When I'm given $sin(x)$, what goes inside of the $x$? [duplicate]












1












$begingroup$



This question already has an answer here:




  • Do you use degrees or radians for trig functions?

    1 answer




I don't know if I'm just randomly blanking or if I never really knew and have just been going with the flow, but I'm not sure what x represents.
In early high school they were degrees, eg. $sin(30)$ which equaled $0.5$.
Later on we learned about radians and $pi$, and how $sin {pi over 3}$ was equal to $sqrt 3 over 2$.



Now I'm in Uni and I'm discovering that I maybe don't know trigonometry as well as I should.



For example, when doing the squeeze theorem, and I'm asked to find the limit of $sin(n) over n$, what is the $n$? Is it in radians? degrees? It's for graphing so what should I visualize? $n$ as just a $x$ value to try and find any $y$??










share|cite|improve this question











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marked as duplicate by Lord Shark the Unknown, user10354138, KReiser, Rebellos, José Carlos Santos calculus
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Dec 6 '18 at 9:15


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    It's radians, as a hint, $|sin(n)|leq 1$.
    $endgroup$
    – Jakobian
    Dec 6 '18 at 0:59






  • 2




    $begingroup$
    If there is no specific information, take $x $ in radians.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 1:00










  • $begingroup$
    The $x$ in $sin(x)$ can be any real number. But you will usually be given a domain, such as $xin[0,2pi]$. In the particular case of $frac{sin(n)}{n}$, the $n$ represents the natural numbers.
    $endgroup$
    – Live Free or π Hard
    Dec 6 '18 at 1:02






  • 1




    $begingroup$
    @Jakobian Actually, if $ninmathbb{N}$, then the strict inequality $|sin(n)| < 1$ holds, as $|sin(x)| = 1 text{ iff } x = frac{(2k+1)pi}{2} notin mathbb{N}$ for any $kinmathbb{Z}$.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 1:20






  • 3




    $begingroup$
    Your Uni lecturer has almost certainly told you whether she wants you to interpret the $x$ in $sin x$ as degrees or as radians, and whatever she says, goes. I hope she has told you it's radians, but you'd best check with her, and not with random people on the internet who haven't been enrolled in the same class as you are.
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:24
















1












$begingroup$



This question already has an answer here:




  • Do you use degrees or radians for trig functions?

    1 answer




I don't know if I'm just randomly blanking or if I never really knew and have just been going with the flow, but I'm not sure what x represents.
In early high school they were degrees, eg. $sin(30)$ which equaled $0.5$.
Later on we learned about radians and $pi$, and how $sin {pi over 3}$ was equal to $sqrt 3 over 2$.



Now I'm in Uni and I'm discovering that I maybe don't know trigonometry as well as I should.



For example, when doing the squeeze theorem, and I'm asked to find the limit of $sin(n) over n$, what is the $n$? Is it in radians? degrees? It's for graphing so what should I visualize? $n$ as just a $x$ value to try and find any $y$??










share|cite|improve this question











$endgroup$



marked as duplicate by Lord Shark the Unknown, user10354138, KReiser, Rebellos, José Carlos Santos calculus
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Dec 6 '18 at 9:15


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    It's radians, as a hint, $|sin(n)|leq 1$.
    $endgroup$
    – Jakobian
    Dec 6 '18 at 0:59






  • 2




    $begingroup$
    If there is no specific information, take $x $ in radians.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 1:00










  • $begingroup$
    The $x$ in $sin(x)$ can be any real number. But you will usually be given a domain, such as $xin[0,2pi]$. In the particular case of $frac{sin(n)}{n}$, the $n$ represents the natural numbers.
    $endgroup$
    – Live Free or π Hard
    Dec 6 '18 at 1:02






  • 1




    $begingroup$
    @Jakobian Actually, if $ninmathbb{N}$, then the strict inequality $|sin(n)| < 1$ holds, as $|sin(x)| = 1 text{ iff } x = frac{(2k+1)pi}{2} notin mathbb{N}$ for any $kinmathbb{Z}$.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 1:20






  • 3




    $begingroup$
    Your Uni lecturer has almost certainly told you whether she wants you to interpret the $x$ in $sin x$ as degrees or as radians, and whatever she says, goes. I hope she has told you it's radians, but you'd best check with her, and not with random people on the internet who haven't been enrolled in the same class as you are.
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:24














1












1








1





$begingroup$



This question already has an answer here:




  • Do you use degrees or radians for trig functions?

    1 answer




I don't know if I'm just randomly blanking or if I never really knew and have just been going with the flow, but I'm not sure what x represents.
In early high school they were degrees, eg. $sin(30)$ which equaled $0.5$.
Later on we learned about radians and $pi$, and how $sin {pi over 3}$ was equal to $sqrt 3 over 2$.



Now I'm in Uni and I'm discovering that I maybe don't know trigonometry as well as I should.



For example, when doing the squeeze theorem, and I'm asked to find the limit of $sin(n) over n$, what is the $n$? Is it in radians? degrees? It's for graphing so what should I visualize? $n$ as just a $x$ value to try and find any $y$??










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Do you use degrees or radians for trig functions?

    1 answer




I don't know if I'm just randomly blanking or if I never really knew and have just been going with the flow, but I'm not sure what x represents.
In early high school they were degrees, eg. $sin(30)$ which equaled $0.5$.
Later on we learned about radians and $pi$, and how $sin {pi over 3}$ was equal to $sqrt 3 over 2$.



Now I'm in Uni and I'm discovering that I maybe don't know trigonometry as well as I should.



For example, when doing the squeeze theorem, and I'm asked to find the limit of $sin(n) over n$, what is the $n$? Is it in radians? degrees? It's for graphing so what should I visualize? $n$ as just a $x$ value to try and find any $y$??





This question already has an answer here:




  • Do you use degrees or radians for trig functions?

    1 answer








calculus trigonometry notation definition






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edited Dec 6 '18 at 4:30









user587192

1,827315




1,827315










asked Dec 6 '18 at 0:57









mingming

3365




3365




marked as duplicate by Lord Shark the Unknown, user10354138, KReiser, Rebellos, José Carlos Santos calculus
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Dec 6 '18 at 9:15


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Lord Shark the Unknown, user10354138, KReiser, Rebellos, José Carlos Santos calculus
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Dec 6 '18 at 9:15


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    It's radians, as a hint, $|sin(n)|leq 1$.
    $endgroup$
    – Jakobian
    Dec 6 '18 at 0:59






  • 2




    $begingroup$
    If there is no specific information, take $x $ in radians.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 1:00










  • $begingroup$
    The $x$ in $sin(x)$ can be any real number. But you will usually be given a domain, such as $xin[0,2pi]$. In the particular case of $frac{sin(n)}{n}$, the $n$ represents the natural numbers.
    $endgroup$
    – Live Free or π Hard
    Dec 6 '18 at 1:02






  • 1




    $begingroup$
    @Jakobian Actually, if $ninmathbb{N}$, then the strict inequality $|sin(n)| < 1$ holds, as $|sin(x)| = 1 text{ iff } x = frac{(2k+1)pi}{2} notin mathbb{N}$ for any $kinmathbb{Z}$.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 1:20






  • 3




    $begingroup$
    Your Uni lecturer has almost certainly told you whether she wants you to interpret the $x$ in $sin x$ as degrees or as radians, and whatever she says, goes. I hope she has told you it's radians, but you'd best check with her, and not with random people on the internet who haven't been enrolled in the same class as you are.
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:24














  • 1




    $begingroup$
    It's radians, as a hint, $|sin(n)|leq 1$.
    $endgroup$
    – Jakobian
    Dec 6 '18 at 0:59






  • 2




    $begingroup$
    If there is no specific information, take $x $ in radians.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 1:00










  • $begingroup$
    The $x$ in $sin(x)$ can be any real number. But you will usually be given a domain, such as $xin[0,2pi]$. In the particular case of $frac{sin(n)}{n}$, the $n$ represents the natural numbers.
    $endgroup$
    – Live Free or π Hard
    Dec 6 '18 at 1:02






  • 1




    $begingroup$
    @Jakobian Actually, if $ninmathbb{N}$, then the strict inequality $|sin(n)| < 1$ holds, as $|sin(x)| = 1 text{ iff } x = frac{(2k+1)pi}{2} notin mathbb{N}$ for any $kinmathbb{Z}$.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 1:20






  • 3




    $begingroup$
    Your Uni lecturer has almost certainly told you whether she wants you to interpret the $x$ in $sin x$ as degrees or as radians, and whatever she says, goes. I hope she has told you it's radians, but you'd best check with her, and not with random people on the internet who haven't been enrolled in the same class as you are.
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:24








1




1




$begingroup$
It's radians, as a hint, $|sin(n)|leq 1$.
$endgroup$
– Jakobian
Dec 6 '18 at 0:59




$begingroup$
It's radians, as a hint, $|sin(n)|leq 1$.
$endgroup$
– Jakobian
Dec 6 '18 at 0:59




2




2




$begingroup$
If there is no specific information, take $x $ in radians.
$endgroup$
– Thomas Shelby
Dec 6 '18 at 1:00




$begingroup$
If there is no specific information, take $x $ in radians.
$endgroup$
– Thomas Shelby
Dec 6 '18 at 1:00












$begingroup$
The $x$ in $sin(x)$ can be any real number. But you will usually be given a domain, such as $xin[0,2pi]$. In the particular case of $frac{sin(n)}{n}$, the $n$ represents the natural numbers.
$endgroup$
– Live Free or π Hard
Dec 6 '18 at 1:02




$begingroup$
The $x$ in $sin(x)$ can be any real number. But you will usually be given a domain, such as $xin[0,2pi]$. In the particular case of $frac{sin(n)}{n}$, the $n$ represents the natural numbers.
$endgroup$
– Live Free or π Hard
Dec 6 '18 at 1:02




1




1




$begingroup$
@Jakobian Actually, if $ninmathbb{N}$, then the strict inequality $|sin(n)| < 1$ holds, as $|sin(x)| = 1 text{ iff } x = frac{(2k+1)pi}{2} notin mathbb{N}$ for any $kinmathbb{Z}$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:20




$begingroup$
@Jakobian Actually, if $ninmathbb{N}$, then the strict inequality $|sin(n)| < 1$ holds, as $|sin(x)| = 1 text{ iff } x = frac{(2k+1)pi}{2} notin mathbb{N}$ for any $kinmathbb{Z}$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:20




3




3




$begingroup$
Your Uni lecturer has almost certainly told you whether she wants you to interpret the $x$ in $sin x$ as degrees or as radians, and whatever she says, goes. I hope she has told you it's radians, but you'd best check with her, and not with random people on the internet who haven't been enrolled in the same class as you are.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:24




$begingroup$
Your Uni lecturer has almost certainly told you whether she wants you to interpret the $x$ in $sin x$ as degrees or as radians, and whatever she says, goes. I hope she has told you it's radians, but you'd best check with her, and not with random people on the internet who haven't been enrolled in the same class as you are.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:24










3 Answers
3






active

oldest

votes


















4












$begingroup$

In a Calculus course it's almost always radians. That's because the limit
$$lim_{xrightarrow 0} frac{sin(x)}{x}=1$$
Is only correct if you take the argument to be given in radians. In particular, the formulas
$$frac{d}{dx} sin x=cos x,$$
and
$$frac{d}{dx} cos x=-sin x$$
(which you'll learn shortly if you haven't already) are also only true when the argument is taken in radians. If you were to use degrees then you'll need to multiply the right hand side of both formulas by $pi/180$ which, of course, looks ugly.



This is actually the main argument for introducing radians in the first place and having everyone coming out of highschool confused about the new way of measuring angles. It's not just because, it actually makes the formulas you use in calculus simpler and more elegant.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The $sin$ function and it's kin can be defined in many ways, such as using Taylor series or differential equations. Like any other function on the reals, they are just a way of turning numbers into numbers. It does have some special properties, like $sin(0) = 0$, $sin(-x) = -sin(x)$, $sin(x+2pi) = sin(x)$, to name a few.



    When defining this function in terms of geometry of circles and angles, we often start with a discussion of degrees, because they are more familiar to us at the time. However, the true definition is in terms of radians. If you define $sintheta$ as the $y$-coordinate of the point on the unit circle reached by traveling a distance of $theta$ counter-clockwise along the circle, starting at $(1,0)$, it will be consistent with the definitions given by other means.



    To make trigonometry work with degrees, we need to scale the inputs to these functions. Since there are $2pi$ radians in a circle and $360$ degrees, converting degrees to radians involves multiplying by $frac{180}{pi}$, so we can define the function $text{sind}(a) = sin(frac{180}{pi}a)$ for an angle $a$ written in degrees.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      First of all this is not a dumb question at all. Hardy also asked and answered the very same question in his A Course of Pure Mathematics. It is rather the typical treatment of trigonometric functions in high school which very smartly avoids any discussion on this topic.



      When you see expressions like $x^2,sqrt{x},log x, 2^x$ do you bother to ask "what is the $x$ in these expressions"? Normally not and you assume that $x$ is a real number in these expression. Getting the value of these algebraic expressions like $x^2,sqrt{x}$, given the value of $x$, is explained in typical courses of algebra. The story with $log x, 2^x, sin x$ is a bit different. Given a real number $x$, it is normally never explained (at least in high school), how to evaluate these expressions in $x$.



      What you are really asking here is a definition of the sine function as a function of a real variable. The $x$ in $sin x$ is a real number number and there are ways to get the value of $sin x$, given the value of $x$. Unfortunately all the ways to do so are difficult. The easiest approach uses the geometric notion of a circle but the underlying assumptions are again difficult to prove.



      So let's assume either of these facts:




      • To every arc of a circle there corresponds a unique real number $L$ called its length.

      • To every sector of a circle corresponds a unique real number $A$ called its area.


      It is sufficient to use any one of the assumptions above and while proving these assumptions it can be established that $L=2A$ if the radius of the circle involved is unity.



      The more popular definition of $sin x$ uses the assumption of length of arc of a circle. Thus let $x$ be any given real number. Consider the unit circle $X^2+Y^2=1$ in $XY$ coordinate plane and let $A=(1, 0), O=(0, 0)$. Start with point $A$ on circle and move on the cirle in counterclockwise (clockwise if $x<0$) direction till you cover the distance $|x|$. Mark this point on circle as $P_x$. Then by definition the point $P_x$ is $(cos x, sin x) $. While using this definition most people say that $x$ is the measure of angle $AOP$ in radians.



      The geometric definition above can be translated into the language of calculus via the equation $$int_{0}^{sin x} frac{dt} {sqrt{1-t^2}}=x$$ for suitable values of $x$. And some university courses use the above integral to define $sin x$. There are other approaches to define $sin x$ which make use of infinite series but they are far removed from its geometrical link.






      share|cite|improve this answer











      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        In a Calculus course it's almost always radians. That's because the limit
        $$lim_{xrightarrow 0} frac{sin(x)}{x}=1$$
        Is only correct if you take the argument to be given in radians. In particular, the formulas
        $$frac{d}{dx} sin x=cos x,$$
        and
        $$frac{d}{dx} cos x=-sin x$$
        (which you'll learn shortly if you haven't already) are also only true when the argument is taken in radians. If you were to use degrees then you'll need to multiply the right hand side of both formulas by $pi/180$ which, of course, looks ugly.



        This is actually the main argument for introducing radians in the first place and having everyone coming out of highschool confused about the new way of measuring angles. It's not just because, it actually makes the formulas you use in calculus simpler and more elegant.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          In a Calculus course it's almost always radians. That's because the limit
          $$lim_{xrightarrow 0} frac{sin(x)}{x}=1$$
          Is only correct if you take the argument to be given in radians. In particular, the formulas
          $$frac{d}{dx} sin x=cos x,$$
          and
          $$frac{d}{dx} cos x=-sin x$$
          (which you'll learn shortly if you haven't already) are also only true when the argument is taken in radians. If you were to use degrees then you'll need to multiply the right hand side of both formulas by $pi/180$ which, of course, looks ugly.



          This is actually the main argument for introducing radians in the first place and having everyone coming out of highschool confused about the new way of measuring angles. It's not just because, it actually makes the formulas you use in calculus simpler and more elegant.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            In a Calculus course it's almost always radians. That's because the limit
            $$lim_{xrightarrow 0} frac{sin(x)}{x}=1$$
            Is only correct if you take the argument to be given in radians. In particular, the formulas
            $$frac{d}{dx} sin x=cos x,$$
            and
            $$frac{d}{dx} cos x=-sin x$$
            (which you'll learn shortly if you haven't already) are also only true when the argument is taken in radians. If you were to use degrees then you'll need to multiply the right hand side of both formulas by $pi/180$ which, of course, looks ugly.



            This is actually the main argument for introducing radians in the first place and having everyone coming out of highschool confused about the new way of measuring angles. It's not just because, it actually makes the formulas you use in calculus simpler and more elegant.






            share|cite|improve this answer









            $endgroup$



            In a Calculus course it's almost always radians. That's because the limit
            $$lim_{xrightarrow 0} frac{sin(x)}{x}=1$$
            Is only correct if you take the argument to be given in radians. In particular, the formulas
            $$frac{d}{dx} sin x=cos x,$$
            and
            $$frac{d}{dx} cos x=-sin x$$
            (which you'll learn shortly if you haven't already) are also only true when the argument is taken in radians. If you were to use degrees then you'll need to multiply the right hand side of both formulas by $pi/180$ which, of course, looks ugly.



            This is actually the main argument for introducing radians in the first place and having everyone coming out of highschool confused about the new way of measuring angles. It's not just because, it actually makes the formulas you use in calculus simpler and more elegant.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 1:23









            Raul GomezRaul Gomez

            1665




            1665























                0












                $begingroup$

                The $sin$ function and it's kin can be defined in many ways, such as using Taylor series or differential equations. Like any other function on the reals, they are just a way of turning numbers into numbers. It does have some special properties, like $sin(0) = 0$, $sin(-x) = -sin(x)$, $sin(x+2pi) = sin(x)$, to name a few.



                When defining this function in terms of geometry of circles and angles, we often start with a discussion of degrees, because they are more familiar to us at the time. However, the true definition is in terms of radians. If you define $sintheta$ as the $y$-coordinate of the point on the unit circle reached by traveling a distance of $theta$ counter-clockwise along the circle, starting at $(1,0)$, it will be consistent with the definitions given by other means.



                To make trigonometry work with degrees, we need to scale the inputs to these functions. Since there are $2pi$ radians in a circle and $360$ degrees, converting degrees to radians involves multiplying by $frac{180}{pi}$, so we can define the function $text{sind}(a) = sin(frac{180}{pi}a)$ for an angle $a$ written in degrees.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  The $sin$ function and it's kin can be defined in many ways, such as using Taylor series or differential equations. Like any other function on the reals, they are just a way of turning numbers into numbers. It does have some special properties, like $sin(0) = 0$, $sin(-x) = -sin(x)$, $sin(x+2pi) = sin(x)$, to name a few.



                  When defining this function in terms of geometry of circles and angles, we often start with a discussion of degrees, because they are more familiar to us at the time. However, the true definition is in terms of radians. If you define $sintheta$ as the $y$-coordinate of the point on the unit circle reached by traveling a distance of $theta$ counter-clockwise along the circle, starting at $(1,0)$, it will be consistent with the definitions given by other means.



                  To make trigonometry work with degrees, we need to scale the inputs to these functions. Since there are $2pi$ radians in a circle and $360$ degrees, converting degrees to radians involves multiplying by $frac{180}{pi}$, so we can define the function $text{sind}(a) = sin(frac{180}{pi}a)$ for an angle $a$ written in degrees.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    The $sin$ function and it's kin can be defined in many ways, such as using Taylor series or differential equations. Like any other function on the reals, they are just a way of turning numbers into numbers. It does have some special properties, like $sin(0) = 0$, $sin(-x) = -sin(x)$, $sin(x+2pi) = sin(x)$, to name a few.



                    When defining this function in terms of geometry of circles and angles, we often start with a discussion of degrees, because they are more familiar to us at the time. However, the true definition is in terms of radians. If you define $sintheta$ as the $y$-coordinate of the point on the unit circle reached by traveling a distance of $theta$ counter-clockwise along the circle, starting at $(1,0)$, it will be consistent with the definitions given by other means.



                    To make trigonometry work with degrees, we need to scale the inputs to these functions. Since there are $2pi$ radians in a circle and $360$ degrees, converting degrees to radians involves multiplying by $frac{180}{pi}$, so we can define the function $text{sind}(a) = sin(frac{180}{pi}a)$ for an angle $a$ written in degrees.






                    share|cite|improve this answer









                    $endgroup$



                    The $sin$ function and it's kin can be defined in many ways, such as using Taylor series or differential equations. Like any other function on the reals, they are just a way of turning numbers into numbers. It does have some special properties, like $sin(0) = 0$, $sin(-x) = -sin(x)$, $sin(x+2pi) = sin(x)$, to name a few.



                    When defining this function in terms of geometry of circles and angles, we often start with a discussion of degrees, because they are more familiar to us at the time. However, the true definition is in terms of radians. If you define $sintheta$ as the $y$-coordinate of the point on the unit circle reached by traveling a distance of $theta$ counter-clockwise along the circle, starting at $(1,0)$, it will be consistent with the definitions given by other means.



                    To make trigonometry work with degrees, we need to scale the inputs to these functions. Since there are $2pi$ radians in a circle and $360$ degrees, converting degrees to radians involves multiplying by $frac{180}{pi}$, so we can define the function $text{sind}(a) = sin(frac{180}{pi}a)$ for an angle $a$ written in degrees.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 6 '18 at 1:29









                    AlexanderJ93AlexanderJ93

                    6,158823




                    6,158823























                        0












                        $begingroup$

                        First of all this is not a dumb question at all. Hardy also asked and answered the very same question in his A Course of Pure Mathematics. It is rather the typical treatment of trigonometric functions in high school which very smartly avoids any discussion on this topic.



                        When you see expressions like $x^2,sqrt{x},log x, 2^x$ do you bother to ask "what is the $x$ in these expressions"? Normally not and you assume that $x$ is a real number in these expression. Getting the value of these algebraic expressions like $x^2,sqrt{x}$, given the value of $x$, is explained in typical courses of algebra. The story with $log x, 2^x, sin x$ is a bit different. Given a real number $x$, it is normally never explained (at least in high school), how to evaluate these expressions in $x$.



                        What you are really asking here is a definition of the sine function as a function of a real variable. The $x$ in $sin x$ is a real number number and there are ways to get the value of $sin x$, given the value of $x$. Unfortunately all the ways to do so are difficult. The easiest approach uses the geometric notion of a circle but the underlying assumptions are again difficult to prove.



                        So let's assume either of these facts:




                        • To every arc of a circle there corresponds a unique real number $L$ called its length.

                        • To every sector of a circle corresponds a unique real number $A$ called its area.


                        It is sufficient to use any one of the assumptions above and while proving these assumptions it can be established that $L=2A$ if the radius of the circle involved is unity.



                        The more popular definition of $sin x$ uses the assumption of length of arc of a circle. Thus let $x$ be any given real number. Consider the unit circle $X^2+Y^2=1$ in $XY$ coordinate plane and let $A=(1, 0), O=(0, 0)$. Start with point $A$ on circle and move on the cirle in counterclockwise (clockwise if $x<0$) direction till you cover the distance $|x|$. Mark this point on circle as $P_x$. Then by definition the point $P_x$ is $(cos x, sin x) $. While using this definition most people say that $x$ is the measure of angle $AOP$ in radians.



                        The geometric definition above can be translated into the language of calculus via the equation $$int_{0}^{sin x} frac{dt} {sqrt{1-t^2}}=x$$ for suitable values of $x$. And some university courses use the above integral to define $sin x$. There are other approaches to define $sin x$ which make use of infinite series but they are far removed from its geometrical link.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          First of all this is not a dumb question at all. Hardy also asked and answered the very same question in his A Course of Pure Mathematics. It is rather the typical treatment of trigonometric functions in high school which very smartly avoids any discussion on this topic.



                          When you see expressions like $x^2,sqrt{x},log x, 2^x$ do you bother to ask "what is the $x$ in these expressions"? Normally not and you assume that $x$ is a real number in these expression. Getting the value of these algebraic expressions like $x^2,sqrt{x}$, given the value of $x$, is explained in typical courses of algebra. The story with $log x, 2^x, sin x$ is a bit different. Given a real number $x$, it is normally never explained (at least in high school), how to evaluate these expressions in $x$.



                          What you are really asking here is a definition of the sine function as a function of a real variable. The $x$ in $sin x$ is a real number number and there are ways to get the value of $sin x$, given the value of $x$. Unfortunately all the ways to do so are difficult. The easiest approach uses the geometric notion of a circle but the underlying assumptions are again difficult to prove.



                          So let's assume either of these facts:




                          • To every arc of a circle there corresponds a unique real number $L$ called its length.

                          • To every sector of a circle corresponds a unique real number $A$ called its area.


                          It is sufficient to use any one of the assumptions above and while proving these assumptions it can be established that $L=2A$ if the radius of the circle involved is unity.



                          The more popular definition of $sin x$ uses the assumption of length of arc of a circle. Thus let $x$ be any given real number. Consider the unit circle $X^2+Y^2=1$ in $XY$ coordinate plane and let $A=(1, 0), O=(0, 0)$. Start with point $A$ on circle and move on the cirle in counterclockwise (clockwise if $x<0$) direction till you cover the distance $|x|$. Mark this point on circle as $P_x$. Then by definition the point $P_x$ is $(cos x, sin x) $. While using this definition most people say that $x$ is the measure of angle $AOP$ in radians.



                          The geometric definition above can be translated into the language of calculus via the equation $$int_{0}^{sin x} frac{dt} {sqrt{1-t^2}}=x$$ for suitable values of $x$. And some university courses use the above integral to define $sin x$. There are other approaches to define $sin x$ which make use of infinite series but they are far removed from its geometrical link.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            First of all this is not a dumb question at all. Hardy also asked and answered the very same question in his A Course of Pure Mathematics. It is rather the typical treatment of trigonometric functions in high school which very smartly avoids any discussion on this topic.



                            When you see expressions like $x^2,sqrt{x},log x, 2^x$ do you bother to ask "what is the $x$ in these expressions"? Normally not and you assume that $x$ is a real number in these expression. Getting the value of these algebraic expressions like $x^2,sqrt{x}$, given the value of $x$, is explained in typical courses of algebra. The story with $log x, 2^x, sin x$ is a bit different. Given a real number $x$, it is normally never explained (at least in high school), how to evaluate these expressions in $x$.



                            What you are really asking here is a definition of the sine function as a function of a real variable. The $x$ in $sin x$ is a real number number and there are ways to get the value of $sin x$, given the value of $x$. Unfortunately all the ways to do so are difficult. The easiest approach uses the geometric notion of a circle but the underlying assumptions are again difficult to prove.



                            So let's assume either of these facts:




                            • To every arc of a circle there corresponds a unique real number $L$ called its length.

                            • To every sector of a circle corresponds a unique real number $A$ called its area.


                            It is sufficient to use any one of the assumptions above and while proving these assumptions it can be established that $L=2A$ if the radius of the circle involved is unity.



                            The more popular definition of $sin x$ uses the assumption of length of arc of a circle. Thus let $x$ be any given real number. Consider the unit circle $X^2+Y^2=1$ in $XY$ coordinate plane and let $A=(1, 0), O=(0, 0)$. Start with point $A$ on circle and move on the cirle in counterclockwise (clockwise if $x<0$) direction till you cover the distance $|x|$. Mark this point on circle as $P_x$. Then by definition the point $P_x$ is $(cos x, sin x) $. While using this definition most people say that $x$ is the measure of angle $AOP$ in radians.



                            The geometric definition above can be translated into the language of calculus via the equation $$int_{0}^{sin x} frac{dt} {sqrt{1-t^2}}=x$$ for suitable values of $x$. And some university courses use the above integral to define $sin x$. There are other approaches to define $sin x$ which make use of infinite series but they are far removed from its geometrical link.






                            share|cite|improve this answer











                            $endgroup$



                            First of all this is not a dumb question at all. Hardy also asked and answered the very same question in his A Course of Pure Mathematics. It is rather the typical treatment of trigonometric functions in high school which very smartly avoids any discussion on this topic.



                            When you see expressions like $x^2,sqrt{x},log x, 2^x$ do you bother to ask "what is the $x$ in these expressions"? Normally not and you assume that $x$ is a real number in these expression. Getting the value of these algebraic expressions like $x^2,sqrt{x}$, given the value of $x$, is explained in typical courses of algebra. The story with $log x, 2^x, sin x$ is a bit different. Given a real number $x$, it is normally never explained (at least in high school), how to evaluate these expressions in $x$.



                            What you are really asking here is a definition of the sine function as a function of a real variable. The $x$ in $sin x$ is a real number number and there are ways to get the value of $sin x$, given the value of $x$. Unfortunately all the ways to do so are difficult. The easiest approach uses the geometric notion of a circle but the underlying assumptions are again difficult to prove.



                            So let's assume either of these facts:




                            • To every arc of a circle there corresponds a unique real number $L$ called its length.

                            • To every sector of a circle corresponds a unique real number $A$ called its area.


                            It is sufficient to use any one of the assumptions above and while proving these assumptions it can be established that $L=2A$ if the radius of the circle involved is unity.



                            The more popular definition of $sin x$ uses the assumption of length of arc of a circle. Thus let $x$ be any given real number. Consider the unit circle $X^2+Y^2=1$ in $XY$ coordinate plane and let $A=(1, 0), O=(0, 0)$. Start with point $A$ on circle and move on the cirle in counterclockwise (clockwise if $x<0$) direction till you cover the distance $|x|$. Mark this point on circle as $P_x$. Then by definition the point $P_x$ is $(cos x, sin x) $. While using this definition most people say that $x$ is the measure of angle $AOP$ in radians.



                            The geometric definition above can be translated into the language of calculus via the equation $$int_{0}^{sin x} frac{dt} {sqrt{1-t^2}}=x$$ for suitable values of $x$. And some university courses use the above integral to define $sin x$. There are other approaches to define $sin x$ which make use of infinite series but they are far removed from its geometrical link.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 6 '18 at 3:23

























                            answered Dec 6 '18 at 3:11









                            Paramanand SinghParamanand Singh

                            49.7k556163




                            49.7k556163















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