# arcsin(x)+arccos(x)

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#1
I was recently helping a student with the following problem:

a) y=arcsin(x), -1<x<1, -pi/2<y<pi/2

write arcos(x) in terms of y

b) find the value of arccos(x)+arcsin(x)

Now here was my train of thought.
a)
y=arcsin(x)
sin(y)=x
cos(y-(pi/2))=x (using knowledge of the trig grapghs)
y-(pi/2)=arccos(x)

b)
arcos(x)+arcsin(x)
y-(pi/2)+y
2y-(pi/2)

Now since checking the solution and though searching this problem I see that:
we can use sin(y)=cos((pi/2)-y)

which leads to the solution arcsin(x)+arccos(x) =pi/2

But my question is what is the problem with doing it the first way? Does the maths still hold out? How are we expected to know to do it one way and not the other?
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#2
I seem to have confused myseld even further..

Originally I defined y=arcsin(x)

Therefore if
arccos(x)+arcsin(x)=2y-pi/2
Then
arccos(x)+arcsin(x)=2arcsin(x)-pi/2
And so we end up with
arcsin(x)-arccos(x)=90,

which is incorrect, isn't it?

Can anybody help me out here and point out my mistake(s)?
0
1 year ago
#3
It’s because of how, by convention, the functions arcsin and arccos are defined.

arcsin: [-1, 1] —> [-pi/2, pi/2],
arccos: [-1, 1] —> [0, pi].

For a function to have an inverse it must have a one-to-one correspondence between elements of the domain and the “range” (strictly speaking, the codomain). So we need to restrict the codomains, and by convention we do this to [-pi/2, pi/2].

So when you do your transformation you need to be very careful that you’re transforming values between the right sets.

We have

-pi/2 <= y = arcsin(x) <= pi/2

and we need

0 <= arccos(x) <= pi

So you can’t use y - pi/2 = arccos(x).

(If the graphs extended to infinity then you could, but then we wouldn’t have “functions” in the traditional sense.)

pi/2 - y = arccos(x) is the correct transformation.

Pretty devilish for an alevel exam question if you ask me.

The take-home message is “use the identities
sin(x) == cos(pi/2 - x)
and
cos(x) == sin(pi/2 - x)

Edited because I got it embarrassingly wrong the first time around.
Last edited by Tommy59375; 1 year ago
2
1 year ago
#4
You have to consider the principle value of arcsin and arccos. If you subtract pi/2 from x, then it falls outside the accepted range. Try drawing the arcsin and arccos graphs between-pi/2 and pi/2. Then apply the transformations, you'll see that doing it your way means some values are mapped outside of the domain -pi/2 to pi/2.
1
#5

So let me see if I've got this straight...

I wasn't thinking of arcsin(x) and arccos(x) as functions? (Further reading would suggest I was considering Arcsin(x) and Arccos(x) which are different)

But anyway, what I was doing would mean that if for example y=0, then I would be considering arccos(x)=-pi/2, which is outside the range of arccos(x). And thus I was not considering arccos(x) the function and that was my error?
0
1 year ago
#6
(Original post by jc768)

So let me see if I've got this straight...

I wasn't thinking of arcsin(x) and arccos(x) as functions? (Further reading would suggest I was considering Arcsin(x) and Arccos(x) which are different)

But anyway, what I was doing would mean that if for example y=0, then I would be considering arccos(x)=-pi/2, which is outside the range of arccos(x). And thus I was not considering arccos(x) the function and that was my error?
essentially, yes.

There is no difference between “arcsin” and “Arcsin”. Similarly for arccos.

I’ll first clarify a condition which needs to be true for an inverse function to exist.

————

The domain is the set of values the function maps from (essentially the set of input values), and the codomain the set of values it maps to (the output values).

A “function”, strictly speaking, must be able to take any value in its domain and map it onto a single value in its codomain. Call this condition “*”, as I’ll refer to it later.

Suppose f is a function with domain A and codomain B.

(img1)

(In the first case, think how the function f(x)=x^2 maps has both f(2)=4 and f(-2)=4.)

Suppose f is a function with domain A and codomain B (as a above).

Then (assuming f^-1 exists), f^-1 will have domain B and codomain A. Think “reversing the arrows” in the picture above.

For f^-1 to exist, each value in B needs to be mapped onto a unique value in A. So the diagram would need to look something like the following:

(img2)

Both f and f^-1 need to be functions so they both need to satisfy “*”.

One can “improve” a function so that it satisfies “*” by restricting the domain and essentially excluding problematic values.

————

Bringing this all back to the question at hand...

arcsin and arccos are the inverse functions of the functions sin and cos respectively.

But if we take sin and cos as one would normally think of them...

(img3)

then they don’t — strictly speaking — have inverses. For example, you’d have arcsin(0) = 0 and arcsin(0) = pi and arcsin(0) = 2pi and ...

And that doesn’t make any sense because 0 =/= pi =/= 2pi =/= ... . (It wouldn’t satisfy “*” above.)

Instead we need to restrict their domains, and the usual way to do this is:

(img4)

So, for sin, we restrict the domain to
-pi/2 <= x <= pi/2,
and for cos, we restrict it to
0 <= x <= pi.

Note. There’s no particular reason why we choose these ranges of values for x, it’s just done this way by convention.

Therefore we have
-pi/2 <= arcsin(sin(x)) <= pi/2
and
0 <= arccos(cos(x)) <= pi.

So if we use “sin(x)=cos(x - pi/2)” then sin(x) won’t be the “improved” function we’ve created. Instead it would be:

(img5)

(classic graph transformations)

Using “sin(x) = cos(pi/2 - x)” works a treat:

(img6)

————

Footnote 1. The conventional use of “sin” and “cos” don’t use these restricted domains. Typing sin(3pi/2) into a calculator gives -1, surprising nobody. But whatever value of x (between -1 and 1) you type in, arcsin(x) will only ever give values between -pi/2 and pi/2.

Footnote 2. There’s technically a difference between the codomain and the range of a function. But here, you may assume that they have equivalent definitions.
Last edited by Tommy59375; 1 year ago
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