Is $[sin(1/2)]^{-1}$ identical to $frac{1}{sin(1/2)}$
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I'm in Grade 12 Advanced Functions and having some trouble with understanding the difference between $sin^{-1}(1/2)$ and $(sin(1/2))^{-1}$. I recognize that the former asks to find an angle whose sine is 1/2, and the latter is just a multiplicative inverse, and I just need this question answered to be absolutely sure in my understanding of what's going on. Thanks.
inverse
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I'm in Grade 12 Advanced Functions and having some trouble with understanding the difference between $sin^{-1}(1/2)$ and $(sin(1/2))^{-1}$. I recognize that the former asks to find an angle whose sine is 1/2, and the latter is just a multiplicative inverse, and I just need this question answered to be absolutely sure in my understanding of what's going on. Thanks.
inverse
1
$sin^{-1}$ is the standard notation for the inverse sine function, which is defined by $sin^{-1} x = y iff sin y = x$ (depending on the domain). This is very different from $(sin x)^{-1}$.
– T. Bongers
Nov 17 at 1:08
So for the quick answer to your question, yes your understanding is correct.
– T. Bongers
Nov 17 at 1:09
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm in Grade 12 Advanced Functions and having some trouble with understanding the difference between $sin^{-1}(1/2)$ and $(sin(1/2))^{-1}$. I recognize that the former asks to find an angle whose sine is 1/2, and the latter is just a multiplicative inverse, and I just need this question answered to be absolutely sure in my understanding of what's going on. Thanks.
inverse
I'm in Grade 12 Advanced Functions and having some trouble with understanding the difference between $sin^{-1}(1/2)$ and $(sin(1/2))^{-1}$. I recognize that the former asks to find an angle whose sine is 1/2, and the latter is just a multiplicative inverse, and I just need this question answered to be absolutely sure in my understanding of what's going on. Thanks.
inverse
inverse
edited Nov 17 at 1:14
Moo
5,4033920
5,4033920
asked Nov 17 at 1:05
Korvexius
72
72
1
$sin^{-1}$ is the standard notation for the inverse sine function, which is defined by $sin^{-1} x = y iff sin y = x$ (depending on the domain). This is very different from $(sin x)^{-1}$.
– T. Bongers
Nov 17 at 1:08
So for the quick answer to your question, yes your understanding is correct.
– T. Bongers
Nov 17 at 1:09
add a comment |
1
$sin^{-1}$ is the standard notation for the inverse sine function, which is defined by $sin^{-1} x = y iff sin y = x$ (depending on the domain). This is very different from $(sin x)^{-1}$.
– T. Bongers
Nov 17 at 1:08
So for the quick answer to your question, yes your understanding is correct.
– T. Bongers
Nov 17 at 1:09
1
1
$sin^{-1}$ is the standard notation for the inverse sine function, which is defined by $sin^{-1} x = y iff sin y = x$ (depending on the domain). This is very different from $(sin x)^{-1}$.
– T. Bongers
Nov 17 at 1:08
$sin^{-1}$ is the standard notation for the inverse sine function, which is defined by $sin^{-1} x = y iff sin y = x$ (depending on the domain). This is very different from $(sin x)^{-1}$.
– T. Bongers
Nov 17 at 1:08
So for the quick answer to your question, yes your understanding is correct.
– T. Bongers
Nov 17 at 1:09
So for the quick answer to your question, yes your understanding is correct.
– T. Bongers
Nov 17 at 1:09
add a comment |
2 Answers
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up vote
0
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Yeah, the notation used in the cases of these are really really dumb, if we're being honest.
Basically, where the exponent is located is important:
$text{sin}^{-1}(x)$ basically denotes the "inverse sine" or "arcsine" function. It is the function you use to find the angle. For example, if you have $text{sin}^{-1}(x)=theta$, you see that
$$text{sin}^{-1}(x)=theta ;;; Rightarrow ;;; text{sin}(text{sin}^{-1} (x))=text{sin}(theta) ;;; Rightarrow ;;; x = text{sin}(theta)$$
$[text{sin}(theta)]^{-1}$ denotes the multiplicative inverse of sine. Note that this isn't the same as the previous "inverse." The previous is an inverse function (more generally, $f^{-1}$ is an inverse function to $f$ if $f(f^{-1}(x)) = f^{-1}(f(x))=x$), whereas this is a multiplicative inverse. A multiplicative inverse to a number $x$ is the number $y$ such that $xcdot y=1$, i.e. $1/x$ is the multiplicative inverse to $x$. In that sense,
$$[text{sin}(theta)]^{-1} = frac{1}{sin(theta)}$$
We note this latter result can go by a different name, the cosecant ($csc(x)$) function.
In writing out these things, to avoid any potential ambiguities, it will be helpful to you to write the first case as $arcsin(x)$ (which is an acceptable notation). The latter case could be written as a fraction or as $csc(x)$, whichever is most convenient.
Just note that, in general, the two are very, very different things that carry very different meanings ... whoever thought up the notation wasn't exactly being helpful.
add a comment |
up vote
0
down vote
Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yeah, the notation used in the cases of these are really really dumb, if we're being honest.
Basically, where the exponent is located is important:
$text{sin}^{-1}(x)$ basically denotes the "inverse sine" or "arcsine" function. It is the function you use to find the angle. For example, if you have $text{sin}^{-1}(x)=theta$, you see that
$$text{sin}^{-1}(x)=theta ;;; Rightarrow ;;; text{sin}(text{sin}^{-1} (x))=text{sin}(theta) ;;; Rightarrow ;;; x = text{sin}(theta)$$
$[text{sin}(theta)]^{-1}$ denotes the multiplicative inverse of sine. Note that this isn't the same as the previous "inverse." The previous is an inverse function (more generally, $f^{-1}$ is an inverse function to $f$ if $f(f^{-1}(x)) = f^{-1}(f(x))=x$), whereas this is a multiplicative inverse. A multiplicative inverse to a number $x$ is the number $y$ such that $xcdot y=1$, i.e. $1/x$ is the multiplicative inverse to $x$. In that sense,
$$[text{sin}(theta)]^{-1} = frac{1}{sin(theta)}$$
We note this latter result can go by a different name, the cosecant ($csc(x)$) function.
In writing out these things, to avoid any potential ambiguities, it will be helpful to you to write the first case as $arcsin(x)$ (which is an acceptable notation). The latter case could be written as a fraction or as $csc(x)$, whichever is most convenient.
Just note that, in general, the two are very, very different things that carry very different meanings ... whoever thought up the notation wasn't exactly being helpful.
add a comment |
up vote
0
down vote
accepted
Yeah, the notation used in the cases of these are really really dumb, if we're being honest.
Basically, where the exponent is located is important:
$text{sin}^{-1}(x)$ basically denotes the "inverse sine" or "arcsine" function. It is the function you use to find the angle. For example, if you have $text{sin}^{-1}(x)=theta$, you see that
$$text{sin}^{-1}(x)=theta ;;; Rightarrow ;;; text{sin}(text{sin}^{-1} (x))=text{sin}(theta) ;;; Rightarrow ;;; x = text{sin}(theta)$$
$[text{sin}(theta)]^{-1}$ denotes the multiplicative inverse of sine. Note that this isn't the same as the previous "inverse." The previous is an inverse function (more generally, $f^{-1}$ is an inverse function to $f$ if $f(f^{-1}(x)) = f^{-1}(f(x))=x$), whereas this is a multiplicative inverse. A multiplicative inverse to a number $x$ is the number $y$ such that $xcdot y=1$, i.e. $1/x$ is the multiplicative inverse to $x$. In that sense,
$$[text{sin}(theta)]^{-1} = frac{1}{sin(theta)}$$
We note this latter result can go by a different name, the cosecant ($csc(x)$) function.
In writing out these things, to avoid any potential ambiguities, it will be helpful to you to write the first case as $arcsin(x)$ (which is an acceptable notation). The latter case could be written as a fraction or as $csc(x)$, whichever is most convenient.
Just note that, in general, the two are very, very different things that carry very different meanings ... whoever thought up the notation wasn't exactly being helpful.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yeah, the notation used in the cases of these are really really dumb, if we're being honest.
Basically, where the exponent is located is important:
$text{sin}^{-1}(x)$ basically denotes the "inverse sine" or "arcsine" function. It is the function you use to find the angle. For example, if you have $text{sin}^{-1}(x)=theta$, you see that
$$text{sin}^{-1}(x)=theta ;;; Rightarrow ;;; text{sin}(text{sin}^{-1} (x))=text{sin}(theta) ;;; Rightarrow ;;; x = text{sin}(theta)$$
$[text{sin}(theta)]^{-1}$ denotes the multiplicative inverse of sine. Note that this isn't the same as the previous "inverse." The previous is an inverse function (more generally, $f^{-1}$ is an inverse function to $f$ if $f(f^{-1}(x)) = f^{-1}(f(x))=x$), whereas this is a multiplicative inverse. A multiplicative inverse to a number $x$ is the number $y$ such that $xcdot y=1$, i.e. $1/x$ is the multiplicative inverse to $x$. In that sense,
$$[text{sin}(theta)]^{-1} = frac{1}{sin(theta)}$$
We note this latter result can go by a different name, the cosecant ($csc(x)$) function.
In writing out these things, to avoid any potential ambiguities, it will be helpful to you to write the first case as $arcsin(x)$ (which is an acceptable notation). The latter case could be written as a fraction or as $csc(x)$, whichever is most convenient.
Just note that, in general, the two are very, very different things that carry very different meanings ... whoever thought up the notation wasn't exactly being helpful.
Yeah, the notation used in the cases of these are really really dumb, if we're being honest.
Basically, where the exponent is located is important:
$text{sin}^{-1}(x)$ basically denotes the "inverse sine" or "arcsine" function. It is the function you use to find the angle. For example, if you have $text{sin}^{-1}(x)=theta$, you see that
$$text{sin}^{-1}(x)=theta ;;; Rightarrow ;;; text{sin}(text{sin}^{-1} (x))=text{sin}(theta) ;;; Rightarrow ;;; x = text{sin}(theta)$$
$[text{sin}(theta)]^{-1}$ denotes the multiplicative inverse of sine. Note that this isn't the same as the previous "inverse." The previous is an inverse function (more generally, $f^{-1}$ is an inverse function to $f$ if $f(f^{-1}(x)) = f^{-1}(f(x))=x$), whereas this is a multiplicative inverse. A multiplicative inverse to a number $x$ is the number $y$ such that $xcdot y=1$, i.e. $1/x$ is the multiplicative inverse to $x$. In that sense,
$$[text{sin}(theta)]^{-1} = frac{1}{sin(theta)}$$
We note this latter result can go by a different name, the cosecant ($csc(x)$) function.
In writing out these things, to avoid any potential ambiguities, it will be helpful to you to write the first case as $arcsin(x)$ (which is an acceptable notation). The latter case could be written as a fraction or as $csc(x)$, whichever is most convenient.
Just note that, in general, the two are very, very different things that carry very different meanings ... whoever thought up the notation wasn't exactly being helpful.
edited Nov 17 at 1:26
answered Nov 17 at 1:14
Eevee Trainer
1,510216
1,510216
add a comment |
add a comment |
up vote
0
down vote
Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$
add a comment |
up vote
0
down vote
Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$
Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$
answered Nov 17 at 1:16
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
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$sin^{-1}$ is the standard notation for the inverse sine function, which is defined by $sin^{-1} x = y iff sin y = x$ (depending on the domain). This is very different from $(sin x)^{-1}$.
– T. Bongers
Nov 17 at 1:08
So for the quick answer to your question, yes your understanding is correct.
– T. Bongers
Nov 17 at 1:09