Mixing
![How to ask for specifics in a negative part of performance review with very defensive boss [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjbpfN9tAutmK93bJRC3ZoROZzi2TJDms5n8_qJuhgE0a9b52OOHayv3NGT8igAdFL7byXNst-_1DZK5SjrIJ28_6RQPUpBROqMs5s6jo-ZsjX8kjDwfxJufIitH3TaQRXWaGSQKRQib-f/s72-c/1.jpg)
Proof for hyperbolic trigonometric identities
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I've been studying hyperbolic functions and was wondering where the following two identities were derived from:
$$sinh(x) = frace^x-e^-x2$$
$$cosh(x) = frace^x+e^-x2$$
I understand how to use these to prove other identities and I understand how to use Euler's formula to find the identities for $sin(x)$ and $cos(x)$ but I am unable to find any proof for these two. Perhaps I am just unsure what to search for, so if there is a proof somewhere already I would love some directions or links.
Thank you.
trigonometry hyperbolic-functions
edited 1 hour ago
N. F. Taussig
40.4k93253
asked 2 hours ago
Daniel
2515
add a comment |Â
up vote
2
down vote
favorite
I've been studying hyperbolic functions and was wondering where the following two identities were derived from:
$$sinh(x) = frace^x-e^-x2$$
$$cosh(x) = frace^x+e^-x2$$
I understand how to use these to prove other identities and I understand how to use Euler's formula to find the identities for $sin(x)$ and $cos(x)$ but I am unable to find any proof for these two. Perhaps I am just unsure what to search for, so if there is a proof somewhere already I would love some directions or links.
Thank you.
trigonometry hyperbolic-functions
edited 1 hour ago
N. F. Taussig
40.4k93253
asked 2 hours ago
Daniel
2515
5
How do you define $sinh$ and $cosh$?
– José Carlos Santos
2 hours ago
Possible duplicate of What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Nosrati
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've been studying hyperbolic functions and was wondering where the following two identities were derived from:
$$sinh(x) = frace^x-e^-x2$$
$$cosh(x) = frace^x+e^-x2$$
I understand how to use these to prove other identities and I understand how to use Euler's formula to find the identities for $sin(x)$ and $cos(x)$ but I am unable to find any proof for these two. Perhaps I am just unsure what to search for, so if there is a proof somewhere already I would love some directions or links.
Thank you.
trigonometry hyperbolic-functions
edited 1 hour ago
N. F. Taussig
40.4k93253
asked 2 hours ago
Daniel
2515
I've been studying hyperbolic functions and was wondering where the following two identities were derived from:
$$sinh(x) = frace^x-e^-x2$$
$$cosh(x) = frace^x+e^-x2$$
I understand how to use these to prove other identities and I understand how to use Euler's formula to find the identities for $sin(x)$ and $cos(x)$ but I am unable to find any proof for these two. Perhaps I am just unsure what to search for, so if there is a proof somewhere already I would love some directions or links.
Thank you.
trigonometry hyperbolic-functions
trigonometry hyperbolic-functions
edited 1 hour ago
N. F. Taussig
40.4k93253
asked 2 hours ago
Daniel
2515
edited 1 hour ago
N. F. Taussig
40.4k93253
asked 2 hours ago
Daniel
2515
edited 1 hour ago
N. F. Taussig
40.4k93253
edited 1 hour ago
N. F. Taussig
40.4k93253
edited 1 hour ago
N. F. Taussig
40.4k93253
40.4k93253
asked 2 hours ago
Daniel
2515
asked 2 hours ago
Daniel
2515
asked 2 hours ago
Daniel
2515
2515
5
How do you define $sinh$ and $cosh$?
– José Carlos Santos
2 hours ago
Possible duplicate of What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Nosrati
1 hour ago
add a comment |Â
5
How do you define $sinh$ and $cosh$?
– José Carlos Santos
2 hours ago
Possible duplicate of What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Nosrati
1 hour ago
5
5
How do you define $sinh$ and $cosh$?
– José Carlos Santos
2 hours ago
How do you define $sinh$ and $cosh$?
– José Carlos Santos
2 hours ago
Possible duplicate of What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Nosrati
1 hour ago
Possible duplicate of What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Nosrati
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
The hyperbolic functions are defined as the even and odd parts of $exp x$ so $exppm x=cosh xpmsinh x$, in analogy with $exppm ix=cos xpm isin x$. Rearranging gives the desired results.
answered 1 hour ago
J.G.
15.4k11828
add a comment |Â
up vote
2
down vote
You can't answer this until you have defined the hyperbolic functions !
One way is via the complex numbers,
$$isinh x=sin ix,
\cosh x=cos ix,$$ giving
$$sin ix=frace^i^2x-e^-i^2x2i,
\cos ix=frace^i^2x+e^-i^2x2,$$
or
$$sin ix=frace^-x-e^x2i,
\cos ix=frace^-x+e^x2.$$
The qualifier hyperbolic comes from the relation
$$cosh^2x-sinh^2x=1$$ or $$u^2-v^2=1,$$ i.e. an equilateral hyperbola. (Compare to $u^2+v^2=1$.)
answered 1 hour ago
Yves Daoust
116k667211
add a comment |Â
up vote
1
down vote
Using the series expansions for $e^x$ and and $e^-x dots$ $$e^x = 1 + dfrac x1! + dfrac x^22! + dfrac x^33! + dfrac x^44! dots$$ $$e^-x = 1 - dfrac x1! + dfrac x^22! - dfrac x^33! + dfrac x^44!dots$$
We add the first two series to get $$e^x + e^-x = 2 + dfrac 2x^22 + dfrac 2x^44 dots$$ or $$dfrac e^x + e^-x2 = 1 + dfrac x^22! + dfrac x^44! dots$$ As the right side is the series for $cos(x)$, we can define $$dfrac e^x + e^-x2 Leftrightarrow cosh (x)$$
Subtracting the two series yields $$e^x - e^-x = dfrac 2x1! + dfrac 2x^33! + dfrac 2x^55!dots$$ or $$dfrac e^x - e^-x2 = x + dfrac x^33! + dfrac x^55! dots$$ As the right side is the series for $sin(x)$, we can define $$dfrac e^x - e^-x2 Leftrightarrow sinh (x)$$
answered 1 hour ago
bjcolby15
9031916
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The hyperbolic functions are defined as the even and odd parts of $exp x$ so $exppm x=cosh xpmsinh x$, in analogy with $exppm ix=cos xpm isin x$. Rearranging gives the desired results.
answered 1 hour ago
J.G.
15.4k11828
add a comment |Â
up vote
3
down vote
The hyperbolic functions are defined as the even and odd parts of $exp x$ so $exppm x=cosh xpmsinh x$, in analogy with $exppm ix=cos xpm isin x$. Rearranging gives the desired results.
answered 1 hour ago
J.G.
15.4k11828
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The hyperbolic functions are defined as the even and odd parts of $exp x$ so $exppm x=cosh xpmsinh x$, in analogy with $exppm ix=cos xpm isin x$. Rearranging gives the desired results.
answered 1 hour ago
J.G.
15.4k11828
The hyperbolic functions are defined as the even and odd parts of $exp x$ so $exppm x=cosh xpmsinh x$, in analogy with $exppm ix=cos xpm isin x$. Rearranging gives the desired results.
answered 1 hour ago
J.G.
15.4k11828
answered 1 hour ago
J.G.
15.4k11828
answered 1 hour ago
J.G.
15.4k11828
answered 1 hour ago
J.G.
15.4k11828
15.4k11828
add a comment |Â
add a comment |Â
up vote
2
down vote
You can't answer this until you have defined the hyperbolic functions !
One way is via the complex numbers,
$$isinh x=sin ix,
\cosh x=cos ix,$$ giving
$$sin ix=frace^i^2x-e^-i^2x2i,
\cos ix=frace^i^2x+e^-i^2x2,$$
or
$$sin ix=frace^-x-e^x2i,
\cos ix=frace^-x+e^x2.$$
The qualifier hyperbolic comes from the relation
$$cosh^2x-sinh^2x=1$$ or $$u^2-v^2=1,$$ i.e. an equilateral hyperbola. (Compare to $u^2+v^2=1$.)
answered 1 hour ago
Yves Daoust
116k667211
add a comment |Â
up vote
2
down vote
You can't answer this until you have defined the hyperbolic functions !
One way is via the complex numbers,
$$isinh x=sin ix,
\cosh x=cos ix,$$ giving
$$sin ix=frace^i^2x-e^-i^2x2i,
\cos ix=frace^i^2x+e^-i^2x2,$$
or
$$sin ix=frace^-x-e^x2i,
\cos ix=frace^-x+e^x2.$$
The qualifier hyperbolic comes from the relation
$$cosh^2x-sinh^2x=1$$ or $$u^2-v^2=1,$$ i.e. an equilateral hyperbola. (Compare to $u^2+v^2=1$.)
answered 1 hour ago
Yves Daoust
116k667211
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You can't answer this until you have defined the hyperbolic functions !
One way is via the complex numbers,
$$isinh x=sin ix,
\cosh x=cos ix,$$ giving
$$sin ix=frace^i^2x-e^-i^2x2i,
\cos ix=frace^i^2x+e^-i^2x2,$$
or
$$sin ix=frace^-x-e^x2i,
\cos ix=frace^-x+e^x2.$$
The qualifier hyperbolic comes from the relation
$$cosh^2x-sinh^2x=1$$ or $$u^2-v^2=1,$$ i.e. an equilateral hyperbola. (Compare to $u^2+v^2=1$.)
answered 1 hour ago
Yves Daoust
116k667211
You can't answer this until you have defined the hyperbolic functions !
One way is via the complex numbers,
$$isinh x=sin ix,
\cosh x=cos ix,$$ giving
$$sin ix=frace^i^2x-e^-i^2x2i,
\cos ix=frace^i^2x+e^-i^2x2,$$
or
$$sin ix=frace^-x-e^x2i,
\cos ix=frace^-x+e^x2.$$
The qualifier hyperbolic comes from the relation
$$cosh^2x-sinh^2x=1$$ or $$u^2-v^2=1,$$ i.e. an equilateral hyperbola. (Compare to $u^2+v^2=1$.)
answered 1 hour ago
Yves Daoust
116k667211
answered 1 hour ago
Yves Daoust
116k667211
answered 1 hour ago
Yves Daoust
116k667211
answered 1 hour ago
Yves Daoust
116k667211
116k667211
add a comment |Â
add a comment |Â
up vote
1
down vote
Using the series expansions for $e^x$ and and $e^-x dots$ $$e^x = 1 + dfrac x1! + dfrac x^22! + dfrac x^33! + dfrac x^44! dots$$ $$e^-x = 1 - dfrac x1! + dfrac x^22! - dfrac x^33! + dfrac x^44!dots$$
We add the first two series to get $$e^x + e^-x = 2 + dfrac 2x^22 + dfrac 2x^44 dots$$ or $$dfrac e^x + e^-x2 = 1 + dfrac x^22! + dfrac x^44! dots$$ As the right side is the series for $cos(x)$, we can define $$dfrac e^x + e^-x2 Leftrightarrow cosh (x)$$
Subtracting the two series yields $$e^x - e^-x = dfrac 2x1! + dfrac 2x^33! + dfrac 2x^55!dots$$ or $$dfrac e^x - e^-x2 = x + dfrac x^33! + dfrac x^55! dots$$ As the right side is the series for $sin(x)$, we can define $$dfrac e^x - e^-x2 Leftrightarrow sinh (x)$$
answered 1 hour ago
bjcolby15
9031916
add a comment |Â
up vote
1
down vote
Using the series expansions for $e^x$ and and $e^-x dots$ $$e^x = 1 + dfrac x1! + dfrac x^22! + dfrac x^33! + dfrac x^44! dots$$ $$e^-x = 1 - dfrac x1! + dfrac x^22! - dfrac x^33! + dfrac x^44!dots$$
We add the first two series to get $$e^x + e^-x = 2 + dfrac 2x^22 + dfrac 2x^44 dots$$ or $$dfrac e^x + e^-x2 = 1 + dfrac x^22! + dfrac x^44! dots$$ As the right side is the series for $cos(x)$, we can define $$dfrac e^x + e^-x2 Leftrightarrow cosh (x)$$
Subtracting the two series yields $$e^x - e^-x = dfrac 2x1! + dfrac 2x^33! + dfrac 2x^55!dots$$ or $$dfrac e^x - e^-x2 = x + dfrac x^33! + dfrac x^55! dots$$ As the right side is the series for $sin(x)$, we can define $$dfrac e^x - e^-x2 Leftrightarrow sinh (x)$$
answered 1 hour ago
bjcolby15
9031916
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using the series expansions for $e^x$ and and $e^-x dots$ $$e^x = 1 + dfrac x1! + dfrac x^22! + dfrac x^33! + dfrac x^44! dots$$ $$e^-x = 1 - dfrac x1! + dfrac x^22! - dfrac x^33! + dfrac x^44!dots$$
We add the first two series to get $$e^x + e^-x = 2 + dfrac 2x^22 + dfrac 2x^44 dots$$ or $$dfrac e^x + e^-x2 = 1 + dfrac x^22! + dfrac x^44! dots$$ As the right side is the series for $cos(x)$, we can define $$dfrac e^x + e^-x2 Leftrightarrow cosh (x)$$
Subtracting the two series yields $$e^x - e^-x = dfrac 2x1! + dfrac 2x^33! + dfrac 2x^55!dots$$ or $$dfrac e^x - e^-x2 = x + dfrac x^33! + dfrac x^55! dots$$ As the right side is the series for $sin(x)$, we can define $$dfrac e^x - e^-x2 Leftrightarrow sinh (x)$$
answered 1 hour ago
bjcolby15
9031916
Using the series expansions for $e^x$ and and $e^-x dots$ $$e^x = 1 + dfrac x1! + dfrac x^22! + dfrac x^33! + dfrac x^44! dots$$ $$e^-x = 1 - dfrac x1! + dfrac x^22! - dfrac x^33! + dfrac x^44!dots$$
We add the first two series to get $$e^x + e^-x = 2 + dfrac 2x^22 + dfrac 2x^44 dots$$ or $$dfrac e^x + e^-x2 = 1 + dfrac x^22! + dfrac x^44! dots$$ As the right side is the series for $cos(x)$, we can define $$dfrac e^x + e^-x2 Leftrightarrow cosh (x)$$
Subtracting the two series yields $$e^x - e^-x = dfrac 2x1! + dfrac 2x^33! + dfrac 2x^55!dots$$ or $$dfrac e^x - e^-x2 = x + dfrac x^33! + dfrac x^55! dots$$ As the right side is the series for $sin(x)$, we can define $$dfrac e^x - e^-x2 Leftrightarrow sinh (x)$$
answered 1 hour ago
bjcolby15
9031916
answered 1 hour ago
bjcolby15
9031916
answered 1 hour ago
bjcolby15
9031916
answered 1 hour ago
bjcolby15
9031916
9031916
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2939161%2fproof-for-hyperbolic-trigonometric-identities%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
5
How do you define $sinh$ and $cosh$?
– José Carlos Santos
2 hours ago
Possible duplicate of What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Nosrati
1 hour ago