How do you prove #Arc cosh(x) = ln(x + (x^2 - 1)^(1/2))#?

Question

37350 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-04T21:41:01

see below

Let #y=cosh^-1x# then by definition

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

#2x=e^y+e^-y#

#e^y-2x+e^-y=0#

#e^y-2x+1/e^y=0#

#e^(2y)-2xe^y+1=0#

Let #u=e^y# then we have

#u^2-2xu+1=0#--> Now use quadratic formula to solve

#u=(2x+-sqrt(4x^2-4))/2#

#e^y = (2x+-sqrt(4x^2-4))/2#

#e^y = (2x+-sqrt(4(x^2-1)))/2#

#2e^y = 2x+-2sqrt(x^2-1)#

#e^y=x+-sqrt(x^2-1)#

#ln e^y = ln (x+-sqrt(x^2-1))#

#y=ln (x+sqrt(x^2-1))#

#cosh ^-1 x = ln (x+sqrt(x^2-1))#