How do you show $(coshx + sinhx)^n = cosh(nx) + sinh(nx)$ for any real number n?

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Answer 1 · 2016-05-04T22:05:54

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Use the definition $cosh x = (e^x+e^-x)/2 and sinh x = (e^x-e^-x)/2$

Left Side: $[(e^x+e^-x)/2 + (e^x-e^-x)/2]^n$

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

$=[(2e^x)/2]^n$

$=e^(xn)$

Right Side: $=(e^(nx)+e^(-nx))/2 +(e^(nx)-e^(-nx))/2$

$=(e^(nx)+e^(-nx) +e^(nx)-e^(-nx))/2$

$=(2e^(nx))/2$

$=e^(nx)$

$=$ Left side

$:. LHS=RHS$

How do you show (coshx + sinhx)^n = cosh(nx) + sinh(nx)(coshx + sinhx)^n = cosh(nx) + sinh(nx) for any real number n?