How to derive the relation between hyperbolic functions and complex trigonometric ratios? [on hold]
$begingroup$
$$cos(ix)=cosh(x) \
-isin(ix)=sinh(x)$$
I have read that this is derived using Euler's formula $a+bi=re^{itheta}$, but how is this actually done?
trigonometry complex-numbers hyperbolic-functions
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
$endgroup$
put on hold as off-topic by user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos Jan 16 at 15:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$cos(ix)=cosh(x) \
-isin(ix)=sinh(x)$$
I have read that this is derived using Euler's formula $a+bi=re^{itheta}$, but how is this actually done?
trigonometry complex-numbers hyperbolic-functions
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
$endgroup$
put on hold as off-topic by user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos Jan 16 at 15:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$cos(ix)=cosh(x) \
-isin(ix)=sinh(x)$$
I have read that this is derived using Euler's formula $a+bi=re^{itheta}$, but how is this actually done?
trigonometry complex-numbers hyperbolic-functions
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
$endgroup$
$$cos(ix)=cosh(x) \
-isin(ix)=sinh(x)$$
I have read that this is derived using Euler's formula $a+bi=re^{itheta}$, but how is this actually done?
trigonometry complex-numbers hyperbolic-functions
trigonometry complex-numbers hyperbolic-functions
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
asked Jan 9 at 19:41
Pancake_SenpaiPancake_Senpai
23515
23515
put on hold as off-topic by user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos Jan 16 at 15:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos Jan 16 at 15:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, user91500, Adrian Keister, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
hint
Euler formulas are
$$cos(z)=frac{e^{iz}+e^{-iz}}{2}$$
and
$$sin(z)=frac{e^{iz}-e^{-iz}}{2i}$$
on the other hand
$$cosh(a)=frac{e^a+e^{-a}}{2}$$
and
$$sinh(a)=frac{e^a-e^{-a}}{2}$$
thus...
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint
Euler formulas are
$$cos(z)=frac{e^{iz}+e^{-iz}}{2}$$
and
$$sin(z)=frac{e^{iz}-e^{-iz}}{2i}$$
on the other hand
$$cosh(a)=frac{e^a+e^{-a}}{2}$$
and
$$sinh(a)=frac{e^a-e^{-a}}{2}$$
thus...
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
add a comment |
$begingroup$
hint
Euler formulas are
$$cos(z)=frac{e^{iz}+e^{-iz}}{2}$$
and
$$sin(z)=frac{e^{iz}-e^{-iz}}{2i}$$
on the other hand
$$cosh(a)=frac{e^a+e^{-a}}{2}$$
and
$$sinh(a)=frac{e^a-e^{-a}}{2}$$
thus...
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
add a comment |
$begingroup$
hint
Euler formulas are
$$cos(z)=frac{e^{iz}+e^{-iz}}{2}$$
and
$$sin(z)=frac{e^{iz}-e^{-iz}}{2i}$$
on the other hand
$$cosh(a)=frac{e^a+e^{-a}}{2}$$
and
$$sinh(a)=frac{e^a-e^{-a}}{2}$$
thus...
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hint
Euler formulas are
$$cos(z)=frac{e^{iz}+e^{-iz}}{2}$$
and
$$sin(z)=frac{e^{iz}-e^{-iz}}{2i}$$
on the other hand
$$cosh(a)=frac{e^a+e^{-a}}{2}$$
and
$$sinh(a)=frac{e^a-e^{-a}}{2}$$
thus...
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:12
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
add a comment |
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
$begingroup$
I don't know why, but I've looked at these equations multiple times and I just didn't 'see' it. It seems painfully obvious now. Thanks.
$endgroup$
– Pancake_Senpai
Jan 9 at 23:38
add a comment |
