Mixing
Prove the size of a hyperbolic angle is twice the area of its hyperbolic sector.
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I'm trying to figure out how the hyperbolic functions are derived using a unit hyperbola.
According to this walkthrough, argument u in (cosh(u), sinh(u)) should be equal to 2A, where A is the area of an intercepted hyperbolic sector from (0,0) to (cosh(u), sinh(u)).
Confused as to why this is defined as such, I found on Wikipedia:
The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.
However, I couldn't find a explanation for this anywhere. Can anyone help show me why this is?
calculus geometry trigonometry hyperbolic-functions
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
$endgroup$
add a comment |
$begingroup$
I'm trying to figure out how the hyperbolic functions are derived using a unit hyperbola.
According to this walkthrough, argument u in (cosh(u), sinh(u)) should be equal to 2A, where A is the area of an intercepted hyperbolic sector from (0,0) to (cosh(u), sinh(u)).
Confused as to why this is defined as such, I found on Wikipedia:
The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.
However, I couldn't find a explanation for this anywhere. Can anyone help show me why this is?
calculus geometry trigonometry hyperbolic-functions
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
$endgroup$
$begingroup$
See the last paragraph of this answer.
$endgroup$
– Blue
Jan 12 at 5:11
$begingroup$
The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started.
$endgroup$
– sqrtpapi2001
Jan 12 at 5:52
add a comment |
$begingroup$
I'm trying to figure out how the hyperbolic functions are derived using a unit hyperbola.
According to this walkthrough, argument u in (cosh(u), sinh(u)) should be equal to 2A, where A is the area of an intercepted hyperbolic sector from (0,0) to (cosh(u), sinh(u)).
Confused as to why this is defined as such, I found on Wikipedia:
The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.
However, I couldn't find a explanation for this anywhere. Can anyone help show me why this is?
calculus geometry trigonometry hyperbolic-functions
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
$endgroup$
I'm trying to figure out how the hyperbolic functions are derived using a unit hyperbola.
According to this walkthrough, argument u in (cosh(u), sinh(u)) should be equal to 2A, where A is the area of an intercepted hyperbolic sector from (0,0) to (cosh(u), sinh(u)).
Confused as to why this is defined as such, I found on Wikipedia:
The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.
However, I couldn't find a explanation for this anywhere. Can anyone help show me why this is?
calculus geometry trigonometry hyperbolic-functions
calculus geometry trigonometry hyperbolic-functions
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
edited Jan 12 at 5:03
sqrtpapi2001
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
asked Jan 5 at 6:04
sqrtpapi2001sqrtpapi2001
112
112
$begingroup$
See the last paragraph of this answer.
$endgroup$
– Blue
Jan 12 at 5:11
$begingroup$
The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started.
$endgroup$
– sqrtpapi2001
Jan 12 at 5:52
add a comment |
$begingroup$
See the last paragraph of this answer.
$endgroup$
– Blue
Jan 12 at 5:11
$begingroup$
The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started.
$endgroup$
– sqrtpapi2001
Jan 12 at 5:52
$begingroup$
See the last paragraph of this answer.
$endgroup$
– Blue
Jan 12 at 5:11
$begingroup$
See the last paragraph of this answer.
$endgroup$
– Blue
Jan 12 at 5:11
$begingroup$
The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started.
$endgroup$
– sqrtpapi2001
Jan 12 at 5:52
$begingroup$
The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started.
$endgroup$
– sqrtpapi2001
Jan 12 at 5:52
add a comment |
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$begingroup$
See the last paragraph of this answer.
$endgroup$
– Blue
Jan 12 at 5:11
$begingroup$
The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started.
$endgroup$
– sqrtpapi2001
Jan 12 at 5:52