Actually, it's the other way around: cosh(ix) = cos(x). But to answer your question, "i" is the imaginary number constant √-1.
As for how you can take the hyperbolic cosine of an imaginary number, well Euler's formula says e^(ix) = cos(x) + i sin(x). You can find a number of pages on-line that show the derivation of this. So starting with the definition of cosh(x), you get:
Well, this 'i' stands for an imaginary number. It is defined to be,
i=sqrt(-1)
Then:
i^2 = -1
To calculate any high power of i, you can convert it to a lower power by taking the closest multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent.
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Actually, it's the other way around: cosh(ix) = cos(x). But to answer your question, "i" is the imaginary number constant √-1.
As for how you can take the hyperbolic cosine of an imaginary number, well Euler's formula says e^(ix) = cos(x) + i sin(x). You can find a number of pages on-line that show the derivation of this. So starting with the definition of cosh(x), you get:
cosh(x) = (e^(x) + e^(-x)) / 2
cosh(ix) = (e^(ix) + e^(-ix)) / 2
cosh(ix) = (e^(ix) + e^(i * -x)) / 2
cosh(ix) = ([cos(x) + isin(x)] + [cos(-x) + isin(-x)]) / 2
cosh(ix) = (cos(x) + isin(x) + cos(x) - isin(x)) / 2
cosh(ix) = 2cos(x) / 2
cosh(ix) = cos(x)
What Is Cosh
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RE:
cosh(x) and cos(x) relationship?
I know that cosh(x) = cos(ix)
but what does the "i" stand for?
Well, this 'i' stands for an imaginary number. It is defined to be,
i=sqrt(-1)
Then:
i^2 = -1
To calculate any high power of i, you can convert it to a lower power by taking the closest multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent.
i = sqrt(-1) an imaginary number.