# Thread: trigonometric simplification

1. ## trigonometric simplification

Hi,

I seem to have forgotten much about trig. identities and so on.

How did the authors reduce the following:

$\displaystyle 32\cos^2{t}\sin^2{t}$

$\displaystyle = 8\sin^2{2t}$

$\displaystyle = 4-4\cos{4t}$

2. The following two identities were used

2sin(x)cos(x) =sin(2x)

sin^2(y) = (1-cos(2y))/2

3. ## Trig Identities

Hello scorpion007
Originally Posted by scorpion007
Hi,

I seem to have forgotten much about trig. identities and so on.

How did the authors reduce the following:

$\displaystyle 32\cos^2{t}\sin^2{t}$

$\displaystyle = 8\sin^2{2t}$

$\displaystyle = 4-4\cos{4t}$
First, 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 $\displaystyle \cos 2t = 1 - 2\sin^2t$

$\displaystyle \Rightarrow \sin^2t = \tfrac12(1-\cos 2t)$

$\displaystyle \Rightarrow \sin^22t= \tfrac12(1-\cos 4t)$

$\displaystyle \Rightarrow 8\sin^22t = 4(1 -\cos 4t)$