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Full explanation on the cancellation property: $;sinbig(arcsin(x)big)=x,; forall; xin [-1,1]$
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Please, I need full explanation on the cancellation property applied to the composition of functions below
$$;arcsinbig(sin(x)big)=x,; forall; xin Big[-fracpi2,fracpi2Big] ,;;;sinbig(arcsin(x)big)=x,; forall; xin [-1,1].$$
and why is $$sinbig(arcsin(x)big)neq x,; forall; xin BbbR?$$
I need some explanation on the kind of maps, how the values are calculated and perhaps, some drawings for clarity!
algebra-precalculus functions trigonometry
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
asked Jul 17 at 20:35
Mike
72912
 |Â
show 7 more comments
up vote
0
down vote
favorite
Please, I need full explanation on the cancellation property applied to the composition of functions below
$$;arcsinbig(sin(x)big)=x,; forall; xin Big[-fracpi2,fracpi2Big] ,;;;sinbig(arcsin(x)big)=x,; forall; xin [-1,1].$$
and why is $$sinbig(arcsin(x)big)neq x,; forall; xin BbbR?$$
I need some explanation on the kind of maps, how the values are calculated and perhaps, some drawings for clarity!
algebra-precalculus functions trigonometry
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
asked Jul 17 at 20:35
Mike
72912
1
As for your last question, you can't take the arcsine of any number not within the interval $[-1,1]$.
– RayDansh
Jul 17 at 20:37
$sin(x) in [-1,1]$
– gd1035
Jul 17 at 20:37
2
I discourage the use of the notation $sin^-1$ as that can cause confusion and frustration across borders as it is ambiguous whether the^-1is intended to mean multiplicative inversion or functional inversion. The notation $arcsin$ is superior.
– JMoravitz
Jul 17 at 20:38
2
You'we mixed them up. Replace $[-1,1]$ and $Big[-fracpi2,fracpi2Big]$
– Zeekless
Jul 17 at 20:38
1
As for how the cancellation property works... just remember how $arcsin(x)$ and $sin(x)$ act. $arcsin(x)$ returns the unique angle within the range $[-fracpi2,fracpi2]$ such that the sine of that angle gives $x$. Note that $-1leq sin(x)leq 1$ for all real $x$, so (the real) $arcsin(x)$ is undefined for $x$ outside of $[-1,1]$ (things get weirder when allowing complex inputs and outputs) and note that there are multiple inputs which map to the same output for $sin$ such as how $sin(0)=sin(2pi)$, so $arcsin$ can't distinguish between them.
– JMoravitz
Jul 17 at 20:46
 |Â
show 7 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Please, I need full explanation on the cancellation property applied to the composition of functions below
$$;arcsinbig(sin(x)big)=x,; forall; xin Big[-fracpi2,fracpi2Big] ,;;;sinbig(arcsin(x)big)=x,; forall; xin [-1,1].$$
and why is $$sinbig(arcsin(x)big)neq x,; forall; xin BbbR?$$
I need some explanation on the kind of maps, how the values are calculated and perhaps, some drawings for clarity!
algebra-precalculus functions trigonometry
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
asked Jul 17 at 20:35
Mike
72912
Please, I need full explanation on the cancellation property applied to the composition of functions below
$$;arcsinbig(sin(x)big)=x,; forall; xin Big[-fracpi2,fracpi2Big] ,;;;sinbig(arcsin(x)big)=x,; forall; xin [-1,1].$$
and why is $$sinbig(arcsin(x)big)neq x,; forall; xin BbbR?$$
I need some explanation on the kind of maps, how the values are calculated and perhaps, some drawings for clarity!
algebra-precalculus functions trigonometry
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
asked Jul 17 at 20:35
Mike
72912
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
edited Jul 17 at 22:10
Benedict Voltaire
1,243725
1,243725
asked Jul 17 at 20:35
Mike
72912
asked Jul 17 at 20:35
Mike
72912
asked Jul 17 at 20:35
Mike
72912
72912
1
As for your last question, you can't take the arcsine of any number not within the interval $[-1,1]$.
– RayDansh
Jul 17 at 20:37
$sin(x) in [-1,1]$
– gd1035
Jul 17 at 20:37
2
I discourage the use of the notation $sin^-1$ as that can cause confusion and frustration across borders as it is ambiguous whether the^-1is intended to mean multiplicative inversion or functional inversion. The notation $arcsin$ is superior.
– JMoravitz
Jul 17 at 20:38
2
You'we mixed them up. Replace $[-1,1]$ and $Big[-fracpi2,fracpi2Big]$
– Zeekless
Jul 17 at 20:38
1
As for how the cancellation property works... just remember how $arcsin(x)$ and $sin(x)$ act. $arcsin(x)$ returns the unique angle within the range $[-fracpi2,fracpi2]$ such that the sine of that angle gives $x$. Note that $-1leq sin(x)leq 1$ for all real $x$, so (the real) $arcsin(x)$ is undefined for $x$ outside of $[-1,1]$ (things get weirder when allowing complex inputs and outputs) and note that there are multiple inputs which map to the same output for $sin$ such as how $sin(0)=sin(2pi)$, so $arcsin$ can't distinguish between them.
– JMoravitz
Jul 17 at 20:46
 |Â
show 7 more comments
1
As for your last question, you can't take the arcsine of any number not within the interval $[-1,1]$.
– RayDansh
Jul 17 at 20:37
$sin(x) in [-1,1]$
– gd1035
Jul 17 at 20:37
2
I discourage the use of the notation $sin^-1$ as that can cause confusion and frustration across borders as it is ambiguous whether the^-1is intended to mean multiplicative inversion or functional inversion. The notation $arcsin$ is superior.
– JMoravitz
Jul 17 at 20:38
2
You'we mixed them up. Replace $[-1,1]$ and $Big[-fracpi2,fracpi2Big]$
– Zeekless
Jul 17 at 20:38
1
As for how the cancellation property works... just remember how $arcsin(x)$ and $sin(x)$ act. $arcsin(x)$ returns the unique angle within the range $[-fracpi2,fracpi2]$ such that the sine of that angle gives $x$. Note that $-1leq sin(x)leq 1$ for all real $x$, so (the real) $arcsin(x)$ is undefined for $x$ outside of $[-1,1]$ (things get weirder when allowing complex inputs and outputs) and note that there are multiple inputs which map to the same output for $sin$ such as how $sin(0)=sin(2pi)$, so $arcsin$ can't distinguish between them.
– JMoravitz
Jul 17 at 20:46
1
1
As for your last question, you can't take the arcsine of any number not within the interval $[-1,1]$.
– RayDansh
Jul 17 at 20:37
As for your last question, you can't take the arcsine of any number not within the interval $[-1,1]$.
– RayDansh
Jul 17 at 20:37
$sin(x) in [-1,1]$
– gd1035
Jul 17 at 20:37
$sin(x) in [-1,1]$
– gd1035
Jul 17 at 20:37
2
2
I discourage the use of the notation $sin^-1$ as that can cause confusion and frustration across borders as it is ambiguous whether the
^-1 is intended to mean multiplicative inversion or functional inversion. The notation $arcsin$ is superior.– JMoravitz
Jul 17 at 20:38
I discourage the use of the notation $sin^-1$ as that can cause confusion and frustration across borders as it is ambiguous whether the
^-1 is intended to mean multiplicative inversion or functional inversion. The notation $arcsin$ is superior.– JMoravitz
Jul 17 at 20:38
2
2
You'we mixed them up. Replace $[-1,1]$ and $Big[-fracpi2,fracpi2Big]$
– Zeekless
Jul 17 at 20:38
You'we mixed them up. Replace $[-1,1]$ and $Big[-fracpi2,fracpi2Big]$
– Zeekless
Jul 17 at 20:38
1
1
As for how the cancellation property works... just remember how $arcsin(x)$ and $sin(x)$ act. $arcsin(x)$ returns the unique angle within the range $[-fracpi2,fracpi2]$ such that the sine of that angle gives $x$. Note that $-1leq sin(x)leq 1$ for all real $x$, so (the real) $arcsin(x)$ is undefined for $x$ outside of $[-1,1]$ (things get weirder when allowing complex inputs and outputs) and note that there are multiple inputs which map to the same output for $sin$ such as how $sin(0)=sin(2pi)$, so $arcsin$ can't distinguish between them.
– JMoravitz
Jul 17 at 20:46
As for how the cancellation property works... just remember how $arcsin(x)$ and $sin(x)$ act. $arcsin(x)$ returns the unique angle within the range $[-fracpi2,fracpi2]$ such that the sine of that angle gives $x$. Note that $-1leq sin(x)leq 1$ for all real $x$, so (the real) $arcsin(x)$ is undefined for $x$ outside of $[-1,1]$ (things get weirder when allowing complex inputs and outputs) and note that there are multiple inputs which map to the same output for $sin$ such as how $sin(0)=sin(2pi)$, so $arcsin$ can't distinguish between them.
– JMoravitz
Jul 17 at 20:46
 |Â
show 7 more comments
4 Answers
4
active
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up vote
2
down vote
accepted
It's all about Domain and Range.
$$ rm sin(x) ;: ;mathbbR rightarrow [-1, 1]$$
$$ rm arcsin(x) ;: ;[-1, 1] rightarrow Big[-fracpi2, fracpi2Big]$$
$$
Downarrow
$$
$$ rm sin(arcsin((x)) ;: ;[-1, 1] xrightarrow[id] [-1, 1]$$
$$ rm arcsin(sin((x)) ;: ;mathbbR rightarrow Big[-fracpi2, fracpi2Big]$$
The last map is not an identity, but its restriction on $[-fracpi2, fracpi2] subset mathbbR$ is:
$$ rm arcsin(sin((x))_[-fracpi2, fracpi2] ;: ;Big[-fracpi2, fracpi2Big] xrightarrow[id] Big[-fracpi2, fracpi2Big].$$
answered Jul 17 at 21:00
Zeekless
31110
add a comment |Â
up vote
1
down vote
For example $$sin(dfracpi4)=sqrt2over 2$$but so does $sin(dfrac3pi4)$ , $sin(dfrac9pi4)$ , $sin(dfrac11pi4)$ , ... so what is the value of $arcsin(x)$? There to avoid ambiguity, $arcsin$ function has bee defines such that for each argument gives a result in $[-dfracpi2,dfracpi2]$ since $sin$ function is bijective at this interval.$$arcsin(x)=thetain[-dfracpi2,dfracpi2]quad,quadsintheta=x$$
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
This is all fairly simple.
By convention $arcsin x$ is a function that takes value between -1 and 1 (inclusive) and returns angle between $-pi/2$ and $+pi/2$. That's it. $arcsin$ will never return an angle outside of this segment.
But you can calculate sine function for any angle, outside of $[-pi/2,+pi/2]$ range. For such angles inverse sine function will never return the original angle.
For example:
$$arcsin(sinfracpi6)=fracpi6$$
$$arcsin(sinfrac5pi6)=fracpi6$$
In other words, identity:
$$arcsin(sin x)=x$$
...is valid only for $xin [-pi/2, +pi/2]$
answered Jul 17 at 20:47
Oldboy
2,6101316
add a comment |Â
up vote
1
down vote
What is the answer to this question: for what $x$ the equation $sin(x)=0$?
You can clearly see that you could say $x=0$ or $x=pi$ as well $x=2pi,3pi,4pidots$ and so on. All this answers are clearly different from each other, so if I don't restrict myself on a certain domain the question $$arcsin(sin(x))=x$$ has more than one answer. For example let us choose $x=piover4$ and $y=3piover4$ and compute $$arcsin(sin(y))=arcsin(fracsqrt22) = piover4 = x$$
but clearly $xneq y$. So in general this equalities hold only on a smaller domain: if you want to invert the function $sin(x)$ you have to do it in $[-piover 2,piover 2]$ and in this case: $$arcsin(x): [-1,1]rightarrow [-piover 2,piover 2]$$ and as well $$sin(x):[-piover 2,piover 2]rightarrow[-1,1]$$ in this cases every element in the domain gets mapped to one element of the codomain, so the answer to that equations is unique
answered Jul 17 at 20:53
Davide Morgante
1,875220
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It's all about Domain and Range.
$$ rm sin(x) ;: ;mathbbR rightarrow [-1, 1]$$
$$ rm arcsin(x) ;: ;[-1, 1] rightarrow Big[-fracpi2, fracpi2Big]$$
$$
Downarrow
$$
$$ rm sin(arcsin((x)) ;: ;[-1, 1] xrightarrow[id] [-1, 1]$$
$$ rm arcsin(sin((x)) ;: ;mathbbR rightarrow Big[-fracpi2, fracpi2Big]$$
The last map is not an identity, but its restriction on $[-fracpi2, fracpi2] subset mathbbR$ is:
$$ rm arcsin(sin((x))_[-fracpi2, fracpi2] ;: ;Big[-fracpi2, fracpi2Big] xrightarrow[id] Big[-fracpi2, fracpi2Big].$$
answered Jul 17 at 21:00
Zeekless
31110
add a comment |Â
up vote
2
down vote
accepted
It's all about Domain and Range.
$$ rm sin(x) ;: ;mathbbR rightarrow [-1, 1]$$
$$ rm arcsin(x) ;: ;[-1, 1] rightarrow Big[-fracpi2, fracpi2Big]$$
$$
Downarrow
$$
$$ rm sin(arcsin((x)) ;: ;[-1, 1] xrightarrow[id] [-1, 1]$$
$$ rm arcsin(sin((x)) ;: ;mathbbR rightarrow Big[-fracpi2, fracpi2Big]$$
The last map is not an identity, but its restriction on $[-fracpi2, fracpi2] subset mathbbR$ is:
$$ rm arcsin(sin((x))_[-fracpi2, fracpi2] ;: ;Big[-fracpi2, fracpi2Big] xrightarrow[id] Big[-fracpi2, fracpi2Big].$$
answered Jul 17 at 21:00
Zeekless
31110
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's all about Domain and Range.
$$ rm sin(x) ;: ;mathbbR rightarrow [-1, 1]$$
$$ rm arcsin(x) ;: ;[-1, 1] rightarrow Big[-fracpi2, fracpi2Big]$$
$$
Downarrow
$$
$$ rm sin(arcsin((x)) ;: ;[-1, 1] xrightarrow[id] [-1, 1]$$
$$ rm arcsin(sin((x)) ;: ;mathbbR rightarrow Big[-fracpi2, fracpi2Big]$$
The last map is not an identity, but its restriction on $[-fracpi2, fracpi2] subset mathbbR$ is:
$$ rm arcsin(sin((x))_[-fracpi2, fracpi2] ;: ;Big[-fracpi2, fracpi2Big] xrightarrow[id] Big[-fracpi2, fracpi2Big].$$
answered Jul 17 at 21:00
Zeekless
31110
It's all about Domain and Range.
$$ rm sin(x) ;: ;mathbbR rightarrow [-1, 1]$$
$$ rm arcsin(x) ;: ;[-1, 1] rightarrow Big[-fracpi2, fracpi2Big]$$
$$
Downarrow
$$
$$ rm sin(arcsin((x)) ;: ;[-1, 1] xrightarrow[id] [-1, 1]$$
$$ rm arcsin(sin((x)) ;: ;mathbbR rightarrow Big[-fracpi2, fracpi2Big]$$
The last map is not an identity, but its restriction on $[-fracpi2, fracpi2] subset mathbbR$ is:
$$ rm arcsin(sin((x))_[-fracpi2, fracpi2] ;: ;Big[-fracpi2, fracpi2Big] xrightarrow[id] Big[-fracpi2, fracpi2Big].$$
answered Jul 17 at 21:00
Zeekless
31110
answered Jul 17 at 21:00
Zeekless
31110
answered Jul 17 at 21:00
Zeekless
31110
answered Jul 17 at 21:00
Zeekless
31110
31110
add a comment |Â
add a comment |Â
up vote
1
down vote
For example $$sin(dfracpi4)=sqrt2over 2$$but so does $sin(dfrac3pi4)$ , $sin(dfrac9pi4)$ , $sin(dfrac11pi4)$ , ... so what is the value of $arcsin(x)$? There to avoid ambiguity, $arcsin$ function has bee defines such that for each argument gives a result in $[-dfracpi2,dfracpi2]$ since $sin$ function is bijective at this interval.$$arcsin(x)=thetain[-dfracpi2,dfracpi2]quad,quadsintheta=x$$
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
For example $$sin(dfracpi4)=sqrt2over 2$$but so does $sin(dfrac3pi4)$ , $sin(dfrac9pi4)$ , $sin(dfrac11pi4)$ , ... so what is the value of $arcsin(x)$? There to avoid ambiguity, $arcsin$ function has bee defines such that for each argument gives a result in $[-dfracpi2,dfracpi2]$ since $sin$ function is bijective at this interval.$$arcsin(x)=thetain[-dfracpi2,dfracpi2]quad,quadsintheta=x$$
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For example $$sin(dfracpi4)=sqrt2over 2$$but so does $sin(dfrac3pi4)$ , $sin(dfrac9pi4)$ , $sin(dfrac11pi4)$ , ... so what is the value of $arcsin(x)$? There to avoid ambiguity, $arcsin$ function has bee defines such that for each argument gives a result in $[-dfracpi2,dfracpi2]$ since $sin$ function is bijective at this interval.$$arcsin(x)=thetain[-dfracpi2,dfracpi2]quad,quadsintheta=x$$
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
For example $$sin(dfracpi4)=sqrt2over 2$$but so does $sin(dfrac3pi4)$ , $sin(dfrac9pi4)$ , $sin(dfrac11pi4)$ , ... so what is the value of $arcsin(x)$? There to avoid ambiguity, $arcsin$ function has bee defines such that for each argument gives a result in $[-dfracpi2,dfracpi2]$ since $sin$ function is bijective at this interval.$$arcsin(x)=thetain[-dfracpi2,dfracpi2]quad,quadsintheta=x$$
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
answered Jul 17 at 20:51
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
up vote
1
down vote
This is all fairly simple.
By convention $arcsin x$ is a function that takes value between -1 and 1 (inclusive) and returns angle between $-pi/2$ and $+pi/2$. That's it. $arcsin$ will never return an angle outside of this segment.
But you can calculate sine function for any angle, outside of $[-pi/2,+pi/2]$ range. For such angles inverse sine function will never return the original angle.
For example:
$$arcsin(sinfracpi6)=fracpi6$$
$$arcsin(sinfrac5pi6)=fracpi6$$
In other words, identity:
$$arcsin(sin x)=x$$
...is valid only for $xin [-pi/2, +pi/2]$
answered Jul 17 at 20:47
Oldboy
2,6101316
add a comment |Â
up vote
1
down vote
This is all fairly simple.
By convention $arcsin x$ is a function that takes value between -1 and 1 (inclusive) and returns angle between $-pi/2$ and $+pi/2$. That's it. $arcsin$ will never return an angle outside of this segment.
But you can calculate sine function for any angle, outside of $[-pi/2,+pi/2]$ range. For such angles inverse sine function will never return the original angle.
For example:
$$arcsin(sinfracpi6)=fracpi6$$
$$arcsin(sinfrac5pi6)=fracpi6$$
In other words, identity:
$$arcsin(sin x)=x$$
...is valid only for $xin [-pi/2, +pi/2]$
answered Jul 17 at 20:47
Oldboy
2,6101316
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is all fairly simple.
By convention $arcsin x$ is a function that takes value between -1 and 1 (inclusive) and returns angle between $-pi/2$ and $+pi/2$. That's it. $arcsin$ will never return an angle outside of this segment.
But you can calculate sine function for any angle, outside of $[-pi/2,+pi/2]$ range. For such angles inverse sine function will never return the original angle.
For example:
$$arcsin(sinfracpi6)=fracpi6$$
$$arcsin(sinfrac5pi6)=fracpi6$$
In other words, identity:
$$arcsin(sin x)=x$$
...is valid only for $xin [-pi/2, +pi/2]$
answered Jul 17 at 20:47
Oldboy
2,6101316
This is all fairly simple.
By convention $arcsin x$ is a function that takes value between -1 and 1 (inclusive) and returns angle between $-pi/2$ and $+pi/2$. That's it. $arcsin$ will never return an angle outside of this segment.
But you can calculate sine function for any angle, outside of $[-pi/2,+pi/2]$ range. For such angles inverse sine function will never return the original angle.
For example:
$$arcsin(sinfracpi6)=fracpi6$$
$$arcsin(sinfrac5pi6)=fracpi6$$
In other words, identity:
$$arcsin(sin x)=x$$
...is valid only for $xin [-pi/2, +pi/2]$
answered Jul 17 at 20:47
Oldboy
2,6101316
edited Jul 17 at 20:54
answered Jul 17 at 20:47
Oldboy
2,6101316
answered Jul 17 at 20:47
Oldboy
2,6101316
answered Jul 17 at 20:47
Oldboy
2,6101316
2,6101316
add a comment |Â
add a comment |Â
up vote
1
down vote
What is the answer to this question: for what $x$ the equation $sin(x)=0$?
You can clearly see that you could say $x=0$ or $x=pi$ as well $x=2pi,3pi,4pidots$ and so on. All this answers are clearly different from each other, so if I don't restrict myself on a certain domain the question $$arcsin(sin(x))=x$$ has more than one answer. For example let us choose $x=piover4$ and $y=3piover4$ and compute $$arcsin(sin(y))=arcsin(fracsqrt22) = piover4 = x$$
but clearly $xneq y$. So in general this equalities hold only on a smaller domain: if you want to invert the function $sin(x)$ you have to do it in $[-piover 2,piover 2]$ and in this case: $$arcsin(x): [-1,1]rightarrow [-piover 2,piover 2]$$ and as well $$sin(x):[-piover 2,piover 2]rightarrow[-1,1]$$ in this cases every element in the domain gets mapped to one element of the codomain, so the answer to that equations is unique
answered Jul 17 at 20:53
Davide Morgante
1,875220
add a comment |Â
up vote
1
down vote
What is the answer to this question: for what $x$ the equation $sin(x)=0$?
You can clearly see that you could say $x=0$ or $x=pi$ as well $x=2pi,3pi,4pidots$ and so on. All this answers are clearly different from each other, so if I don't restrict myself on a certain domain the question $$arcsin(sin(x))=x$$ has more than one answer. For example let us choose $x=piover4$ and $y=3piover4$ and compute $$arcsin(sin(y))=arcsin(fracsqrt22) = piover4 = x$$
but clearly $xneq y$. So in general this equalities hold only on a smaller domain: if you want to invert the function $sin(x)$ you have to do it in $[-piover 2,piover 2]$ and in this case: $$arcsin(x): [-1,1]rightarrow [-piover 2,piover 2]$$ and as well $$sin(x):[-piover 2,piover 2]rightarrow[-1,1]$$ in this cases every element in the domain gets mapped to one element of the codomain, so the answer to that equations is unique
answered Jul 17 at 20:53
Davide Morgante
1,875220
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What is the answer to this question: for what $x$ the equation $sin(x)=0$?
You can clearly see that you could say $x=0$ or $x=pi$ as well $x=2pi,3pi,4pidots$ and so on. All this answers are clearly different from each other, so if I don't restrict myself on a certain domain the question $$arcsin(sin(x))=x$$ has more than one answer. For example let us choose $x=piover4$ and $y=3piover4$ and compute $$arcsin(sin(y))=arcsin(fracsqrt22) = piover4 = x$$
but clearly $xneq y$. So in general this equalities hold only on a smaller domain: if you want to invert the function $sin(x)$ you have to do it in $[-piover 2,piover 2]$ and in this case: $$arcsin(x): [-1,1]rightarrow [-piover 2,piover 2]$$ and as well $$sin(x):[-piover 2,piover 2]rightarrow[-1,1]$$ in this cases every element in the domain gets mapped to one element of the codomain, so the answer to that equations is unique
answered Jul 17 at 20:53
Davide Morgante
1,875220
What is the answer to this question: for what $x$ the equation $sin(x)=0$?
You can clearly see that you could say $x=0$ or $x=pi$ as well $x=2pi,3pi,4pidots$ and so on. All this answers are clearly different from each other, so if I don't restrict myself on a certain domain the question $$arcsin(sin(x))=x$$ has more than one answer. For example let us choose $x=piover4$ and $y=3piover4$ and compute $$arcsin(sin(y))=arcsin(fracsqrt22) = piover4 = x$$
but clearly $xneq y$. So in general this equalities hold only on a smaller domain: if you want to invert the function $sin(x)$ you have to do it in $[-piover 2,piover 2]$ and in this case: $$arcsin(x): [-1,1]rightarrow [-piover 2,piover 2]$$ and as well $$sin(x):[-piover 2,piover 2]rightarrow[-1,1]$$ in this cases every element in the domain gets mapped to one element of the codomain, so the answer to that equations is unique
answered Jul 17 at 20:53
Davide Morgante
1,875220
edited Jul 17 at 21:23
answered Jul 17 at 20:53
Davide Morgante
1,875220
answered Jul 17 at 20:53
Davide Morgante
1,875220
answered Jul 17 at 20:53
Davide Morgante
1,875220
1,875220
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1
As for your last question, you can't take the arcsine of any number not within the interval $[-1,1]$.
– RayDansh
Jul 17 at 20:37
$sin(x) in [-1,1]$
– gd1035
Jul 17 at 20:37
2
I discourage the use of the notation $sin^-1$ as that can cause confusion and frustration across borders as it is ambiguous whether the
^-1is intended to mean multiplicative inversion or functional inversion. The notation $arcsin$ is superior.– JMoravitz
Jul 17 at 20:38
2
You'we mixed them up. Replace $[-1,1]$ and $Big[-fracpi2,fracpi2Big]$
– Zeekless
Jul 17 at 20:38
1
As for how the cancellation property works... just remember how $arcsin(x)$ and $sin(x)$ act. $arcsin(x)$ returns the unique angle within the range $[-fracpi2,fracpi2]$ such that the sine of that angle gives $x$. Note that $-1leq sin(x)leq 1$ for all real $x$, so (the real) $arcsin(x)$ is undefined for $x$ outside of $[-1,1]$ (things get weirder when allowing complex inputs and outputs) and note that there are multiple inputs which map to the same output for $sin$ such as how $sin(0)=sin(2pi)$, so $arcsin$ can't distinguish between them.
– JMoravitz
Jul 17 at 20:46