Hyperbolic function

Question

Find the numerical value of the expression:

cosh^(-1) (3)

no idea how to do please help

Answer 1

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

We solve (e^x+e^-x)/2 = 3

e^x+e^-x = 6

Let e^x = y

e^-x = 1/y

y + 1/y = 6

Multiplying through by y,

y^2 + 1 = 6y

y^2 - 6y + 1 = 0

y = 3 ± 2√2

e^x = y

x = ln(y)

x = ln(3 - 2√2), ln(3 + 2√2)

Note that ln(3 - 2√2) = - ln(3 + 2√2) because (3+2√2)(3-2√2) = 1. We should expect this result, as cosh is an even function

Thus, if you prefer, you could write the solution as - ln(3 + 2√2), ln(3 + 2√2)

The standard way to define the above so that we have a unique inverse is to use the positive value, ln(3 + 2√2)

If you want the numeric value of x, you can calculate the above, or use ACOSH(3) in Excel