How do you prove #sinh^-1 (t) = ln(t+ sqrt(t^2 + 1))#?

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Answer 1 · 2016-05-04T21:32:49

see below

Let #y=sinh^-1t# then by definition

#t=sinh y =(e^y-e^-y)/2#

#2t=e^y-e^-y#

#e^y-2t-e^-y=0#

#e^y-2t-1/e^y=0#

#e^(2y)-2te^y-1=0#

Let #x=e^y# then we have

#x^2-2tx-1=0#--> Now use quadratic formula to solve

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

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

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

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

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

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

#y=ln (t+sqrt(t^2+1))#

#sinh ^-1 t = ln (t+sqrt(t^2+1))#

#=#Right side