Proving Hyperbolic Identity Using Osborn's Rule

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The discussion focuses on proving the hyperbolic identity for cosh(x+y) using Osborn's rule and the definitions of hyperbolic functions. Participants clarify that Osborn's rule allows for the transformation of the trigonometric identity cos(x+y) into its hyperbolic equivalent, leading to cosh(x+y) = cosh(x)cosh(y) + sinh(x)sinh(y). The proof involves substituting the definitions of cosh and sinh in terms of exponential functions and performing algebraic manipulations. One participant highlights the need to ensure accuracy in the expression, particularly in maintaining the correct terms. The conversation emphasizes the importance of understanding both the application of Osborn's rule and the algebraic proof process.
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Homework Statement



Given the trigonometric identity cos(x+y)... use Osborn's rule to write down the corresponding identity for cosh(x+y)... Use the definitionis of the hyperbolic functions to prove this identity

Homework Equations





The Attempt at a Solution



I can use Osborns rule to find the hyperbolic equivilent of the identity, however, I don't understand how I am to prove this identity...
 
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cos(x+y)=cosxcosy-sinxsiny

Osborn said that when you have the product of two sines, you replace the sines with sinh and a negative sign.

so therefore cosh(x+y)=cosxcosy-(-sinhxsinhy)
 
rock.freak missed the h in his last bit for the 2 cos terms on the rhs.
<br /> cosh(x+y)=coshxcoshy-(-sinhxsinhy)<br />
 
hex.halo said:

Homework Statement



Given the trigonometric identity cos(x+y)... use Osborn's rule to write down the corresponding identity for cosh(x+y)... Use the definitionis of the hyperbolic functions to prove this identity

Homework Equations





The Attempt at a Solution



I can use Osborns rule to find the hyperbolic equivilent of the identity, however, I don't understand how I am to prove this identity...

Okay, you already know that cosh(x+ y)= cosh(x)cosh(y)+ sinh(x)sinh(y). Now replace cosh(x) by (e^x+ e^{-x})/2, replace sinh(x)= (e^x- e{-x})/2[/tex], the corresponding things for cosh(y) and sinh(y) and do the algebra. What do you get when you multiply<br /> \frac{e^x+ e^{-x}}{2}\frac{e^y+ e^{-y}}{2}
 
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