theta) so \cos(2\theta)-1 = -2 \sin^2(\theta) ... so your original formula is correct except for a missing square.

[Peon666]

Thanks a lot. :)