Trigonometric and Hyperbolic Function?

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

2 Answers
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

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