square root of trig functions.

[maeveoneill]

What are the steps you take to get from

squareroot of all of [2e^2t (cost2t - sin2t)]^2 + [2e^2t (sin2t + cost 2t)]^2

to squareroot of 8e^4t

 

[galactus]

$\displaystyle (2e^{2t})^{2}=4e^{4t}$. Factor that out.

Then you have $\displaystyle (cos(2t)-sin(2t))^{2}-(sin(2t)+cos(2t))^{2}$, which boils down to 2.

 

[maeveoneill]

how does it boil down to 2?

 

[galactus]

Do the math. Expand it out and see. Remember the identity $\displaystyle sin^{2}x+cos^{2}x=1$