Specialist's Specialist Thread

[Hancock]

What is the implied domain of the function y = arccos(sin(3x))?
I thought the answer was (-∞, ∞), but other people keep saying it's [-pi/6, pi/6].
Could someone please explain?

arccos' domain is restricted from [-1,1]. If that was an x in there instead of sin(3x) we would say that. However, we must now solve the inequality of -1 =< sin(3x) =< 1, because the x from arccos(x) has been replaced with sin(3x).

So:
-1 =< sin(3x) =< 1
arcsin(-1) =< 3x =< arcsin(1)     (arcsin function output limited from -pi/2 to pi/2)
-pi/2 =< 3x =< pi/2
-pi/6 =< x =< pi/6

So the domain of arccos(sin(3x)) is [-pi/6, pi/6].

[Special At Specialist]

What is the implied domain of the function y = arccos(sin(3x))?
I thought the answer was (-∞, ∞), but other people keep saying it's [-pi/6, pi/6].
Could someone please explain?

arccos' domain is restricted from [-1,1]. If that was an x in there instead of sin(3x) we would say that. However, we must now solve the inequality of -1 =< sin(3x) =< 1, because the x from arccos(x) has been replaced with sin(3x).

So:
-1 =< sin(3x) =< 1

I got up to this part, and then I deduced that -∞ < x < ∞. Is there anything wrong with that logic?

Also, is this correct:
y = arccos(sin(3x))
since sin(u) = cos(pi/2 - u) then sin(3x) = cos(pi/2 - 3x)
y = arccos(cos(pi/2 - 3x))
y = pi/2 - 3x

[Hancock]

The problem with saying it's neg infinity to positive infinity is because of the restriction on the sin(x) in order to be able to make it have an inverse (hence why arcsin(x) has an restricted range/output of -pi/2 to pi/2 as a function itself). So arcsin(1) is only ever pi/2, because of that specific restriction .

And as for the function F = pi/2 - 3x, it is equal to arccos(sin(3x)) only in it's defined domain of -pi/6 to pi/6. Good job on that one btw

[Lasercookie]

I'm sorry to be stubborn, but I still don't have my head completely around this.

I can see for something like we have to restrict the domain since the inequality will not be true for all x. If you tried to substitute in , then you'd end up with which we know is outside the restricted domain of arccos(x).

However with this example of , and the inequality we can let x be any real number, and we'll have a number that satisfies the inequality.

Let's say we have

So and therefore which we know to be inside the domain of arccos(x).

[Hancock]

I'm sorry to be stubborn, but I still don't have my head completely around this.

I can see for something like we have to restrict the domain since the inequality will not be true for all x. If you tried to substitute in , then you'd end up with which we know is outside the restricted domain of arccos(x).

However with this example of , and the inequality we can let x be any real number, and we'll have a number that satisfies the inequality.

Let's say we have

I see what you mean, however the I believe that the inequality of is just a stepping stone to the "actual" inequality of , from which the restriction of the arcsin function limits the domain of the arccos(sin(3x)).

[brightsky]

i'm pretty sure the domain is R. you can sub in any x and the function f(x)= arccos(sin(3x)) will spit out a value.

@Hancock - how would you solve sin(3x) = 1? 3x = arcsin(3x) is only one solution. the inequality -1 =< sin(...) =< 1 always holds provided sin(...) has a large enough domain, which in this case it has.

[Jenny_2108]

i'm pretty sure the domain is R. you can sub in any x and the function f(x)= arccos(sin(3x)) will spit out a value.

@Hancock - how would you solve sin(3x) = 1? 3x = arcsin(3x) is only one solution. the inequality -1 =< sin(...) =< 1 always holds provided sin(...) has a large enough domain, which in this case it has.

I think if we solve sin(3x)=1 => 3x+2npi=arcsin(1) (n is integers)
                                                  3x= pi/2-2npi
                                                    x=pi/6-2npi/3
Thus the implied domain is R.
If we put in CAS the function arccos(sin3x), the function is defined with all values of x

-1 =< sin(3x) =< 1
arcsin(-1) =< 3x =< arcsin(1)     (arcsin function output limited from -pi/2 to pi/2)
-pi/2 =< 3x =< pi/2
-pi/6 =< x =< pi/6

So the domain of arccos(sin(3x)) is [-pi/6, pi/6].

If you say the range of arcsin function is -pi/2 to pi/2 so
arcsin(-1)=< 3x+2npi=<arcsin(1)
-pi/2-2npi=< 3x=<pi/2-2npi
-pi/6-2npi/3=<x=<pi/6-2npi/3

Therefore, the implied domain is R

[Hancock]

I'm actually not too sure what to think now. Granted that -1 =< sin(3x) =< 1 due to the obvious state of the trig function. Maybe you're right.

[Special At Specialist]

Let's start with a much simpler problem of the same form:

What is the domain of y = arcsin(sin(x)) ?

If I assume that y = x, then I end up with a straight line, rather than a zig-zag line. The line y = x is the same as y = arcsin(sin(x)) only in the domain [-pi/2, pi/2]. Everything outside that domain zig-zags up and down, whereas the y = x just continues in a straight line.

[Lasercookie]

I'm pretty much convinced now that the maximal domain would be R.

From what I see, it seems all you're doing there is solving for there, you're not looking at solely.

The implied domain is "the set of all real numbers for which the expression is defined.", and we can see that the function is defined outside of [-pi/2,pi/2]

If you're solving the inequality don't forget to account for other periods with the general solution (like Jenny did in her post earlier).

With the not ignoring the other periods thing, it's also comes up with what you did with , it's also equivalent every now and then to ?

Edit: You can look at an alternate problem, of a different form where the maximal domain would not be R, e.g. , or

[Jenny_2108]

I think Laseredd is right.
Considering the trig function, the maximal domain of arcsin(sinx) is still R because its defined with all values of x
But arcsin(sinx) is only equal x when x E [-pi/2,pi/2]
Outside this domain, arcsin(sinx) is still defined but not equal to x.

Does it make sense, Special at Specialist?

[Special At Specialist]

Yes, all of that makes sense, thank you. I thought that the answer was something like that. Now I have a different problem:
Are these steps mathematically valid (ie. would they score full marks if written on a VCAA exam):
dy/dx = 2x
∫dy = ∫2x dx
y = x^2 + C

As opposed to:
dy/dx = 2x
y = ∫2x dx
y = x^2 + C

Basically, is it correct to split it up into 2 different integrals, or do I have to go from dy/dx = ... to y = ... dx?

[#1procrastinator]

^ The first one doesn't make sense to me (but I'm not comfortable with differentials). I think the second one's okay.

But a question like that would be one mark, so do you really have to show the steps?

[Special At Specialist]

Well that was just an example. The real question was more complicated than that and it was worth 3 marks. I know that VCAA are quite pedantic markers so I'd like to make sure that I get every line of working out correct to ensure that I get 100% on the exam (since most of the marks I lose are from petty things like this or not crossing out false solutions).

[BubbleWrapMan]

It's an indefinite integral I do this:









If it's a definite integral I do this:

with when