# Thread: easy trigo identity, not getting answer in solutions..

1. ## easy trigo identity, not getting answer in solutions..

I really just want to verify this for peace of mind, as I am beyond certain of my answer (I even ran it through C++ to verify my answer):

$\displaystyle \sqrt{(-2sin2t)^2 + (2cos2t)^2} = 2$, simple enough right?

however the solutions shows it to be equal to 4.

Surely this is a mistake unless I'm completely missing something here.

2. You are correct; it's 2.

3. $\displaystyle \displaystyle \sqrt{(-2\sin{2t})^2+(2\cos{2t})^2} = \sqrt{4\sin^2{2t}+4\cos^2{2t}} = \sqrt{4\left(\sin^2{2t}+\cos^2{2t}\right)} = \sqrt{4} = 2.$

Ohms 1, book 0.

4. That's what I figured.. I should really have more confidence in my brain. I'm one of those people that has the bad habit of double checking simple arithmetic (like 13 + 9) on a calculator.. this is coming from a 2nd year mechanical engineering student (no joke)

thanks

5. Originally Posted by ohms
That's what I figured.. I should really have more confidence in my brain. I'm one of those people that has the bad habit of double checking simple arithmetic (like 13 + 9) on a calculator.. this is coming from a 2nd year mechanical engineering student (no joke)

thanks
what is so bad about that habit? beats someone who don't know where he went wrong when the mistake is so obvious. it pays to be careful maybe u should keep that habit