Trigonometric and Hyperbolic Function?
Prove: arctan(sinh(x)) = arcsin(tanh(x))
Prove: arctan(sinh(x)) = arcsin(tanh(x))
2 Answers
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^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
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
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