Square root promblem

**purplerain** · November 21st 2009, 06:53 AM

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.

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

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

**Grandad** · November 21st 2009, 07:08 AM

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.

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

I understand that $sint^2\ and\ cost^2 = 1$ but thought the answer should be $sqrt(32)$ . 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 $4dt$ . It's $\sin^2t + \cos^2t = 1$ , not $-\sin^2t+\cos^2t$ . And in any event, when you take the square root, you'll get $2$ , not $4$ .

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

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

Grandad

**purplerain** · November 21st 2009, 07:17 AM

Originally Posted by Grandad

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

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

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

Grandad

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.

**e^(i*pi)** · November 21st 2009, 07:46 AM

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.

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

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

Be aware that you may be expected to simplify your answer as $\sqrt{32}$ is not fully simplified $\sqrt{32} = 4\sqrt{2}$

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

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

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

$cos(2t)$

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

**Grandad** · November 21st 2009, 08:04 AM

Hello everyone

Originally Posted by e^(i*pi)

...

$cos^2t-sin^2t = cos^2(2t)$ ...

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

Grandad

**e^(i*pi)** · November 21st 2009, 08:12 AM

Originally Posted by Grandad

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

Grandad

How could I not notice that

*edited above*

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