trigonometric simplification

**scorpion007** · April 30th 2009, 05:33 AM

Hi,

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

How did the authors reduce the following:

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

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

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

**ADARSH** · April 30th 2009, 05:37 AM

The following two identities were used

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

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

**Grandad** · April 30th 2009, 05:39 AM

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:

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

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

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

First, note the identity: $\sin2t = 2\sin t\cos t$

$\Rightarrow 8\sin^22t = 8(2\sin t \cos t)^2 = 32\sin^2t\cos^2t$

and then the identity $\cos 2t = 1 - 2\sin^2t$

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

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

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

Grandad

**scorpion007** · April 30th 2009, 05:50 AM

Ah, thank you both very much for the explanation!

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