4/t [cos(wt/2)-1] = -8/t sin(wt/4)
This equation simply is not true. For t >0, the LHS is always <= 0, while the RHS becomes both positive and negative.
double-angle formula, [itex]\cos(2\theta) = 1-2 \sin^2(\theta)[/itex] so [itex]\cos(2\theta)-1 = -2 \sin^2(\theta)[/itex] ... so your original formula is correct except for a missing square.
Thanks a lot. :)
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