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.










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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















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.










share|cite|improve this question




















  • 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













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.










share|cite|improve this question















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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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














  • 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










2 Answers
2






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0
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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.






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    Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$






    share|cite|improve this answer





















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      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.






      share|cite|improve this answer



























        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.






        share|cite|improve this answer

























          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 17 at 1:26

























          answered Nov 17 at 1:14









          Eevee Trainer

          1,510216




          1,510216






















              up vote
              0
              down vote













              Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$






                  share|cite|improve this answer












                  Yes, your answer is correct. $$sin ^{-1} (1/2) =pi/6 $$ and $$[ sin (1/2)]^{-1} = frac {1}{sin (1/2)} = csc (1/2)$$







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                  answered Nov 17 at 1:16









                  Mohammad Riazi-Kermani

                  40.3k41958




                  40.3k41958






























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