Geometric proof for trigonometric angle sum formulas.
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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 $alpha-beta$ :
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
$endgroup$
|
show 4 more comments
$begingroup$
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 $alpha-beta$ :
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
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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 $alpha-beta$ :
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
$endgroup$
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 $alpha-beta$ :
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
trigonometry
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
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$begingroup$
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