Simplifying an equation using trig identities

**centenial** · October 28th 2010, 04:33 PM

Hi,

I need to simplify this equation:

$\sqrt{(-2\sin{2t})^2 + (2\cos{2t})^2}$

I know that this equals 2 (by plugging into a calculator). However, I can't figure out the steps to simplify it. This is what I have... could someone show me what I'm doing wrong?

$\sqrt{-4\sin^2{2t} + 4\cos^2{2t}}$

$\sqrt{4(\cos^2{2t} - \sin^2{2t})}$

$2 \sqrt{1 - 2\sin^2{2t}}$

but I'm not sure where to go from here...

**Prove It** · October 28th 2010, 04:36 PM

$(-2\sin{2t})^2 = 4\sin^2{2t}$ .

So you should actually have

$\sqrt{(-2\sin{2t})^2 + (2\cos{2t})^2} = \sqrt{4\sin^2{2t} + 4\cos{2t}}$

$= \sqrt{4(\sin^2{2t} + \cos^2{2t})}$ .

Go from here.

**centenial** · October 28th 2010, 04:40 PM

Oh... then I just use cos^2 + sin^2 = 1, and it's easy.

Thanks so much for showing me my mistake!

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