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/t*sin(t/4)+and+(4/t)*(cos(t/2)-1)

 

[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. :)