Hi,
I seem to have forgotten much about trig. identities and so on.
How did the authors reduce the following:
$\displaystyle 32\cos^2{t}\sin^2{t}$
$\displaystyle = 8\sin^2{2t}$
$\displaystyle = 4-4\cos{4t}$
Hello scorpion007First, note the identity: $\displaystyle \sin2t = 2\sin t\cos t$
$\displaystyle \Rightarrow 8\sin^22t = 8(2\sin t \cos t)^2 = 32\sin^2t\cos^2t$
and then the identity $\displaystyle \cos 2t = 1 - 2\sin^2t$
$\displaystyle \Rightarrow \sin^2t = \tfrac12(1-\cos 2t)$
$\displaystyle \Rightarrow \sin^22t= \tfrac12(1-\cos 4t)$
$\displaystyle \Rightarrow 8\sin^22t = 4(1 -\cos 4t)$
Grandad