Trigonometric and Hyperbolic Function?

Prove: arctan(sinh(x)) = arcsin(tanh(x))

2 Answers
Steve M
Aug 8, 2017

We are asked to verify that:

# arctan(sinh(x)) = arcsin(tanh(x)) #

There are shorter ways to prove this result, but referring back to the hyperbolic definitions is a fool-proof method that always works:

Consider the LHS first:

Let # alpha = arctan(sinh(x)) #
# => tan alpha = sinhx # ..... [A]

Now, consider the RHS:

Let #beta = arcsin(tanh(x)) #
# => sin beta = tanh(x) #

# :. sin beta = sinh(x)/cosh(x) #

# " " = (1/2(e^x-e^-x))/(1/2(e^x+e^-x)) #

# " " = (e^x-e^-x)/(e^x+e^-x) #

And using the trig identity #sin^2A+cos^2A-=1# we can also find an expression for #cos beta#

# sin^2 beta +cos^2 beta = 1 #
# :. cos^2 beta = 1 - ((e^x-e^-x)/(e^x+e^-x))^2#

# " " = 1 - ((e^x-e^-x)^2)/((e^x+e^-x)^2)#

# " " = ((e^x+e^-x)^2 - (e^x-e^-x)^2) / (e^x+e^-x)^2#

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

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

# :. cos beta = 2 / (e^x+e^-x) #

And now we have expression for #sin beta# and #cos beta# we can form an expression for 'tan beta#, thus:

# tan beta = sin beta / cos beta #

# " " = ( (e^x-e^-x)/(e^x+e^-x) ) / (2 / (e^x+e^-x)) #

# " " = ( (e^x-e^-x)/(e^x+e^-x) ) * ((e^x+e^-x)/2) #

# " " = (e^x-e^-x)/2 #
# " " = sinh x # ... [B]

Comparing [B] with [A], we have:

# tan beta = sinhx \ \ \ # and # \ \ \ tan alpha = sinhx #

Hence:

# tan alpha = tan beta => alpha = beta #

And so we can conclude that:

# alpha = arctan(sinh(x)) = beta = arcsin(tanh(x)) #

Hence:

# arctan(sinh(x)) = arcsin(tanh(x)) \ \ \ # QED

Steve M
Aug 8, 2017

We are asked to verify that:

# arctan(sinh(x)) = arcsin(tanh(x)) #

Shorter Method:

Consider the LHS first:

Let # alpha = arctan(sinh(x)) #
# => tan alpha = sinhx # ..... [A]

Now, consider the RHS:

Let #beta = arcsin(tanh(x)) #
# => sin beta = tanh(x) #

Now:

# tan beta = (sin beta)/(cos beta) #

# " " = (tanhx)/(sqrt(1-sin^2 beta)) #

# " " = (tanhx)/(sqrt(1-tanh^2 x)) #

# " " = (tanhx)/(sqrt(sech^2 x)) #

# " " = (tanhx)/(sech x) #

# " " = (sinhx/cosh)/(1/cosh x) #

# " " = sinhx #

# " " = tan alpha #

And so we conclude that:

# alpha = beta => arctan(sinh(x)) = arcsin(tanh(x)) \ \ \ # QED

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