Hi, I know its probably something really simple but does anyone know how I would go aboutprovingthatcosh(x+y)=coshxcoshy+sinhxsinhy?

I would really appreciate a fast answer!

:confused:

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- October 10th 2005, 07:21 AMOverwhelmed[SOLVED] Proof of Double Angle
Hi, I know its probably something really simple but does anyone know how I would go about

**proving**that**cosh(x+y)=coshxcoshy+sinhxsinhy**?

I would really appreciate a fast answer!

:confused: - October 12th 2005, 03:02 AMticbol
Prove cosh(x+y) = cosh(x)cosh(y) +sinh(x)sinh(y).

Righthand Side =

= [(e^x +e^-x)/2]*[e^y +e^-y)/2] +[(e^x -e^-x)/2]*[e^y -e^-y)/2]

= (1/4)[e^(x+y) +e^(x-y) +e^(-x+y) +e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) -e^(-x+y) +e^(-x-y)]

= (1/4)[e^(x+y) +e^(x-y) +e^(-x+y) +e^(-x-y) +e^(x+y) -e^(x-y) -e^(-x+y) +e^(-x-y)]

= (1/4)[2e^(x+y) +2e^(-x-y)]

= (1/4)[2e^(x+y) +2e^-(x+y)]

= (1/2)[e^(x+y) +e^-(x+y)]

= [e^(x+y) +e^-(x+y)]/2

= cosh(x+y)

= Lefthand Side.

Proven.

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Remember,

cosh(u) = (e^u +e^-u)/2

sinh(u) = (e^u -e^-u)/2