Proving Trigonometric “Definitions” [on hold] The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Derive the Trigonometric FunctionsDifferent definitions of trigonometric functionsProving simple Trigonometric identityProving simple trigonometric identityProving a trigonometric equationTrigonometric Ratios for angles greater than 90 degrees and the Unit CircleProving this trigonometric resultProblem proving a trigonometric result.Trigonometric Ratios for angles greater than 90 degrees in unit circleWhy do we need so many trigonometric definitions?Proving trigonometric problem from given trigonometric equations

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Proving Trigonometric “Definitions” [on hold]



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Derive the Trigonometric FunctionsDifferent definitions of trigonometric functionsProving simple Trigonometric identityProving simple trigonometric identityProving a trigonometric equationTrigonometric Ratios for angles greater than 90 degrees and the Unit CircleProving this trigonometric resultProblem proving a trigonometric result.Trigonometric Ratios for angles greater than 90 degrees in unit circleWhy do we need so many trigonometric definitions?Proving trigonometric problem from given trigonometric equations










6












$begingroup$


The expression trigonometric “definitions” refers here, rather narrowly, to statements expressing stable relations between the sides of right triangle.



Thus, for instance, the traditional definition of sine supposes that one has a demonstration that the ratios between opposite side and hypothenuses are independent of the size of the right triangle and are dependent instead on the amplitude of the angles of the right triangle.



My question is the following: How these stable relations, simply assumed by most trigonometric “definitions,” can actually be demonstrated?



A very similar question has been posted here but I remain unsatisfied by most answers — which seem to convoke either unnecessary complex mathematical objects or simplistic historical accounts.



I guess, what I am looking for, is something like a geometrical proof — but am open to others!










share|cite|improve this question











$endgroup$



put on hold as unclear what you're asking by Lord Shark the Unknown, Adrian Keister, José Carlos Santos, Alexander Gruber yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















  • $begingroup$
    You'd like to see a geometric proof of what exactly? The sine and cosine of angles are defined geometrically (with either triangles or a unit circle). You can't really "prove" this, since it's a definition. You can prove relations between them, though.
    $endgroup$
    – John Doe
    yesterday






  • 1




    $begingroup$
    @JohnDoe The OP is asking for a proof that the trig functions are well defined (do not depend on the triangle you use)
    $endgroup$
    – Ovi
    yesterday










  • $begingroup$
    @Ovi, Indeed. Thanks. I reframed the question.
    $endgroup$
    – Touki
    yesterday










  • $begingroup$
    It sounds like you doubt the obvious properties of similar triangles
    $endgroup$
    – MPW
    yesterday






  • 2




    $begingroup$
    I have nominated this question to be re-opened. Given an upvoted and accepted answer, demonstrating that the question was sufficiently clear, closing it seems to have been pointless.
    $endgroup$
    – Lee Mosher
    yesterday















6












$begingroup$


The expression trigonometric “definitions” refers here, rather narrowly, to statements expressing stable relations between the sides of right triangle.



Thus, for instance, the traditional definition of sine supposes that one has a demonstration that the ratios between opposite side and hypothenuses are independent of the size of the right triangle and are dependent instead on the amplitude of the angles of the right triangle.



My question is the following: How these stable relations, simply assumed by most trigonometric “definitions,” can actually be demonstrated?



A very similar question has been posted here but I remain unsatisfied by most answers — which seem to convoke either unnecessary complex mathematical objects or simplistic historical accounts.



I guess, what I am looking for, is something like a geometrical proof — but am open to others!










share|cite|improve this question











$endgroup$



put on hold as unclear what you're asking by Lord Shark the Unknown, Adrian Keister, José Carlos Santos, Alexander Gruber yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















  • $begingroup$
    You'd like to see a geometric proof of what exactly? The sine and cosine of angles are defined geometrically (with either triangles or a unit circle). You can't really "prove" this, since it's a definition. You can prove relations between them, though.
    $endgroup$
    – John Doe
    yesterday






  • 1




    $begingroup$
    @JohnDoe The OP is asking for a proof that the trig functions are well defined (do not depend on the triangle you use)
    $endgroup$
    – Ovi
    yesterday










  • $begingroup$
    @Ovi, Indeed. Thanks. I reframed the question.
    $endgroup$
    – Touki
    yesterday










  • $begingroup$
    It sounds like you doubt the obvious properties of similar triangles
    $endgroup$
    – MPW
    yesterday






  • 2




    $begingroup$
    I have nominated this question to be re-opened. Given an upvoted and accepted answer, demonstrating that the question was sufficiently clear, closing it seems to have been pointless.
    $endgroup$
    – Lee Mosher
    yesterday













6












6








6





$begingroup$


The expression trigonometric “definitions” refers here, rather narrowly, to statements expressing stable relations between the sides of right triangle.



Thus, for instance, the traditional definition of sine supposes that one has a demonstration that the ratios between opposite side and hypothenuses are independent of the size of the right triangle and are dependent instead on the amplitude of the angles of the right triangle.



My question is the following: How these stable relations, simply assumed by most trigonometric “definitions,” can actually be demonstrated?



A very similar question has been posted here but I remain unsatisfied by most answers — which seem to convoke either unnecessary complex mathematical objects or simplistic historical accounts.



I guess, what I am looking for, is something like a geometrical proof — but am open to others!










share|cite|improve this question











$endgroup$




The expression trigonometric “definitions” refers here, rather narrowly, to statements expressing stable relations between the sides of right triangle.



Thus, for instance, the traditional definition of sine supposes that one has a demonstration that the ratios between opposite side and hypothenuses are independent of the size of the right triangle and are dependent instead on the amplitude of the angles of the right triangle.



My question is the following: How these stable relations, simply assumed by most trigonometric “definitions,” can actually be demonstrated?



A very similar question has been posted here but I remain unsatisfied by most answers — which seem to convoke either unnecessary complex mathematical objects or simplistic historical accounts.



I guess, what I am looking for, is something like a geometrical proof — but am open to others!







trigonometry proof-writing definition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 23 hours ago







Touki

















asked yesterday









ToukiTouki

1364




1364




put on hold as unclear what you're asking by Lord Shark the Unknown, Adrian Keister, José Carlos Santos, Alexander Gruber yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









put on hold as unclear what you're asking by Lord Shark the Unknown, Adrian Keister, José Carlos Santos, Alexander Gruber yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    You'd like to see a geometric proof of what exactly? The sine and cosine of angles are defined geometrically (with either triangles or a unit circle). You can't really "prove" this, since it's a definition. You can prove relations between them, though.
    $endgroup$
    – John Doe
    yesterday






  • 1




    $begingroup$
    @JohnDoe The OP is asking for a proof that the trig functions are well defined (do not depend on the triangle you use)
    $endgroup$
    – Ovi
    yesterday










  • $begingroup$
    @Ovi, Indeed. Thanks. I reframed the question.
    $endgroup$
    – Touki
    yesterday










  • $begingroup$
    It sounds like you doubt the obvious properties of similar triangles
    $endgroup$
    – MPW
    yesterday






  • 2




    $begingroup$
    I have nominated this question to be re-opened. Given an upvoted and accepted answer, demonstrating that the question was sufficiently clear, closing it seems to have been pointless.
    $endgroup$
    – Lee Mosher
    yesterday
















  • $begingroup$
    You'd like to see a geometric proof of what exactly? The sine and cosine of angles are defined geometrically (with either triangles or a unit circle). You can't really "prove" this, since it's a definition. You can prove relations between them, though.
    $endgroup$
    – John Doe
    yesterday






  • 1




    $begingroup$
    @JohnDoe The OP is asking for a proof that the trig functions are well defined (do not depend on the triangle you use)
    $endgroup$
    – Ovi
    yesterday










  • $begingroup$
    @Ovi, Indeed. Thanks. I reframed the question.
    $endgroup$
    – Touki
    yesterday










  • $begingroup$
    It sounds like you doubt the obvious properties of similar triangles
    $endgroup$
    – MPW
    yesterday






  • 2




    $begingroup$
    I have nominated this question to be re-opened. Given an upvoted and accepted answer, demonstrating that the question was sufficiently clear, closing it seems to have been pointless.
    $endgroup$
    – Lee Mosher
    yesterday















$begingroup$
You'd like to see a geometric proof of what exactly? The sine and cosine of angles are defined geometrically (with either triangles or a unit circle). You can't really "prove" this, since it's a definition. You can prove relations between them, though.
$endgroup$
– John Doe
yesterday




$begingroup$
You'd like to see a geometric proof of what exactly? The sine and cosine of angles are defined geometrically (with either triangles or a unit circle). You can't really "prove" this, since it's a definition. You can prove relations between them, though.
$endgroup$
– John Doe
yesterday




1




1




$begingroup$
@JohnDoe The OP is asking for a proof that the trig functions are well defined (do not depend on the triangle you use)
$endgroup$
– Ovi
yesterday




$begingroup$
@JohnDoe The OP is asking for a proof that the trig functions are well defined (do not depend on the triangle you use)
$endgroup$
– Ovi
yesterday












$begingroup$
@Ovi, Indeed. Thanks. I reframed the question.
$endgroup$
– Touki
yesterday




$begingroup$
@Ovi, Indeed. Thanks. I reframed the question.
$endgroup$
– Touki
yesterday












$begingroup$
It sounds like you doubt the obvious properties of similar triangles
$endgroup$
– MPW
yesterday




$begingroup$
It sounds like you doubt the obvious properties of similar triangles
$endgroup$
– MPW
yesterday




2




2




$begingroup$
I have nominated this question to be re-opened. Given an upvoted and accepted answer, demonstrating that the question was sufficiently clear, closing it seems to have been pointless.
$endgroup$
– Lee Mosher
yesterday




$begingroup$
I have nominated this question to be re-opened. Given an upvoted and accepted answer, demonstrating that the question was sufficiently clear, closing it seems to have been pointless.
$endgroup$
– Lee Mosher
yesterday










2 Answers
2






active

oldest

votes


















9












$begingroup$

For proving that the sine function is well-defined, one uses two theorems from Euclidean geometry combined with a tiny bit of algebra. The two theorems are:




Theorem 1: In any triangle, the sum of the angles equals $pi$.




I don't actually care about the numerical value of the sum, so perhaps one can state Theorem 1 more classically: the sum of the angles of any triangle is equal to the sum of two right angles. In any case, all that I'll use is that the sums of the angles of any two triangles are equal.




Theorem 2 (The "Angle-Angle-Angle" theorem): For any two triangles $triangle ABC$ and $triangle A'B'C'$, if $angle ABC = angle A'B'C'$ are congruent, and if $angle BCA = angle B'C'A'$ are congruent, and if $angle CAB = angle C'A'B'$ are congruent, then the triangles are similar. In more detail, this means that we have equality of ratios
$$textLength(overlineAB) bigm/ textLength(overlineA'B') = textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
$$




So now let's consider two right triangles $triangle ABC$ and $triangle A'B'C'$, such that the $angle ABC$ and $angle A'B'C'$ are right angles. It follows that $angle ABC = angle A'B'C'$.



Suppose also that $angle CAB = angle C'A'B'$. By applying Theorem 1, it follows that $angle BCA = angle B'C'A'$. The hypotheses of Theorem 2 have therefore been verified, so its conclusions are true. From the equation
$$textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
$$

we deduce, by a tiny bit of algebra, that
$$textLength(overlineBC) bigm/ textLength(overlineCA) = textLength(overlineB'C') bigm/ textLength(overlineC'A')
$$

In words, this says that if in triangle $ABC$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, and in triangle $A'B'C'$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, we get the same number. That number is the sine of the angle $A$.



This proves that the sine of an angle is well-defined no matter what right triangle we use for its calculation.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    This makes the case! Thanks @Lee Mosher.
    $endgroup$
    – Touki
    23 hours ago










  • $begingroup$
    By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
    $endgroup$
    – Lee Mosher
    14 hours ago


















1












$begingroup$

You should look up similarity. In particular, similarity of triangles, and especially right triangles.



It is because any two right triangles containing the same angle are similar that the functions depend only on the angle in question.



As for why right triangles. Why not, say, isosceles triangles, since these also allow one to define a certain trig function of the vertex angle uniquely, especially as they are laterally symmetric? Well, there's no reason why one couldn't do that; indeed, the first compiler of trig tables compiled them for isosceles triangles, and called them chords of the vertex angle. However, one can argue that that would not be as fundamental as using right triangles for the simple reason that whereas not all triangles can be divided into two isosceles triangles, all can always be partitioned into two right triangles. This keeps things simpler. NB. What was called the chord of $x$ is now double the sine of $x/2.$






share|cite|improve this answer











$endgroup$



















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    For proving that the sine function is well-defined, one uses two theorems from Euclidean geometry combined with a tiny bit of algebra. The two theorems are:




    Theorem 1: In any triangle, the sum of the angles equals $pi$.




    I don't actually care about the numerical value of the sum, so perhaps one can state Theorem 1 more classically: the sum of the angles of any triangle is equal to the sum of two right angles. In any case, all that I'll use is that the sums of the angles of any two triangles are equal.




    Theorem 2 (The "Angle-Angle-Angle" theorem): For any two triangles $triangle ABC$ and $triangle A'B'C'$, if $angle ABC = angle A'B'C'$ are congruent, and if $angle BCA = angle B'C'A'$ are congruent, and if $angle CAB = angle C'A'B'$ are congruent, then the triangles are similar. In more detail, this means that we have equality of ratios
    $$textLength(overlineAB) bigm/ textLength(overlineA'B') = textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$




    So now let's consider two right triangles $triangle ABC$ and $triangle A'B'C'$, such that the $angle ABC$ and $angle A'B'C'$ are right angles. It follows that $angle ABC = angle A'B'C'$.



    Suppose also that $angle CAB = angle C'A'B'$. By applying Theorem 1, it follows that $angle BCA = angle B'C'A'$. The hypotheses of Theorem 2 have therefore been verified, so its conclusions are true. From the equation
    $$textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$

    we deduce, by a tiny bit of algebra, that
    $$textLength(overlineBC) bigm/ textLength(overlineCA) = textLength(overlineB'C') bigm/ textLength(overlineC'A')
    $$

    In words, this says that if in triangle $ABC$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, and in triangle $A'B'C'$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, we get the same number. That number is the sine of the angle $A$.



    This proves that the sine of an angle is well-defined no matter what right triangle we use for its calculation.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      This makes the case! Thanks @Lee Mosher.
      $endgroup$
      – Touki
      23 hours ago










    • $begingroup$
      By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
      $endgroup$
      – Lee Mosher
      14 hours ago















    9












    $begingroup$

    For proving that the sine function is well-defined, one uses two theorems from Euclidean geometry combined with a tiny bit of algebra. The two theorems are:




    Theorem 1: In any triangle, the sum of the angles equals $pi$.




    I don't actually care about the numerical value of the sum, so perhaps one can state Theorem 1 more classically: the sum of the angles of any triangle is equal to the sum of two right angles. In any case, all that I'll use is that the sums of the angles of any two triangles are equal.




    Theorem 2 (The "Angle-Angle-Angle" theorem): For any two triangles $triangle ABC$ and $triangle A'B'C'$, if $angle ABC = angle A'B'C'$ are congruent, and if $angle BCA = angle B'C'A'$ are congruent, and if $angle CAB = angle C'A'B'$ are congruent, then the triangles are similar. In more detail, this means that we have equality of ratios
    $$textLength(overlineAB) bigm/ textLength(overlineA'B') = textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$




    So now let's consider two right triangles $triangle ABC$ and $triangle A'B'C'$, such that the $angle ABC$ and $angle A'B'C'$ are right angles. It follows that $angle ABC = angle A'B'C'$.



    Suppose also that $angle CAB = angle C'A'B'$. By applying Theorem 1, it follows that $angle BCA = angle B'C'A'$. The hypotheses of Theorem 2 have therefore been verified, so its conclusions are true. From the equation
    $$textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$

    we deduce, by a tiny bit of algebra, that
    $$textLength(overlineBC) bigm/ textLength(overlineCA) = textLength(overlineB'C') bigm/ textLength(overlineC'A')
    $$

    In words, this says that if in triangle $ABC$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, and in triangle $A'B'C'$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, we get the same number. That number is the sine of the angle $A$.



    This proves that the sine of an angle is well-defined no matter what right triangle we use for its calculation.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      This makes the case! Thanks @Lee Mosher.
      $endgroup$
      – Touki
      23 hours ago










    • $begingroup$
      By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
      $endgroup$
      – Lee Mosher
      14 hours ago













    9












    9








    9





    $begingroup$

    For proving that the sine function is well-defined, one uses two theorems from Euclidean geometry combined with a tiny bit of algebra. The two theorems are:




    Theorem 1: In any triangle, the sum of the angles equals $pi$.




    I don't actually care about the numerical value of the sum, so perhaps one can state Theorem 1 more classically: the sum of the angles of any triangle is equal to the sum of two right angles. In any case, all that I'll use is that the sums of the angles of any two triangles are equal.




    Theorem 2 (The "Angle-Angle-Angle" theorem): For any two triangles $triangle ABC$ and $triangle A'B'C'$, if $angle ABC = angle A'B'C'$ are congruent, and if $angle BCA = angle B'C'A'$ are congruent, and if $angle CAB = angle C'A'B'$ are congruent, then the triangles are similar. In more detail, this means that we have equality of ratios
    $$textLength(overlineAB) bigm/ textLength(overlineA'B') = textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$




    So now let's consider two right triangles $triangle ABC$ and $triangle A'B'C'$, such that the $angle ABC$ and $angle A'B'C'$ are right angles. It follows that $angle ABC = angle A'B'C'$.



    Suppose also that $angle CAB = angle C'A'B'$. By applying Theorem 1, it follows that $angle BCA = angle B'C'A'$. The hypotheses of Theorem 2 have therefore been verified, so its conclusions are true. From the equation
    $$textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$

    we deduce, by a tiny bit of algebra, that
    $$textLength(overlineBC) bigm/ textLength(overlineCA) = textLength(overlineB'C') bigm/ textLength(overlineC'A')
    $$

    In words, this says that if in triangle $ABC$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, and in triangle $A'B'C'$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, we get the same number. That number is the sine of the angle $A$.



    This proves that the sine of an angle is well-defined no matter what right triangle we use for its calculation.






    share|cite|improve this answer









    $endgroup$



    For proving that the sine function is well-defined, one uses two theorems from Euclidean geometry combined with a tiny bit of algebra. The two theorems are:




    Theorem 1: In any triangle, the sum of the angles equals $pi$.




    I don't actually care about the numerical value of the sum, so perhaps one can state Theorem 1 more classically: the sum of the angles of any triangle is equal to the sum of two right angles. In any case, all that I'll use is that the sums of the angles of any two triangles are equal.




    Theorem 2 (The "Angle-Angle-Angle" theorem): For any two triangles $triangle ABC$ and $triangle A'B'C'$, if $angle ABC = angle A'B'C'$ are congruent, and if $angle BCA = angle B'C'A'$ are congruent, and if $angle CAB = angle C'A'B'$ are congruent, then the triangles are similar. In more detail, this means that we have equality of ratios
    $$textLength(overlineAB) bigm/ textLength(overlineA'B') = textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$




    So now let's consider two right triangles $triangle ABC$ and $triangle A'B'C'$, such that the $angle ABC$ and $angle A'B'C'$ are right angles. It follows that $angle ABC = angle A'B'C'$.



    Suppose also that $angle CAB = angle C'A'B'$. By applying Theorem 1, it follows that $angle BCA = angle B'C'A'$. The hypotheses of Theorem 2 have therefore been verified, so its conclusions are true. From the equation
    $$textLength(overlineBC) bigm/ textLength(overlineB'C') = textLength(overlineCA) bigm/ textLength(overlineC'A')
    $$

    we deduce, by a tiny bit of algebra, that
    $$textLength(overlineBC) bigm/ textLength(overlineCA) = textLength(overlineB'C') bigm/ textLength(overlineC'A')
    $$

    In words, this says that if in triangle $ABC$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, and in triangle $A'B'C'$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, we get the same number. That number is the sine of the angle $A$.



    This proves that the sine of an angle is well-defined no matter what right triangle we use for its calculation.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    Lee MosherLee Mosher

    52.1k33890




    52.1k33890











    • $begingroup$
      This makes the case! Thanks @Lee Mosher.
      $endgroup$
      – Touki
      23 hours ago










    • $begingroup$
      By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
      $endgroup$
      – Lee Mosher
      14 hours ago
















    • $begingroup$
      This makes the case! Thanks @Lee Mosher.
      $endgroup$
      – Touki
      23 hours ago










    • $begingroup$
      By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
      $endgroup$
      – Lee Mosher
      14 hours ago















    $begingroup$
    This makes the case! Thanks @Lee Mosher.
    $endgroup$
    – Touki
    23 hours ago




    $begingroup$
    This makes the case! Thanks @Lee Mosher.
    $endgroup$
    – Touki
    23 hours ago












    $begingroup$
    By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
    $endgroup$
    – Lee Mosher
    14 hours ago




    $begingroup$
    By the way, something tells me that this may be the oldest argument I have ever written out explicitly. I suspect that this argument was known to the Babylonians.
    $endgroup$
    – Lee Mosher
    14 hours ago











    1












    $begingroup$

    You should look up similarity. In particular, similarity of triangles, and especially right triangles.



    It is because any two right triangles containing the same angle are similar that the functions depend only on the angle in question.



    As for why right triangles. Why not, say, isosceles triangles, since these also allow one to define a certain trig function of the vertex angle uniquely, especially as they are laterally symmetric? Well, there's no reason why one couldn't do that; indeed, the first compiler of trig tables compiled them for isosceles triangles, and called them chords of the vertex angle. However, one can argue that that would not be as fundamental as using right triangles for the simple reason that whereas not all triangles can be divided into two isosceles triangles, all can always be partitioned into two right triangles. This keeps things simpler. NB. What was called the chord of $x$ is now double the sine of $x/2.$






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      You should look up similarity. In particular, similarity of triangles, and especially right triangles.



      It is because any two right triangles containing the same angle are similar that the functions depend only on the angle in question.



      As for why right triangles. Why not, say, isosceles triangles, since these also allow one to define a certain trig function of the vertex angle uniquely, especially as they are laterally symmetric? Well, there's no reason why one couldn't do that; indeed, the first compiler of trig tables compiled them for isosceles triangles, and called them chords of the vertex angle. However, one can argue that that would not be as fundamental as using right triangles for the simple reason that whereas not all triangles can be divided into two isosceles triangles, all can always be partitioned into two right triangles. This keeps things simpler. NB. What was called the chord of $x$ is now double the sine of $x/2.$






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        You should look up similarity. In particular, similarity of triangles, and especially right triangles.



        It is because any two right triangles containing the same angle are similar that the functions depend only on the angle in question.



        As for why right triangles. Why not, say, isosceles triangles, since these also allow one to define a certain trig function of the vertex angle uniquely, especially as they are laterally symmetric? Well, there's no reason why one couldn't do that; indeed, the first compiler of trig tables compiled them for isosceles triangles, and called them chords of the vertex angle. However, one can argue that that would not be as fundamental as using right triangles for the simple reason that whereas not all triangles can be divided into two isosceles triangles, all can always be partitioned into two right triangles. This keeps things simpler. NB. What was called the chord of $x$ is now double the sine of $x/2.$






        share|cite|improve this answer











        $endgroup$



        You should look up similarity. In particular, similarity of triangles, and especially right triangles.



        It is because any two right triangles containing the same angle are similar that the functions depend only on the angle in question.



        As for why right triangles. Why not, say, isosceles triangles, since these also allow one to define a certain trig function of the vertex angle uniquely, especially as they are laterally symmetric? Well, there's no reason why one couldn't do that; indeed, the first compiler of trig tables compiled them for isosceles triangles, and called them chords of the vertex angle. However, one can argue that that would not be as fundamental as using right triangles for the simple reason that whereas not all triangles can be divided into two isosceles triangles, all can always be partitioned into two right triangles. This keeps things simpler. NB. What was called the chord of $x$ is now double the sine of $x/2.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        AllawonderAllawonder

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