Where to take square root "positive, negative" and where to take it only "positive"??

• Nov 3rd 2010, 06:39 AM
khamaar
Where to take square root "positive, negative" and where to take it only "positive"??
I wrote in my Notebook a few days ago..

cos(x)= [1-sin^(2)x]^(1/2) {derived from "cos^(2)x + sin^(2)x = 1 ")

But the teacher said i was wrong cuz i shoud have written it like this....

cos(x)= +[1-sin^(2)x]^(1/2) , cos(x)= -[1-sin^(2)x]^(1/2)

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Today we came across an equation

y= cos(x)
and the teacher wrote..

y= [cos^(2)x]^1/2

and then he wrote...

y= [1-sin^(2)x]^(1/2) (since cos^(2)x= 1 - sin^(2)x)

now im really confused....i dont know where to take square root positive and where to take it both positive and negative at the same time.

• Nov 3rd 2010, 08:30 AM
Ackbeet
If you start from $\displaystyle \cos^{2}(x)+\sin^{2}(x)=1,$ then I agree with your teacher that $\displaystyle \cos(x)=\pm\sqrt{1-\sin^{2}(x)}.$ However, if you start from $\displaystyle y=\cos(x),$ then you can't say that $\displaystyle y=\sqrt{\cos^{2}(x)}$ unless, in addition, you know that $\displaystyle y$ is always positive.

The rules come down to context. From where do you start? Is there some physical principle involved that allows you to rule out one sign possibility? In general, if you have the equation $\displaystyle a^{2}=b^{2},$ then without any other context, you must always allow the solutions $\displaystyle a=\pm b.$ However, if you know ahead of time, say, that $\displaystyle a>0$ and $\displaystyle b>0,$ then you can rule out the negative solution and say that $\displaystyle a=b.$

Hope that helps.
• Nov 3rd 2010, 08:59 AM
HallsofIvy
Strictly speaking, a square root is always positive. $\displaystyle \sqrt{4}= 2$ only but the equation $\displaystyle x^2= 4$ has two solutions, x= 2 and x= -2. Solving an equation is different from "taking the square root".

In general, the solution to $\displaystyle x^2= a$ is $\displaystyle x= \pm\sqrt{a}$. We need the "$\displaystyle \pm$" there because just "$\displaystyle x= \sqrt{a}$" is not enough- it gives only the positive solution to the equation.

Once again, solving the equation $\displaystyle x^2= a$ is different from just taking the square root, $\displaystyle \sqrt{a}$. The square root, like any real valued function, has only one value but an equation may have many solutions.
• Nov 3rd 2010, 10:52 AM
khamaar
I still dont get it... is , cos(x)= +[1-sin^(2)x]^(1/2) ?? or is it also equal to cos(x)= -[1-sin^(2)x]^(1/2)

I was reading an article on wikipedia and they have also say that... cos(x) is only equal to "+[1-sin^(2)x]^(1/2)".....

They have used it in getting the differential of inverse trignometric function.

Differentiation of trigonometric functions - Wikipedia, the free encyclopedia

I am still confused in where i should take it "only positive" and where i should take it "positive and negative at the same time"
• Nov 3rd 2010, 11:12 AM
Ackbeet
The equation $\displaystyle \cos(x)=\sqrt{1-\sin^{2}(x)}$ does not hold if

$\displaystyle \dfrac{\pi}{2}<x<\dfrac{3\pi}{2},$

because the LHS is negative, whereas the RHS is always positive.

In the link to which you referred, the wiki was restricting the domain of the cos function to $\displaystyle -\pi/2\le y\le\pi/2.$ In this region, it is true that $\displaystyle \cos(y)=\sqrt{1-\sin^{2}(y)}$. So they are justified in their substitution.

Like I said, you have to know by the context. That's usually stated, but if it's not, you can either leave both solutions in to be safe, or you can try to clarify the question. Make sense?