Why is this wrong?

  • Thread starter heshbon
  • Start date
  • #1
27
0

Main Question or Discussion Point

Why is this wrong???

arcos(x)= ∫(-1)/√(1-x) dx = -∫1/√(1-x) dx = -arcsin(x)

⇒ arcos(x)=-arcsin(x)

i know this is wrong, i think its got something to do with constant of integration, could someone pls point out my mistake. thnks
 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
It's wrong because you forgot the +C at the end of both integrals.

[tex]\int\frac{-1}{\sqrt{1-x^2}}dx=\arccos(x)+C_1[/tex]

[tex]-\int\frac{1}{\sqrt{1-x^2}}dx=-\arcsin(x)+C_2[/tex]

Therefore,

[tex]\arccos(x)+C_1=-\arcsin(x)+C_2[/tex]

[tex]\arcsin(x)+\arccos(x)=C_2-C_1[/tex]

Since [itex]C_1[/itex] doesn't necessarily equal [itex]C_2[/itex], the two inverse trig functions don't necessarily add up to 0.
 
  • #3
27
0
but cant c1, c2 be what ever i choose them to be, say 0?
 
  • #4
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
No. They are arbitrary constants.
 
  • #5
27
0
oh, and also, why is arcsin(x) + arcos(x) = pi/2
 
  • #6
rock.freak667
Homework Helper
6,230
31
oh, and also, why is arcsin(x) + arcos(x) = pi/2
Let A=arcsin(x) and B=arccos(x)

so that sinA=x and cosB=x

and the equation is A+B, then consider what sin(A+B) is and check when A and B are both acute.
 
  • #7
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
Here's one way to show that. Let [itex]x\in[-1,1][/itex]

Then:

[tex]x=x[/tex]

[tex]x=\cos\left(\frac{\pi}{2}\right)cos(x)+\sin\left(\frac{\pi}{2}\right)\sin\sin^{-1}(x)[/tex]

[tex]x=\cos\left(\frac{pi}{2}-sin^{-1}(x)\right)[/tex]

(Here I used the trig identity cos(x-y)=cos(x)cos(y)+sin(x)sin(y))

Take the inverse cos of both sides, and conclude.
 
  • #8
Gib Z
Homework Helper
3,346
5
We could have cloncluded that they only differ by a constant from the integrals we just saw: The functions can only differ by a constant. Let x=0, since the function (arcsin x - arccos x) is a constant, it takes that value for all points.
 
  • #9
261
0
We could have cloncluded that they only differ by a constant from the integrals we just saw: The functions can only differ by a constant. Let x=0, since the function (arcsin x - arccos x) is a constant, it takes that value for all points.

I don't know gib z, the constants are arbitrary in order to integrate since the derivative of a constant is zero

if we take sinx - cosx for several values for x, like the most simple 0 and 1, sinx - cosx will not be constant
 
  • #10
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
But it's not sin(x)-cos(x) we're interested in. It's arcsin(x)-arccos(x).
 
  • #11
261
0
But it's not sin(x)-cos(x) we're interested in. It's arcsin(x)-arccos(x).
since arc is the inverse function I tought that arcsin(x) +- arccos(x) = constant for several values of x <==> sin(x) +- cos(x) = variable for several values of x ==> absurdum

so we can use the trig. functions in our analisys, what do you think Tom?
 
Last edited:
  • #12
Gib Z
Homework Helper
3,346
5
So you don't like my proof >.<" ?
 
  • #13
261
0
So you don't like my proof >.<" ?
I dont have nothing against :smile:, I just think that it is wrong :shy:, but I may be wrong, let's wait for tom mattson opinion
 
  • #15
Mute
Homework Helper
1,388
10
since arc is the inverse function I tought that arcsin(x) +- arccos(x) = constant for several values of x <==> sin(x) +- cos(x) = variable for several values of x ==> absurdum
That is incorrect. arcsin(x) +/- arccos(x) = constant in no way implies sin(x) +/- cos(x) is a constant. I'm not sure, but it looks like in your first sentence you interpret "arc" to be some sort of operator such that

arc(cos(x)) := arccos(x)
arc(sin(x)) := arcsin(x)

and then you conclude that arc(cos(x) +/- sin(x)) = arccos(x) +\- arcsin(x), which isn't the case at all. arcsine and arccosine are simply functions with the property such that

arccos(cos(x)) = x
cos(arccos(x)) = x

and similarly for arcsine. (Note that the above first expression is really only 100% true in the first quandrant of the cartesian plane. There will be additional constants for the other quadrants.)

As GibZ stated, the integral calculations above clearly show arcsin(x) and arccos(x) differ only by a constant. Choose any value of x and evaluate the expression - it will be the same no matter what value of x you choose.
 
  • #16
261
0
I meant: if you are considering that arcsin(x) +- arccos(x) = constant for several values of x and we know that sin(x) +- cos(x) = variable for several values of x, I think both situations imply an absurd

arcsin(1/2) = 30, arccos(1/2) = 60 ==> arcsin(0.5) - arccos(0.5) = -30

arcsin(sqrt(2)/2) = 45, arccos(sqrt(2)/2) = 45 ==> arcsin(sqrt(2)/2) - arccos(sqrt(2)/2) = 0

so the constant C is not allways the same if you input the same value of x in both functions for all cases of x

why this conclusion is wrong? I think I gave a concrete counter-proof, or not?
 
Last edited:
  • #17
261
0
That is incorrect. arcsin(x) +/- arccos(x) = constant in no way implies sin(x) +/- cos(x) is a constant.

this is exactly the conclusion I was trying to show, but my english is not so good :smile:
 
  • #18
Mute
Homework Helper
1,388
10
I meant: if you are considering that arcsin(x) +- arccos(x) = constant for several values of x and we know that sin(x) +- cos(x) = variable for several values of x, I think both situations imply an absurd

arcsin(1/2) = 30, arccos(1/2) = 60 ==> arcsin(0.5) - arccos(0.5) = -30

arcsin(sqrt(2)/2) = 45, arccos(sqrt(2)/2) = 45 ==> arcsin(sqrt(2)/2) - arccos(sqrt(2)/2) = 0
The expression in question is arccos(x) + arcsin(x), which is pi/2 always (within the domain of definition of arcos and arcsin). arccos(x) - arcsin(x) need not be (and, indeed is not) a constant.

Think about it geometrically, if you like:

For a right-angled triangle with hypotenuse C, base A and height B, let x be the angle between the base and the hypotenuse and y be the angle between the vertical piece of the hypotenuse. By geometry, x + y = pi/2 radians (90 degrees).

Now, arcsin(A/C) = x, and arccos(A/C) = y. Hence, arcsin(A/C) + arccos(A/C) = x + y = pi/2.
 
  • #19
Gib Z
Homework Helper
3,346
5
Instead of taking the integral approach, I'll do the differentiation proof.

Let [itex]f(x) = \arcsin x + \arccos x[/itex].

Then,
[tex]\frac{df(x)}{x} = \frac{1}{\sqrt{1-x^2}} + \frac{-1}{\sqrt{1-x^2}} = 0[/tex].

Since the derivative is 0 everywhere, the value of f(x) never changes. Letting x= [itex]pi/2[/itex], [itex]f(x) = \pi /2[/itex]. Q.E.D.
 
  • #20
261
0
The expression in question is arccos(x) + arcsin(x)

Read carefully and you'll see that I was talking about this GIBz statement:

Gib Z said:
We could have cloncluded that they only differ by a constant from the integrals we just saw: The functions can only differ by a constant. Let x=0, since the function (arcsin x - arccos x) is a constant, it takes that value for all points.

This is the expression in question in my messages

But you're right in one point: I've tried to generalize to +-, and this is a mistake.
 
Last edited:
  • #21
Gib Z
Homework Helper
3,346
5
I really still don't see why you think my proof is invalid :( Please go through it step by step and tell me what raises your objection.
 
  • #22
261
0
I really still don't see why you think my proof is invalid :( Please go through it step by step and tell me what raises your objection.
I think we are thinking on two different statements.

You said that "the function (arcsin x - arccos x) is a constant, it takes that value for all points"

My interpretation of you statement: you're saying that (arcsin x - arccos x) = C for all possible values of x, [-1,1]

for instance if x = 1/2, (arcsin x - arccos x) = 30 - 60 = -30
for instance if x = [itex]\sqrt{2}/2[/itex], (arcsin x - arccos x) = 45 - 45 = 0

THAT statement is wrong if you really stated in those terms.
 
  • #23
Gib Z
Homework Helper
3,346
5
DAMN IT!! Typo, i meant arcsin x PLUS arccos x :( As evident from post 19 :P
 
  • #24
261
0
DAMN IT!! Typo, i meant arcsin x PLUS arccos x :( As evident from post 19 :P
hehe dont worry, this things happens all the time :wink:
 
  • #25
Mute
Homework Helper
1,388
10
I think we are thinking on two different statements.

You said that "the function (arcsin x - arccos x) is a constant, it takes that value for all points"

My interpretation of you statement: you're saying that (arcsin x - arccos x) = C for all possible values of x, [-1,1]

for instance if x = 1/2, (arcsin x - arccos x) = 30 - 60 = -30
for instance if x = [itex]\sqrt{2}/2[/itex], (arcsin x - arccos x) = 45 - 45 = 0

THAT statement is wrong if you really stated in those terms.
Okay, now it makes sense what you were trying to do. However, do realize that trying to argue that arccos(x) - arcsin(x) cannot be a constant for all x by referring to the fact that cos(x) - sin(x) is not a constant is not going to help you establish what you sought to, because one expression implies nothing about the other. Since you apparently accepted that arccos(x) + arcsin(x) = pi/2, a better way to prove that arccos(x) - arcsin(x) can't be constant for all x is the following:

We know
[tex]\arccos(x) + \arcsin(x) = \pi/2[/tex]
for all x. Now, suppose

[tex]\arccos(x) - \arcsin(x) = C[/tex]
for all x, where C is a consant. If this expression is true, then we can add the two expressions together to get

[tex]\arccos(x) = \frac{\pi}{4}+\frac{C}{2}[/tex]
[tex]\arcsin(x) = \frac{\pi}{4}-\frac{C}{2}[/tex]

But this implies that x must be a constant, as cos(pi/4+C/2) and sin(pi/4-C/2) are constant, and so our original assumption is false: \arccos(x) - \arcsin(x) is not a constant.
 

Related Threads on Why is this wrong?

Replies
2
Views
540
Replies
2
Views
597
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
7
Views
2K
Replies
6
Views
2K
Replies
5
Views
2K
Replies
3
Views
978
Replies
1
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
17
Views
16K
Top