Proving Hyperbolic Identity Using Osborn's Rule

hex.halo
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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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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