Geometric proof for trigonometric angle sum formulas.












3












$begingroup$


For some reason I'm starting to get nit picky about everything I have learned. In Geometry/ Algebra I learned how to prove the angle sum formulas, which follow from this picture (and a slightly different picture with $a$ and $b$ instead of $beta$ and $alpha-beta$ :



enter image description here



But now that I realize it this argument only works if the angle sum $x$ follows the inequality $0<x<frac{pi}{2}$. And every angle is larger than $0$, in radians.



How can we geometrically prove (to a high school student taking algebra/geometry) the trigonometric sum formulas but for all angles?



Definition



Here is what I'll take to be the definition of sine and cosine (if you know of another definition I'd appreciate it if you share):



Let $(x,y)$ be a point in the Cartesian plane, let $theta$ be the counterclockwise angle in radians from the positive $x$ axis to the segment connecting $(x,y)$ to the origin. Let a clockwise angle of $theta$ be equivalent to a counterclockwise angle of $-theta$, and vice versa. Take "sin" and "cos" to be $2pi$ periodic functions. Let $r$ be the distance from $(x,y)$ to the origin. In other words $r=sqrt{x^2+y^2}$. Then define:



$$sin (theta)=frac{y}{r}$$



$$cos (theta)=frac{x}{r}$$










share|cite|improve this question











$endgroup$












  • $begingroup$
    This is very simple to do, if you are willing to accept identities like $sin(90-x)=sin(90+x)$ and such. If you aren't, the proof for this is very straightforward after drawing the unit circle.
    $endgroup$
    – Anonymous Pi
    Mar 21 '17 at 21:58






  • 1




    $begingroup$
    You have to go beyond that definition because if you write $sin theta=y/r$ you are saying that $sin theta >0$ and then you can't prove anything for an angle $pi <theta <2pi$
    $endgroup$
    – Arnaldo
    Mar 21 '17 at 22:04








  • 1




    $begingroup$
    I have added my definitions please let me know if anything is unnecessary to be defined, and if I need to add anything @AnonymousPi
    $endgroup$
    – Ahmed S. Attaalla
    Mar 21 '17 at 22:40






  • 1




    $begingroup$
    This answer of mine ---and this one, on which it is based--- may be helpful.
    $endgroup$
    – Blue
    Mar 21 '17 at 22:41








  • 2




    $begingroup$
    Not to go off on too much of a tangent, but I've seen some authors (Rudin, I believe) define $cos(x)$ and $sin(x)$ functions as the real and complex part, respectively, of $e^{ix}$ (or the series representation of such). I am not aware how much that helps here, but I would be interested to know how many trig identities could be shown by this route.
    $endgroup$
    – Thomas Rasberry
    Mar 21 '17 at 22:51
















3












$begingroup$


For some reason I'm starting to get nit picky about everything I have learned. In Geometry/ Algebra I learned how to prove the angle sum formulas, which follow from this picture (and a slightly different picture with $a$ and $b$ instead of $beta$ and $alpha-beta$ :



enter image description here



But now that I realize it this argument only works if the angle sum $x$ follows the inequality $0<x<frac{pi}{2}$. And every angle is larger than $0$, in radians.



How can we geometrically prove (to a high school student taking algebra/geometry) the trigonometric sum formulas but for all angles?



Definition



Here is what I'll take to be the definition of sine and cosine (if you know of another definition I'd appreciate it if you share):



Let $(x,y)$ be a point in the Cartesian plane, let $theta$ be the counterclockwise angle in radians from the positive $x$ axis to the segment connecting $(x,y)$ to the origin. Let a clockwise angle of $theta$ be equivalent to a counterclockwise angle of $-theta$, and vice versa. Take "sin" and "cos" to be $2pi$ periodic functions. Let $r$ be the distance from $(x,y)$ to the origin. In other words $r=sqrt{x^2+y^2}$. Then define:



$$sin (theta)=frac{y}{r}$$



$$cos (theta)=frac{x}{r}$$










share|cite|improve this question











$endgroup$












  • $begingroup$
    This is very simple to do, if you are willing to accept identities like $sin(90-x)=sin(90+x)$ and such. If you aren't, the proof for this is very straightforward after drawing the unit circle.
    $endgroup$
    – Anonymous Pi
    Mar 21 '17 at 21:58






  • 1




    $begingroup$
    You have to go beyond that definition because if you write $sin theta=y/r$ you are saying that $sin theta >0$ and then you can't prove anything for an angle $pi <theta <2pi$
    $endgroup$
    – Arnaldo
    Mar 21 '17 at 22:04








  • 1




    $begingroup$
    I have added my definitions please let me know if anything is unnecessary to be defined, and if I need to add anything @AnonymousPi
    $endgroup$
    – Ahmed S. Attaalla
    Mar 21 '17 at 22:40






  • 1




    $begingroup$
    This answer of mine ---and this one, on which it is based--- may be helpful.
    $endgroup$
    – Blue
    Mar 21 '17 at 22:41








  • 2




    $begingroup$
    Not to go off on too much of a tangent, but I've seen some authors (Rudin, I believe) define $cos(x)$ and $sin(x)$ functions as the real and complex part, respectively, of $e^{ix}$ (or the series representation of such). I am not aware how much that helps here, but I would be interested to know how many trig identities could be shown by this route.
    $endgroup$
    – Thomas Rasberry
    Mar 21 '17 at 22:51














3












3








3





$begingroup$


For some reason I'm starting to get nit picky about everything I have learned. In Geometry/ Algebra I learned how to prove the angle sum formulas, which follow from this picture (and a slightly different picture with $a$ and $b$ instead of $beta$ and $alpha-beta$ :



enter image description here



But now that I realize it this argument only works if the angle sum $x$ follows the inequality $0<x<frac{pi}{2}$. And every angle is larger than $0$, in radians.



How can we geometrically prove (to a high school student taking algebra/geometry) the trigonometric sum formulas but for all angles?



Definition



Here is what I'll take to be the definition of sine and cosine (if you know of another definition I'd appreciate it if you share):



Let $(x,y)$ be a point in the Cartesian plane, let $theta$ be the counterclockwise angle in radians from the positive $x$ axis to the segment connecting $(x,y)$ to the origin. Let a clockwise angle of $theta$ be equivalent to a counterclockwise angle of $-theta$, and vice versa. Take "sin" and "cos" to be $2pi$ periodic functions. Let $r$ be the distance from $(x,y)$ to the origin. In other words $r=sqrt{x^2+y^2}$. Then define:



$$sin (theta)=frac{y}{r}$$



$$cos (theta)=frac{x}{r}$$










share|cite|improve this question











$endgroup$




For some reason I'm starting to get nit picky about everything I have learned. In Geometry/ Algebra I learned how to prove the angle sum formulas, which follow from this picture (and a slightly different picture with $a$ and $b$ instead of $beta$ and $alpha-beta$ :



enter image description here



But now that I realize it this argument only works if the angle sum $x$ follows the inequality $0<x<frac{pi}{2}$. And every angle is larger than $0$, in radians.



How can we geometrically prove (to a high school student taking algebra/geometry) the trigonometric sum formulas but for all angles?



Definition



Here is what I'll take to be the definition of sine and cosine (if you know of another definition I'd appreciate it if you share):



Let $(x,y)$ be a point in the Cartesian plane, let $theta$ be the counterclockwise angle in radians from the positive $x$ axis to the segment connecting $(x,y)$ to the origin. Let a clockwise angle of $theta$ be equivalent to a counterclockwise angle of $-theta$, and vice versa. Take "sin" and "cos" to be $2pi$ periodic functions. Let $r$ be the distance from $(x,y)$ to the origin. In other words $r=sqrt{x^2+y^2}$. Then define:



$$sin (theta)=frac{y}{r}$$



$$cos (theta)=frac{x}{r}$$







trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 '17 at 22:42







Ahmed S. Attaalla

















asked Mar 21 '17 at 21:56









Ahmed S. AttaallaAhmed S. Attaalla

14.8k12049




14.8k12049












  • $begingroup$
    This is very simple to do, if you are willing to accept identities like $sin(90-x)=sin(90+x)$ and such. If you aren't, the proof for this is very straightforward after drawing the unit circle.
    $endgroup$
    – Anonymous Pi
    Mar 21 '17 at 21:58






  • 1




    $begingroup$
    You have to go beyond that definition because if you write $sin theta=y/r$ you are saying that $sin theta >0$ and then you can't prove anything for an angle $pi <theta <2pi$
    $endgroup$
    – Arnaldo
    Mar 21 '17 at 22:04








  • 1




    $begingroup$
    I have added my definitions please let me know if anything is unnecessary to be defined, and if I need to add anything @AnonymousPi
    $endgroup$
    – Ahmed S. Attaalla
    Mar 21 '17 at 22:40






  • 1




    $begingroup$
    This answer of mine ---and this one, on which it is based--- may be helpful.
    $endgroup$
    – Blue
    Mar 21 '17 at 22:41








  • 2




    $begingroup$
    Not to go off on too much of a tangent, but I've seen some authors (Rudin, I believe) define $cos(x)$ and $sin(x)$ functions as the real and complex part, respectively, of $e^{ix}$ (or the series representation of such). I am not aware how much that helps here, but I would be interested to know how many trig identities could be shown by this route.
    $endgroup$
    – Thomas Rasberry
    Mar 21 '17 at 22:51


















  • $begingroup$
    This is very simple to do, if you are willing to accept identities like $sin(90-x)=sin(90+x)$ and such. If you aren't, the proof for this is very straightforward after drawing the unit circle.
    $endgroup$
    – Anonymous Pi
    Mar 21 '17 at 21:58






  • 1




    $begingroup$
    You have to go beyond that definition because if you write $sin theta=y/r$ you are saying that $sin theta >0$ and then you can't prove anything for an angle $pi <theta <2pi$
    $endgroup$
    – Arnaldo
    Mar 21 '17 at 22:04








  • 1




    $begingroup$
    I have added my definitions please let me know if anything is unnecessary to be defined, and if I need to add anything @AnonymousPi
    $endgroup$
    – Ahmed S. Attaalla
    Mar 21 '17 at 22:40






  • 1




    $begingroup$
    This answer of mine ---and this one, on which it is based--- may be helpful.
    $endgroup$
    – Blue
    Mar 21 '17 at 22:41








  • 2




    $begingroup$
    Not to go off on too much of a tangent, but I've seen some authors (Rudin, I believe) define $cos(x)$ and $sin(x)$ functions as the real and complex part, respectively, of $e^{ix}$ (or the series representation of such). I am not aware how much that helps here, but I would be interested to know how many trig identities could be shown by this route.
    $endgroup$
    – Thomas Rasberry
    Mar 21 '17 at 22:51
















$begingroup$
This is very simple to do, if you are willing to accept identities like $sin(90-x)=sin(90+x)$ and such. If you aren't, the proof for this is very straightforward after drawing the unit circle.
$endgroup$
– Anonymous Pi
Mar 21 '17 at 21:58




$begingroup$
This is very simple to do, if you are willing to accept identities like $sin(90-x)=sin(90+x)$ and such. If you aren't, the proof for this is very straightforward after drawing the unit circle.
$endgroup$
– Anonymous Pi
Mar 21 '17 at 21:58




1




1




$begingroup$
You have to go beyond that definition because if you write $sin theta=y/r$ you are saying that $sin theta >0$ and then you can't prove anything for an angle $pi <theta <2pi$
$endgroup$
– Arnaldo
Mar 21 '17 at 22:04






$begingroup$
You have to go beyond that definition because if you write $sin theta=y/r$ you are saying that $sin theta >0$ and then you can't prove anything for an angle $pi <theta <2pi$
$endgroup$
– Arnaldo
Mar 21 '17 at 22:04






1




1




$begingroup$
I have added my definitions please let me know if anything is unnecessary to be defined, and if I need to add anything @AnonymousPi
$endgroup$
– Ahmed S. Attaalla
Mar 21 '17 at 22:40




$begingroup$
I have added my definitions please let me know if anything is unnecessary to be defined, and if I need to add anything @AnonymousPi
$endgroup$
– Ahmed S. Attaalla
Mar 21 '17 at 22:40




1




1




$begingroup$
This answer of mine ---and this one, on which it is based--- may be helpful.
$endgroup$
– Blue
Mar 21 '17 at 22:41






$begingroup$
This answer of mine ---and this one, on which it is based--- may be helpful.
$endgroup$
– Blue
Mar 21 '17 at 22:41






2




2




$begingroup$
Not to go off on too much of a tangent, but I've seen some authors (Rudin, I believe) define $cos(x)$ and $sin(x)$ functions as the real and complex part, respectively, of $e^{ix}$ (or the series representation of such). I am not aware how much that helps here, but I would be interested to know how many trig identities could be shown by this route.
$endgroup$
– Thomas Rasberry
Mar 21 '17 at 22:51




$begingroup$
Not to go off on too much of a tangent, but I've seen some authors (Rudin, I believe) define $cos(x)$ and $sin(x)$ functions as the real and complex part, respectively, of $e^{ix}$ (or the series representation of such). I am not aware how much that helps here, but I would be interested to know how many trig identities could be shown by this route.
$endgroup$
– Thomas Rasberry
Mar 21 '17 at 22:51










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