# Whether root should be positive or negative

I'm trying to find ##\sin (\arccos x)##. I let ##\theta = \arccos x## and then use ##\sin ^2 \theta + \cos ^2 \theta = 1##, I get ##\sin (\arccos x) = \pm \sqrt{1 - x^2}##. I'm not sure whether to take the positive or negative root. On Wolfram Alpha is shows that the result is the positive root, but I'm not sure why...

mfb
Mentor
##0 \leq \arccos(x) \leq \pi##, which means the sine of it is always positive.

Mr Davis 97
For x>0 you have arccos(x)<pi/2, which means the sine is positive. For x<0 you have arccos(x)>pi/2, which means the sine is negative.
Pick the sign depending on the sign of x.
What if I want a general expression and don't know the sign of x? For example, I am trying to find the antiderivative of ##\arcsin x##, and I found that ##\int \arcsin x dx = x \arcsin x + \cos (\arcsin x) + c##. So can I not simplify this further, since I don't know the sign of ##x## beforehand?

mfb
Mentor
Forget that old post, I thought about cos instead of sin. I fixed it.

If you need a range where the sign changes then treat the cases separately. Or see where the expression comes from, sometimes the inverse functions are not even what you actually want.

Forget that old post, I thought about cos instead of sin. I fixed it.

If you need a range where the sign changes then treat the cases separately. Or see where the expression comes from, sometimes the inverse functions are not even what you actually want.
I'm still confused... On various tables of integrals I see that ##\int \arcsin x dx = x \arcsin x + \sqrt{1-x^2}+ c##, which means that ##\cos (\arcsin x) = \sqrt{1-x^2}##. I don't see why they're choosing the positive root over the negative root...

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