Equivalence of two methods to obtain $sin theta geq 1$ with a complex $theta$
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When evaluating the function $sin theta$ with a complex angle $theta$, a real value $A geq 1$ can be obtained in two ways.
- Considering $theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right]$, as stated in this page. After the substitution of $theta$,
$$sin theta = frac{e^{i theta} - e^{-i theta}}{2i} = A$$
- Considering $theta = frac{pi}{2} - iB$. This way,
$$sin theta = cosh B$$
I want to prove that the two methods are equivalent.
My attempt:
I applied the inverse hyperbolic cosine definition:
$$cosh B = A Rightarrow B = mathrm{arccosh} A = log left[ A + sqrt{A^2 - 1} right]$$
$$theta = frac{pi}{2} - ilog left[ A + sqrt{A^2 - 1} right] label{a} tag{1}$$
In method 1, $theta$ can be rewritten as
$$theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right] =\
= i log left( frac{A + sqrt{A^2 - 1}}{j} right) = i left[ log left( A + sqrt{A^2 - 1} right) - log i right] = \
= i left[ log left( A + sqrt{A^2 - 1} right) - i frac{pi}{2} right] =$$
$$=i log left( A + sqrt{A^2 - 1} right) + frac{pi}{2} label{b} tag{2}$$
My question:
In my attempt, ref{b} doesn't match ref{a}. Why? What is wrong?
This question is related to a Physics question about Snell's law.
complex-analysis trigonometry proof-verification complex-numbers hyperbolic-functions
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
asked Nov 23 at 11:16
BowPark
549720
add a comment |
up vote
0
down vote
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When evaluating the function $sin theta$ with a complex angle $theta$, a real value $A geq 1$ can be obtained in two ways.
- Considering $theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right]$, as stated in this page. After the substitution of $theta$,
$$sin theta = frac{e^{i theta} - e^{-i theta}}{2i} = A$$
- Considering $theta = frac{pi}{2} - iB$. This way,
$$sin theta = cosh B$$
I want to prove that the two methods are equivalent.
My attempt:
I applied the inverse hyperbolic cosine definition:
$$cosh B = A Rightarrow B = mathrm{arccosh} A = log left[ A + sqrt{A^2 - 1} right]$$
$$theta = frac{pi}{2} - ilog left[ A + sqrt{A^2 - 1} right] label{a} tag{1}$$
In method 1, $theta$ can be rewritten as
$$theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right] =\
= i log left( frac{A + sqrt{A^2 - 1}}{j} right) = i left[ log left( A + sqrt{A^2 - 1} right) - log i right] = \
= i left[ log left( A + sqrt{A^2 - 1} right) - i frac{pi}{2} right] =$$
$$=i log left( A + sqrt{A^2 - 1} right) + frac{pi}{2} label{b} tag{2}$$
My question:
In my attempt, ref{b} doesn't match ref{a}. Why? What is wrong?
This question is related to a Physics question about Snell's law.
complex-analysis trigonometry proof-verification complex-numbers hyperbolic-functions
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
asked Nov 23 at 11:16
BowPark
549720
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When evaluating the function $sin theta$ with a complex angle $theta$, a real value $A geq 1$ can be obtained in two ways.
- Considering $theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right]$, as stated in this page. After the substitution of $theta$,
$$sin theta = frac{e^{i theta} - e^{-i theta}}{2i} = A$$
- Considering $theta = frac{pi}{2} - iB$. This way,
$$sin theta = cosh B$$
I want to prove that the two methods are equivalent.
My attempt:
I applied the inverse hyperbolic cosine definition:
$$cosh B = A Rightarrow B = mathrm{arccosh} A = log left[ A + sqrt{A^2 - 1} right]$$
$$theta = frac{pi}{2} - ilog left[ A + sqrt{A^2 - 1} right] label{a} tag{1}$$
In method 1, $theta$ can be rewritten as
$$theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right] =\
= i log left( frac{A + sqrt{A^2 - 1}}{j} right) = i left[ log left( A + sqrt{A^2 - 1} right) - log i right] = \
= i left[ log left( A + sqrt{A^2 - 1} right) - i frac{pi}{2} right] =$$
$$=i log left( A + sqrt{A^2 - 1} right) + frac{pi}{2} label{b} tag{2}$$
My question:
In my attempt, ref{b} doesn't match ref{a}. Why? What is wrong?
This question is related to a Physics question about Snell's law.
complex-analysis trigonometry proof-verification complex-numbers hyperbolic-functions
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
asked Nov 23 at 11:16
BowPark
549720
When evaluating the function $sin theta$ with a complex angle $theta$, a real value $A geq 1$ can be obtained in two ways.
- Considering $theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right]$, as stated in this page. After the substitution of $theta$,
$$sin theta = frac{e^{i theta} - e^{-i theta}}{2i} = A$$
- Considering $theta = frac{pi}{2} - iB$. This way,
$$sin theta = cosh B$$
I want to prove that the two methods are equivalent.
My attempt:
I applied the inverse hyperbolic cosine definition:
$$cosh B = A Rightarrow B = mathrm{arccosh} A = log left[ A + sqrt{A^2 - 1} right]$$
$$theta = frac{pi}{2} - ilog left[ A + sqrt{A^2 - 1} right] label{a} tag{1}$$
In method 1, $theta$ can be rewritten as
$$theta = i log left[ -i left( A + sqrt{A^2 - 1} right) right] =\
= i log left( frac{A + sqrt{A^2 - 1}}{j} right) = i left[ log left( A + sqrt{A^2 - 1} right) - log i right] = \
= i left[ log left( A + sqrt{A^2 - 1} right) - i frac{pi}{2} right] =$$
$$=i log left( A + sqrt{A^2 - 1} right) + frac{pi}{2} label{b} tag{2}$$
My question:
In my attempt, ref{b} doesn't match ref{a}. Why? What is wrong?
This question is related to a Physics question about Snell's law.
complex-analysis trigonometry proof-verification complex-numbers hyperbolic-functions
complex-analysis trigonometry proof-verification complex-numbers hyperbolic-functions
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
asked Nov 23 at 11:16
BowPark
549720
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
asked Nov 23 at 11:16
BowPark
549720
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
edited Nov 23 at 16:53
José Carlos Santos
147k22117218
147k22117218
asked Nov 23 at 11:16
BowPark
549720
asked Nov 23 at 11:16
BowPark
549720
asked Nov 23 at 11:16
BowPark
549720
549720
add a comment |
add a comment |
1 Answer
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Yes, they match. Just note that$$(forall zinmathbb{C}):sinleft(fracpi2-zright)=sinleft(fracpi2+zright)=cos(z).$$
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, they match. Just note that$$(forall zinmathbb{C}):sinleft(fracpi2-zright)=sinleft(fracpi2+zright)=cos(z).$$
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
add a comment |
up vote
2
down vote
accepted
Yes, they match. Just note that$$(forall zinmathbb{C}):sinleft(fracpi2-zright)=sinleft(fracpi2+zright)=cos(z).$$
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, they match. Just note that$$(forall zinmathbb{C}):sinleft(fracpi2-zright)=sinleft(fracpi2+zright)=cos(z).$$
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
Yes, they match. Just note that$$(forall zinmathbb{C}):sinleft(fracpi2-zright)=sinleft(fracpi2+zright)=cos(z).$$
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
answered Nov 23 at 11:19
José Carlos Santos
147k22117218
147k22117218
add a comment |
add a comment |
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