arccos(x)=-arcsin(x)

A

Ahmed-

New member
Joined
Feb 28, 2022
Messages
4
Hi,
I was doing int of
777.PNG


if i put x=a.cos(u)
the result i'll be:
arcos(x/a)

now if i put x=a.sin(u)
the result i'll be:
=-arcsin(x/a)

so how can be
arcos(x/a)=-arcsin(x/a)
 
Ahmed- said:
Hi,
I was doing int of
777.PNG


if i put x=a.cos(u)
the result i'll be:
arcos(x/a)

now if i put x=a.sin(u)
the result i'll be:
=-arcsin(x/a)

so how can be
arcos(x/a)=-arcsin(x/a)
Click to expand...
You know:

sin(Θ) = cos (π/2 - Θ) and so on.....
 
Ahmed- said:
can you explaine by more details, ty.
Click to expand...
It will help a lot if you can show us the problem. You apparently tried to attach it, but it didn't work.

You are apparently trying to solve [imath]\arccos(\frac{x}{a})=-\arcsin(\frac{x}{a})[/imath] for x, given a fixed value of a.

Suppose that [imath]\arccos(\frac{x}{a})= \theta[/imath] and [imath]-\arcsin(\frac{x}{a}) = \theta[/imath]. Rewrite each equation in terms of trig functions rather than inverse trig functions, and see what you can do with that.

On the other hand, I suspect you are not starting with the inverse functions, but with the trig functions. In that case, don't use inverse trig functions at all! If you want to solve [imath]a\cos(u)=-a\sin(u)[/imath], then just divide both sides by [imath]a\cos(u)[/imath].

Anything you can show of your own thinking will help us help you.
 
Ahmed- said:
I just wanna to prove that arcos(x/a)=-arcsin(x/a).
Click to expand...
That's not thinking; that's wanting. Please show us your thinking, as I asked.

It's also impossible; those functions are not always equal. In fact, they are never equal.

Here is the graph that BBB suggested you make, since you don't appear very willing to try things:

1646078913277.png​

Are you wrongly thinking that they are equal, because of something in the problem you are refusing to show us?

I'm starting to realize that you have done an integral in two different ways, and got those answers. If so, think about the constant of integration, which you are ignoring.
 
Dr.Peterson said:
I'm starting to realize that you have done an integral in two different ways, and got those answers. If so, think about the constant of integration, which you are ignoring.
Click to expand...
Yes, I got confused when I did the integral in two different ways, i expected the same results, but as you can see.
1. I don't think i made mistakes while doing the integrales. if so please correct me.
2. wich of the result i can take it as answer ?
3. constant of integration C1&C2 should be equial to Pi/2 ,To be right ? "arccos(x)+arcsin(x)= Pi/2"
 
Ahmed- said:
Yes, I got confused when I did the integral in two different ways, i expected the same results, but as you can see.
1. I don't think i made mistakes while doing the integrales. if so please correct me.
2. wich of the result i can take it as answer ?
3. constant of integration C1&C2 should be equial to Pi/2 ,To be right ? "arccos(x)+arcsin(x)= Pi/2"
Click to expand...
Please, please, please, show us the actual problem!!!

1. You made no mistake in integrating, except for a possible sign error if you were finding the indefinite integral of 1/sqrt(a^2-x^2).

2. Neither result is the correct answer, if you don't include an arbitrary constant of integration, "+ C".

3. If you are find an indefinite integral, then you can't set a specific value for C; and they would never be equal. And you are wrong that "arccos(x)+arcsin(x)= Pi/2".

Again, we can't really answer you without seeing the problem. Are you not aware that you didn't attach an image in the original post?
 
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