Question

How do you prove `sin^-1(tanh(x)) = tan^-1(sinh(x))`?

Answer 1

The way that one proves an identity is to make substitutions to only one side until it is identical to the other side:

Explanation:

Given: `sin^-1(tanh(x)) = tan^-1(sinh(x))`

Use the property `u = sin^-1(sin(u))` on the right side and mark as equation [1]:

`sin^-1(tanh(x)) = sin^-1(sin(tan^-1(sinh(x)))" [1]"`

Digress and prove that `sin(tan^-1(sinh(x)) = tanh(x)`

An alternate form for `tan^-1(u) = i/2ln(1-iu)-i/2ln(1+iu)`

Substitute `sinh(x)` for u:

`tan^-1(sinh(x)) = i/2ln(1-isinh(x))-i/2ln(1+isinh(x))`

An alternate form for `sin(v) = i/2e^(-iv)-i/2e^(iv)`

`sin(tan^-1(sinh(x))) = i/2e^(-i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x)))-i/2e^(i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x))))`

Distribute through `-i`:

`sin(tan^-1(sinh(x))) = i/2e^((-i^2/2ln(1-isinh(x))+i^2/2ln(1+isinh(x))))-i/2e^(i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x))))`

Distribute through `i`:

`sin(tan^-1(sinh(x))) = i/2e^((-i^2/2ln(1-isinh(x))+i^2/2ln(1+i(sinh(x))))-i/2e^((i^2/2ln(1-isinh(x))-i^2/2ln(1+isinh(x))))`

use the property `i^2 = -1`:

`sin(tan^-1(sinh(x))) = i/2e^((1/2ln(1-isinh(x))-1/2ln(1+isinh(x))))-i/2e^((-1/2ln(1-isinh(x))+1/2ln(1+isinh(x))))`

Write the `1/2` in the exponent as a square root:

`sin(tan^-1(sinh(x))) = i/2e^((ln(sqrt(1-isinh(x)))-ln(sqrt(1+isinh(x)))))-i/2e^((-ln(sqrt(1-isinh(x)))+ln(sqrt(1+isinh(x)))))`

Factor out `i/2`:

`sin(tan^-1(sinh(x))) = i/2{e^((ln(sqrt(1-isinh(x)))-ln(sqrt(1+isinh(x))))-e^((-ln(sqrt(1-isinh(x)))+ln(sqrt(1+isinh(x)))))}`

Use the property of logarithms `ln(a) - ln(b) = ln(a/b)`:

`sin(tan^-1(sinh(x))) = i/2{e^((ln((sqrt(1-isinh(x)))/(sqrt(1+isinh(x))))))-e^((ln((sqrt(1+isinh(x)))/(sqrt(1-isinh(x))))))}`

Use the property `e^ln(u) = u`:

`sin(tan^-1(sinh(x))) = i/2{(sqrt(1-isinh(x)))/(sqrt(1+isinh(x)))-(sqrt(1+isinh(x)))/(sqrt(1-isinh(x)))}`

When we make a common denominator we obtain:

`sin(tan^-1(sinh(x))) = i/2{(1-isinh(x))/(sqrt(1+sinh^2(x)))-(1+isinh(x))/(sqrt(1+sinh^2(x)))}`

Combine over the common denominator:

`sin(tan^-1(sinh(x))) = i/2{(-2isinh(x))/(sqrt(1+sinh^2(x)))}`

The leading coefficient multiplied into the numerator becomes 1:

`sin(tan^-1(sinh(x))) = sinh(x)/(sqrt(1+sinh^2(x)))`

Use the identity `1 + sinh^2(x) = cosh^2(x)`:

`sin(tan^-1(sinh(x))) = sinh(x)/(sqrt(cosh^2(x)))`

`sin(tan^-1(sinh(x))) = sinh(x)/cosh(x)`

`sin(tan^-1(sinh(x))) = tanh(x)`

Substitute this into equation [1]:

`sin^-1(tanh(x)) = sin^-1(tanh(x))` Q.E.D.