Math problem

arccos x + arcsin x = ? 

Find the answer . First that give the solution and in end the right answer , i will give a nice chess.com trophy .

Pi

Assuming x is real, then

arccos(x) + arcsin(x) = pi/2, -1 <= x <= 1.

Since arcsin is defined only on this interval, and is one-to-one here, we can choose alpha = arcsin(x).  Then

arccos(x) + arcsin(x) = arccos(sin(alpha)) + arcsin(sin(alpha)) = arccos(cos(pi/2 - alpha)) + alpha = pi/2.

I'm too lazy to check whether this works for complex numbers .

oh oh i no!! its 0!!! lol dont worry my answer is completely wrong so dont listen to me hahaha

cos^(-1) 0 = pi/2

sin^(-1)0 = 0

it's pi/2, not pi. Sorry.

balifid is the winner and i am surprised with the elegant solution .Well done !

I have another solution :

d arcsin(x) /dx = 1/sqrt 1-x2   and d arccos(x)/dx = - 1/sqrt 1-x2  so

 d arcsin(x) / dx + d arccos(x)=1/sqrt 1-x2 + ( - 1/sqrt 1-x2 ) =0 so

d ( arcsin(x)+arccos(x) ) = 0 so

arcsin(x)+arccos(x) = constant so

Now the graph of this function is a straight line parallel to ox . Both functions goes to [-1 ;1] . We need to find the other side of the rectangle and we can find through this integral :

integral from -1 to +1 of arcsin(x)+arccos(x) dx is = pi / 2

You can chek alone as homework  .

I just realized that my solution hand-waved that arcsin(sin(x)) = x and arccos(cos(x)) = x.  Actually f(x) = arcsin(sin(x)) is a triangle wave with period 2pi, same goes for cos/arccos.  They just happen to add up properly.

Your solution would be watertight except that the function is not differentiable for |x| = 1, so in addition to testing points inside and outside that circle (in the complex plane), we would have to test the circle itself, and I can't think of a good way to do that.

Yeah my solution is more complex and more inaccurate and modest against your solution is more simple and more easy , accurate and more correct. As you say we have to check for point that are differentiable and some other exceptions.

What I'm saying is that neither proof seems complete, if we wanted to show that this holds for complex numbers.  arcsin and arccos are defined on all complex numbers, and in fact return complex numbers for most real choices of x, such as arccos(2) = i log(2 + sqrt(3)).  So I think it would be appropriate to consider them.

That being said, my proof is only complete if we were to find an algebraic expression for arcsin(sin(x)) and arccos(cos(x)).  Unfortunately I think that would be the bulk of the work right there, because by the time we found those expressions, the result would be obvious.

I think your proof is more promising.  In fact once we know that (d/dx)(arccos(x) + arcsin(x)) = 0 for all x, |x| ≠ 1, then we only need to test three regions: inside, outside, and on |x| = 1.  For the inner and outer regions we only need to test one point each, but the difficulty is testing points on the circle.  The only way I can think to do that is proof by Wikipedia .  The series forms given here apply to points on the boundary, and they clearly add up to pi/2.

Of course, somebody else has already done the task of deriving those crazy things.  If we're willing to use them in our proof, we might as well just use the very first identity on the page .