Mixing
What is the value of $cosh(sqrt{i})$?
$begingroup$
I am puzzling about the value of
$$cosh(sqrt{i})$$
I know that
$$sqrt{i} = sqrt{frac{1}{2}}+isqrt{frac{1}{2}}$$
But how to go on with that? Are there also multiple values?
Thank you all in advance!
hyperbolic-functions
edited Jan 30 at 17:25
Blue
49.4k870157
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
$endgroup$
add a comment |
$begingroup$
I am puzzling about the value of
$$cosh(sqrt{i})$$
I know that
$$sqrt{i} = sqrt{frac{1}{2}}+isqrt{frac{1}{2}}$$
But how to go on with that? Are there also multiple values?
Thank you all in advance!
hyperbolic-functions
edited Jan 30 at 17:25
Blue
49.4k870157
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
$endgroup$
$begingroup$
We need some more context. $cosh(sqrt{i})$ is an expression: are you trying to find its value? If so, you don't use the language of "solution", but of "evaluating". On the other hand, if there's an equals sign, and you're trying to find the value of a variable, you "solve it", you do not "evaluate it".
$endgroup$
– Adrian Keister
Jan 30 at 16:09
$begingroup$
Do you know the formula for $cosh?$ If so, what difficulty are you having? If by $sqrt{i}$ you mean the principal value, then there is only one value of the expression.
$endgroup$
– saulspatz
Jan 30 at 16:12
add a comment |
$begingroup$
I am puzzling about the value of
$$cosh(sqrt{i})$$
I know that
$$sqrt{i} = sqrt{frac{1}{2}}+isqrt{frac{1}{2}}$$
But how to go on with that? Are there also multiple values?
Thank you all in advance!
hyperbolic-functions
edited Jan 30 at 17:25
Blue
49.4k870157
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
$endgroup$
I am puzzling about the value of
$$cosh(sqrt{i})$$
I know that
$$sqrt{i} = sqrt{frac{1}{2}}+isqrt{frac{1}{2}}$$
But how to go on with that? Are there also multiple values?
Thank you all in advance!
hyperbolic-functions
hyperbolic-functions
edited Jan 30 at 17:25
Blue
49.4k870157
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
edited Jan 30 at 17:25
Blue
49.4k870157
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
edited Jan 30 at 17:25
Blue
49.4k870157
edited Jan 30 at 17:25
Blue
49.4k870157
edited Jan 30 at 17:25
Blue
49.4k870157
49.4k870157
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
asked Jan 30 at 16:07
Max MustermannMax Mustermann
1
1
$begingroup$
We need some more context. $cosh(sqrt{i})$ is an expression: are you trying to find its value? If so, you don't use the language of "solution", but of "evaluating". On the other hand, if there's an equals sign, and you're trying to find the value of a variable, you "solve it", you do not "evaluate it".
$endgroup$
– Adrian Keister
Jan 30 at 16:09
$begingroup$
Do you know the formula for $cosh?$ If so, what difficulty are you having? If by $sqrt{i}$ you mean the principal value, then there is only one value of the expression.
$endgroup$
– saulspatz
Jan 30 at 16:12
add a comment |
$begingroup$
We need some more context. $cosh(sqrt{i})$ is an expression: are you trying to find its value? If so, you don't use the language of "solution", but of "evaluating". On the other hand, if there's an equals sign, and you're trying to find the value of a variable, you "solve it", you do not "evaluate it".
$endgroup$
– Adrian Keister
Jan 30 at 16:09
$begingroup$
Do you know the formula for $cosh?$ If so, what difficulty are you having? If by $sqrt{i}$ you mean the principal value, then there is only one value of the expression.
$endgroup$
– saulspatz
Jan 30 at 16:12
$begingroup$
We need some more context. $cosh(sqrt{i})$ is an expression: are you trying to find its value? If so, you don't use the language of "solution", but of "evaluating". On the other hand, if there's an equals sign, and you're trying to find the value of a variable, you "solve it", you do not "evaluate it".
$endgroup$
– Adrian Keister
Jan 30 at 16:09
$begingroup$
We need some more context. $cosh(sqrt{i})$ is an expression: are you trying to find its value? If so, you don't use the language of "solution", but of "evaluating". On the other hand, if there's an equals sign, and you're trying to find the value of a variable, you "solve it", you do not "evaluate it".
$endgroup$
– Adrian Keister
Jan 30 at 16:09
$begingroup$
Do you know the formula for $cosh?$ If so, what difficulty are you having? If by $sqrt{i}$ you mean the principal value, then there is only one value of the expression.
$endgroup$
– saulspatz
Jan 30 at 16:12
$begingroup$
Do you know the formula for $cosh?$ If so, what difficulty are you having? If by $sqrt{i}$ you mean the principal value, then there is only one value of the expression.
$endgroup$
– saulspatz
Jan 30 at 16:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that there are two possible values for $sqrt{i}$, namely, $pmleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$. But since $cosh$ is an even function, it doesn't matter which we choose. So we will work with finding $coshleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$.
One easily verifies that $cosh(a+bi)=cosh a cos b + isinh asin b$. Thus
$$coshleft(tfrac{1}{sqrt{2}}+tfrac{i}{sqrt{2}}right) = boxed{cosh tfrac{1}{sqrt{2}} cos tfrac{1}{sqrt{2}} + isinh tfrac{1}{sqrt{2}}sin tfrac{1}{sqrt{2}}} approx 0.9584 + 0.4986i.$$
Addendum: The rules for handling $f(-t)$ for trig functions $f$ have analogs for handling expressions like $f(it)$; moreover, the same rules apply to hyperbolic and trig functions alike. Let's explore them briefly.
We will distinguish between "c" and "s" functions. The familiar rules are:
- A factor of $-1$ can be pulled out of c-functions and disappears
- A factor of $-1$ can be pulled out of s-functions and remains
So:
$cos(-t)=cos t$, and $cosh(-t)=cosh t$;
$sin(-t)=-sin t$, and $sinh(-t)=-sinh t$.
The new rules are only slightly different:
- A factor of $i$ can be pulled out of c-functions and disappears, transforming the function
- A factor of $i$ can be pulled out of s-functions and remains, transforming the function
(here, "transforming" means changing trig to hyperbolic and vice versa).
So:
$cos(it)=cosh t$, and $cosh(it)=cos t$; $sin(it)=isinh t$, and $sinh(it)=isin t$.
Now it is easy to see what we said above:
$$cosh(a+bi)=cosh acosh(bi) + sinh asinh(bi)$$
$$= (cosh a)( cos b) + (sinh a) (isin b)$$
$$= cosh a cos b + isinh asin b$$
as desired.
answered Jan 30 at 16:23
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
You can use the fact that:
$$cosh(x)=frac{e^x+e^{-x}}{2}$$
And
$$cos(x)=frac{e^{ix}+e^{-ix}}{2}$$
Also make use of compound angle formula
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
$endgroup$
add a comment |
$begingroup$
Use the definition $cosh(x) = frac{1}{2}(e^{x}+e^{-x})$ and the identity $cosh(x+y) = cosh(x)cosh(y)+sinh(x)cosh(y)$ so
$$cosh(sqrt{i}) = cosh(sqrt{1/2})cosh(isqrt{1/2}) + sinh(sqrt{1/2})sinh(isqrt{1/2})$$
$$= cosh(sqrt{1/2})cos(sqrt{1/2}) + isinh(sqrt{1/2})sin(sqrt{1/2})$$
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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votes
$begingroup$
Note that there are two possible values for $sqrt{i}$, namely, $pmleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$. But since $cosh$ is an even function, it doesn't matter which we choose. So we will work with finding $coshleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$.
One easily verifies that $cosh(a+bi)=cosh a cos b + isinh asin b$. Thus
$$coshleft(tfrac{1}{sqrt{2}}+tfrac{i}{sqrt{2}}right) = boxed{cosh tfrac{1}{sqrt{2}} cos tfrac{1}{sqrt{2}} + isinh tfrac{1}{sqrt{2}}sin tfrac{1}{sqrt{2}}} approx 0.9584 + 0.4986i.$$
Addendum: The rules for handling $f(-t)$ for trig functions $f$ have analogs for handling expressions like $f(it)$; moreover, the same rules apply to hyperbolic and trig functions alike. Let's explore them briefly.
We will distinguish between "c" and "s" functions. The familiar rules are:
- A factor of $-1$ can be pulled out of c-functions and disappears
- A factor of $-1$ can be pulled out of s-functions and remains
So:
$cos(-t)=cos t$, and $cosh(-t)=cosh t$;
$sin(-t)=-sin t$, and $sinh(-t)=-sinh t$.
The new rules are only slightly different:
- A factor of $i$ can be pulled out of c-functions and disappears, transforming the function
- A factor of $i$ can be pulled out of s-functions and remains, transforming the function
(here, "transforming" means changing trig to hyperbolic and vice versa).
So:
$cos(it)=cosh t$, and $cosh(it)=cos t$; $sin(it)=isinh t$, and $sinh(it)=isin t$.
Now it is easy to see what we said above:
$$cosh(a+bi)=cosh acosh(bi) + sinh asinh(bi)$$
$$= (cosh a)( cos b) + (sinh a) (isin b)$$
$$= cosh a cos b + isinh asin b$$
as desired.
answered Jan 30 at 16:23
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
Note that there are two possible values for $sqrt{i}$, namely, $pmleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$. But since $cosh$ is an even function, it doesn't matter which we choose. So we will work with finding $coshleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$.
One easily verifies that $cosh(a+bi)=cosh a cos b + isinh asin b$. Thus
$$coshleft(tfrac{1}{sqrt{2}}+tfrac{i}{sqrt{2}}right) = boxed{cosh tfrac{1}{sqrt{2}} cos tfrac{1}{sqrt{2}} + isinh tfrac{1}{sqrt{2}}sin tfrac{1}{sqrt{2}}} approx 0.9584 + 0.4986i.$$
Addendum: The rules for handling $f(-t)$ for trig functions $f$ have analogs for handling expressions like $f(it)$; moreover, the same rules apply to hyperbolic and trig functions alike. Let's explore them briefly.
We will distinguish between "c" and "s" functions. The familiar rules are:
- A factor of $-1$ can be pulled out of c-functions and disappears
- A factor of $-1$ can be pulled out of s-functions and remains
So:
$cos(-t)=cos t$, and $cosh(-t)=cosh t$;
$sin(-t)=-sin t$, and $sinh(-t)=-sinh t$.
The new rules are only slightly different:
- A factor of $i$ can be pulled out of c-functions and disappears, transforming the function
- A factor of $i$ can be pulled out of s-functions and remains, transforming the function
(here, "transforming" means changing trig to hyperbolic and vice versa).
So:
$cos(it)=cosh t$, and $cosh(it)=cos t$; $sin(it)=isinh t$, and $sinh(it)=isin t$.
Now it is easy to see what we said above:
$$cosh(a+bi)=cosh acosh(bi) + sinh asinh(bi)$$
$$= (cosh a)( cos b) + (sinh a) (isin b)$$
$$= cosh a cos b + isinh asin b$$
as desired.
answered Jan 30 at 16:23
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
Note that there are two possible values for $sqrt{i}$, namely, $pmleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$. But since $cosh$ is an even function, it doesn't matter which we choose. So we will work with finding $coshleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$.
One easily verifies that $cosh(a+bi)=cosh a cos b + isinh asin b$. Thus
$$coshleft(tfrac{1}{sqrt{2}}+tfrac{i}{sqrt{2}}right) = boxed{cosh tfrac{1}{sqrt{2}} cos tfrac{1}{sqrt{2}} + isinh tfrac{1}{sqrt{2}}sin tfrac{1}{sqrt{2}}} approx 0.9584 + 0.4986i.$$
Addendum: The rules for handling $f(-t)$ for trig functions $f$ have analogs for handling expressions like $f(it)$; moreover, the same rules apply to hyperbolic and trig functions alike. Let's explore them briefly.
We will distinguish between "c" and "s" functions. The familiar rules are:
- A factor of $-1$ can be pulled out of c-functions and disappears
- A factor of $-1$ can be pulled out of s-functions and remains
So:
$cos(-t)=cos t$, and $cosh(-t)=cosh t$;
$sin(-t)=-sin t$, and $sinh(-t)=-sinh t$.
The new rules are only slightly different:
- A factor of $i$ can be pulled out of c-functions and disappears, transforming the function
- A factor of $i$ can be pulled out of s-functions and remains, transforming the function
(here, "transforming" means changing trig to hyperbolic and vice versa).
So:
$cos(it)=cosh t$, and $cosh(it)=cos t$; $sin(it)=isinh t$, and $sinh(it)=isin t$.
Now it is easy to see what we said above:
$$cosh(a+bi)=cosh acosh(bi) + sinh asinh(bi)$$
$$= (cosh a)( cos b) + (sinh a) (isin b)$$
$$= cosh a cos b + isinh asin b$$
as desired.
answered Jan 30 at 16:23
MPWMPW
31k12157
$endgroup$
Note that there are two possible values for $sqrt{i}$, namely, $pmleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$. But since $cosh$ is an even function, it doesn't matter which we choose. So we will work with finding $coshleft(frac{1}{sqrt{2}}+frac{i}{sqrt{2}}right)$.
One easily verifies that $cosh(a+bi)=cosh a cos b + isinh asin b$. Thus
$$coshleft(tfrac{1}{sqrt{2}}+tfrac{i}{sqrt{2}}right) = boxed{cosh tfrac{1}{sqrt{2}} cos tfrac{1}{sqrt{2}} + isinh tfrac{1}{sqrt{2}}sin tfrac{1}{sqrt{2}}} approx 0.9584 + 0.4986i.$$
Addendum: The rules for handling $f(-t)$ for trig functions $f$ have analogs for handling expressions like $f(it)$; moreover, the same rules apply to hyperbolic and trig functions alike. Let's explore them briefly.
We will distinguish between "c" and "s" functions. The familiar rules are:
- A factor of $-1$ can be pulled out of c-functions and disappears
- A factor of $-1$ can be pulled out of s-functions and remains
So:
$cos(-t)=cos t$, and $cosh(-t)=cosh t$;
$sin(-t)=-sin t$, and $sinh(-t)=-sinh t$.
The new rules are only slightly different:
- A factor of $i$ can be pulled out of c-functions and disappears, transforming the function
- A factor of $i$ can be pulled out of s-functions and remains, transforming the function
(here, "transforming" means changing trig to hyperbolic and vice versa).
So:
$cos(it)=cosh t$, and $cosh(it)=cos t$; $sin(it)=isinh t$, and $sinh(it)=isin t$.
Now it is easy to see what we said above:
$$cosh(a+bi)=cosh acosh(bi) + sinh asinh(bi)$$
$$= (cosh a)( cos b) + (sinh a) (isin b)$$
$$= cosh a cos b + isinh asin b$$
as desired.
answered Jan 30 at 16:23
MPWMPW
31k12157
edited Jan 30 at 17:17
answered Jan 30 at 16:23
MPWMPW
31k12157
answered Jan 30 at 16:23
MPWMPW
31k12157
answered Jan 30 at 16:23
MPWMPW
31k12157
31k12157
add a comment |
add a comment |
$begingroup$
You can use the fact that:
$$cosh(x)=frac{e^x+e^{-x}}{2}$$
And
$$cos(x)=frac{e^{ix}+e^{-ix}}{2}$$
Also make use of compound angle formula
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
$endgroup$
add a comment |
$begingroup$
You can use the fact that:
$$cosh(x)=frac{e^x+e^{-x}}{2}$$
And
$$cos(x)=frac{e^{ix}+e^{-ix}}{2}$$
Also make use of compound angle formula
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
$endgroup$
add a comment |
$begingroup$
You can use the fact that:
$$cosh(x)=frac{e^x+e^{-x}}{2}$$
And
$$cos(x)=frac{e^{ix}+e^{-ix}}{2}$$
Also make use of compound angle formula
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
$endgroup$
You can use the fact that:
$$cosh(x)=frac{e^x+e^{-x}}{2}$$
And
$$cos(x)=frac{e^{ix}+e^{-ix}}{2}$$
Also make use of compound angle formula
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
answered Jan 30 at 16:12
Henry LeeHenry Lee
2,201319
2,201319
add a comment |
add a comment |
$begingroup$
Use the definition $cosh(x) = frac{1}{2}(e^{x}+e^{-x})$ and the identity $cosh(x+y) = cosh(x)cosh(y)+sinh(x)cosh(y)$ so
$$cosh(sqrt{i}) = cosh(sqrt{1/2})cosh(isqrt{1/2}) + sinh(sqrt{1/2})sinh(isqrt{1/2})$$
$$= cosh(sqrt{1/2})cos(sqrt{1/2}) + isinh(sqrt{1/2})sin(sqrt{1/2})$$
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
$endgroup$
add a comment |
$begingroup$
Use the definition $cosh(x) = frac{1}{2}(e^{x}+e^{-x})$ and the identity $cosh(x+y) = cosh(x)cosh(y)+sinh(x)cosh(y)$ so
$$cosh(sqrt{i}) = cosh(sqrt{1/2})cosh(isqrt{1/2}) + sinh(sqrt{1/2})sinh(isqrt{1/2})$$
$$= cosh(sqrt{1/2})cos(sqrt{1/2}) + isinh(sqrt{1/2})sin(sqrt{1/2})$$
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
$endgroup$
add a comment |
$begingroup$
Use the definition $cosh(x) = frac{1}{2}(e^{x}+e^{-x})$ and the identity $cosh(x+y) = cosh(x)cosh(y)+sinh(x)cosh(y)$ so
$$cosh(sqrt{i}) = cosh(sqrt{1/2})cosh(isqrt{1/2}) + sinh(sqrt{1/2})sinh(isqrt{1/2})$$
$$= cosh(sqrt{1/2})cos(sqrt{1/2}) + isinh(sqrt{1/2})sin(sqrt{1/2})$$
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
$endgroup$
Use the definition $cosh(x) = frac{1}{2}(e^{x}+e^{-x})$ and the identity $cosh(x+y) = cosh(x)cosh(y)+sinh(x)cosh(y)$ so
$$cosh(sqrt{i}) = cosh(sqrt{1/2})cosh(isqrt{1/2}) + sinh(sqrt{1/2})sinh(isqrt{1/2})$$
$$= cosh(sqrt{1/2})cos(sqrt{1/2}) + isinh(sqrt{1/2})sin(sqrt{1/2})$$
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
answered Jan 30 at 16:17
Nicholas ParrisNicholas Parris
1563
1563
add a comment |
add a comment |
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$begingroup$
We need some more context. $cosh(sqrt{i})$ is an expression: are you trying to find its value? If so, you don't use the language of "solution", but of "evaluating". On the other hand, if there's an equals sign, and you're trying to find the value of a variable, you "solve it", you do not "evaluate it".
$endgroup$
– Adrian Keister
Jan 30 at 16:09
$begingroup$
Do you know the formula for $cosh?$ If so, what difficulty are you having? If by $sqrt{i}$ you mean the principal value, then there is only one value of the expression.
$endgroup$
– saulspatz
Jan 30 at 16:12