Is arcsin(sin) = 1? | Understanding the Relationship between Arcsin and Sin

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The discussion clarifies that arcsin(sin(y)) does not equal y due to the periodic nature of the sine function, which has infinitely many solutions. Arcsin is defined to return a principal value within the range of -π/2 to π/2. When applying arcsin to both sides of an equation like sin(2x) = √3/2, it leads to arcsin(sin(2x)) = arcsin(√3/2), but the result is constrained to the principal value. Therefore, understanding the limitations of arcsin is crucial for solving trigonometric equations correctly. The conversation emphasizes the importance of recognizing the specific output range of the arcsin function.
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Homework Statement



arcsin(sin) = 1 right?



Homework Equations





The Attempt at a Solution



Basically, I see arcsin as 1/sin

is this correct?
 
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1/sin(x) = csc(x)

arcsin is the function such that sin(arcsin(x)) = x
 
Oh I think I get it.

So, I can take the arcsine of both sides in a problem such as:

sin(2x) = (Root3 )/2

and I would get arcsin(sin(2x)) = arcsin ((root3)/2)

Which would get me to 2x = arcsin ((root3)/2)?

Correct?
 
I'm said:
Oh I think I get it.

So, I can take the arcsine of both sides in a problem such as:

sin(2x) = (Root3 )/2

and I would get arcsin(sin(2x)) = arcsin ((root3)/2)

Which would get me to 2x = arcsin ((root3)/2)?

Correct?
Yes or no, depending on what you literally mean.

The big overwhelming obstacle that you need to make sure you understand is that the equation
sin(y)=x​
has infinitely many solutions. (or zero solutions, if |x| > 1)

If I'm to define a function Arcsin(x) that gives a solution to sin(y)=x, I can only pick one of them. (The solution lying in -\pi/2 \leq y \leq \pi/2 is traditional)

So if I want all solutions to sin(y)=x, I have more work to do because Arcsin(x) gives me one of them. Fortunately, knowing one solution, it's easy to find all of the others. (If it's not obvious, study the graph of sin(y)=x for a while...)

In otherwords, Arcsin(sin(y)) is not y. It is "the number in [-\pi/2 , \pi/2] that is related to y".
 
Hurkyl said:
Yes or no, depending on what you literally mean.

The big overwhelming obstacle that you need to make sure you understand is that the equation
sin(y)=x​
has infinitely many solutions. (or zero solutions, if |x| > 1)

If I'm to define a function Arcsin(x) that gives a solution to sin(y)=x, I can only pick one of them. (The solution lying in -\pi/2 \leq y \leq \pi/2 is traditional)

So if I want all solutions to sin(y)=x, I have more work to do because Arcsin(x) gives me one of them. Fortunately, knowing one solution, it's easy to find all of the others. (If it's not obvious, study the graph of sin(y)=x for a while...)

In otherwords, Arcsin(sin(y)) is not y. It is "the number in [-\pi/2 , \pi/2] that is related to y".

I'm said:
Oh I think I get it.

So, I can take the arcsine of both sides in a problem such as:

sin(2x) = (Root3 )/2

and I would get arcsin(sin(2x)) = arcsin ((root3)/2)

Which would get me to 2x = arcsin ((root3)/2)?

Correct?

so in this case would it be arcsin(sin(60)) = Arcsin ((\sqrt{3}/2
?

Can you give me a problem that displays what you have just told me? I'd really like to see one ( as I have not been told that in my Precalculus class).

Thanks.
 
I'm said:
so in this case would it be arcsin(sin(60)) = Arcsin ((\sqrt{3}/2
?

Can you give me a problem that displays what you have just told me? I'd really like to see one ( as I have not been told that in my Precalculus class).

Thanks.

\sin{0}=\sin{\pi}=0 but \pi \not= 0. A function can only map one output to a given input, so we have to specify which solution we want when we say Arcsin(0). The solutions which are typically used are the ones between -\pi/2 and \pi/2
 
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