1. ## Square root promblem

Err, my algera is bad. Im looking over a line integral promblem, and i understand everything, even the calculus, but im stuck on the algebra. here is the promblem.

$\displaystyle \sqrt{((-4sint^2)+(4cost^2))}dt = 4dt$

I understand that $\displaystyle sint^2\ and\ cost^2 = 1$ but thought the answer should be $\displaystyle sqrt(32)$. Any step by step explanations would be appreciated, thanks.

Hello purplerain
Originally Posted by purplerain
Err, my algera is bad. Im looking over a line integral promblem, and i understand everything, even the calculus, but im stuck on the algebra. here is the promblem.

$\displaystyle \sqrt{((-4sint^2)+(4cost^2))}dt = 4dt$

I understand that $\displaystyle sint^2\ and\ cost^2 = 1$ but thought the answer should be $\displaystyle sqrt(32)$. Any step by step explanations would be appreciated, thanks.
You will have to set the question out more carefully if you want help here. I assume that this is an integration question (in which case it should be in the Calculus Form), but your use of notation is very unclear.

I can't see how you get $\displaystyle 4dt$. It's $\displaystyle \sin^2t + \cos^2t = 1$, not $\displaystyle -\sin^2t+\cos^2t$. And in any event, when you take the square root, you'll get $\displaystyle 2$, not $\displaystyle 4$.

The only way you can get $\displaystyle 4dt$ is if the expression is $\displaystyle \sqrt{(-4\sin t)^2+(4\cos t)^2}\;dt$. Is that what you mean?

The answer you quote, $\displaystyle \sqrt{32}$, suggests it may be a definite integral. Can we see the original question, please?

Hello purplerainYou will have to set the question out more carefully if you want help here. I assume that this is an integration question (in which case it should be in the Calculus Form), but your use of notation is very unclear.

I can't see how you get $\displaystyle 4dt$. It's $\displaystyle \sin^2t + \cos^2t = 1$, not $\displaystyle -\sin^2t+\cos^2t$. And in any event, when you take the square root, you'll get $\displaystyle 2$, not $\displaystyle 4$.

The only way you can get $\displaystyle 4dt$ is if the expression is $\displaystyle \sqrt{(-4\sin t)^2+(4\cos t)^2}\;dt$. Is that what you mean?

The answer you quote, $\displaystyle \sqrt{32}$, suggests it may be a definite integral. Can we see the original question, please?

Hi, yes, i it was an integration question, but as you know, a lot of calculus is algebra. I just had a question about the algebra, but i will post the full question in the appropriate forum.

4. Originally Posted by purplerain
Err, my algera is bad. Im looking over a line integral promblem, and i understand everything, even the calculus, but im stuck on the algebra. here is the promblem.

$\displaystyle \sqrt{((-4sint^2)+(4cost^2))}dt = 4dt$

I understand that $\displaystyle sint^2\ and\ cost^2 = 1$ but thought the answer should be $\displaystyle sqrt(32)$. Any step by step explanations would be appreciated, thanks.
Be aware that you may be expected to simplify your answer as$\displaystyle \sqrt{32}$ is not fully simplified $\displaystyle \sqrt{32} = 4\sqrt{2}$

Because of your identity I will assume you mean $\displaystyle sin^2(t)$ not $\displaystyle sin(t^2)$

$\displaystyle 4cos^2t-4sin^2t = 4(cos^2t-sin^2t)$

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

$\displaystyle cos(2t)$

$\displaystyle \sqrt{4} \sqrt{cos(2t)} = 2\sqrt{cos(2t)}$

5. Hello everyone
Originally Posted by e^(i*pi)
...

$\displaystyle cos^2t-sin^2t = cos^2(2t)$...
In case you need to use this expression, note that $\displaystyle \cos^2t - \sin^2t = \cos2t$, not $\displaystyle \cos^22t$.

Hello everyoneIn case you need to use this expression, note that $\displaystyle \cos^2t - \sin^2t = \cos2t$, not $\displaystyle \cos^22t$.