Can the sinh and cosh functions be expressed in terms of each other?

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The discussion explores the relationship between the sine and cosine functions, establishing that sine can be expressed in terms of cosine and vice versa. It extends this concept to hyperbolic functions, showing that sinh and cosh can also be expressed using imaginary units. However, it is noted that there is no identity for expressing sinh and cosh without involving the imaginary unit, as their definitions are tied to complex numbers. The identity cosh²(x) - sinh²(x) = 1 is acknowledged as a fundamental relationship, but it does not meet the criteria for the desired transformation. The conversation highlights the distinct nature of hyperbolic functions compared to their trigonometric counterparts.
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Is possible express the sine in terms of cosine and vice-versa:

##\sin(x) = \cos(x-\frac{\pi}{2})##

##\cos(x) = \sin(x+\frac{\pi}{2})##

So, of some way, is possible express the sinh in terms of cosh too and vice-versa?
 
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sure

##
\sinh(x)=-\imath \, \cosh(x+\pi\,\imath/2)
\\
\cosh(x)=-\imath \, \sinh(x+\pi\,\imath/2)
##
 
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lurflurf said:
sure

##
\sinh(x)=-\imath \, \cosh(x+\pi\,\imath/2)
\\
\cosh(x)=-\imath \, \sinh(x+\pi\,\imath/2)
##

No exist an identity that no involve imaginary unit?
 
No, because that is the whole point of the hyperbolic functions: sinh(x)= i sin(ix) and cosh(x)= cos(ix).
 
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Of course, there's the 'obvious' one which you can find by rearranging ##\cosh^2 x - \sinh^2 x = 1##, but I don't think that's what you're going for, is it?
 
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Curious3141 said:
Of course, there's the 'obvious' one which you can find by rearranging ##\cosh^2 x - \sinh^2 x = 1##, but I don't think that's what you're going for, is it?

no, but thanks
 
Considering the graphs of sinh(x) and cosh(x), they don't have the shift property that sin(x) and cos(x) have for x in set of real numbers.
 
cosh2(x) - sinh2(x) = 1.
 
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