can anyone solve this? i.sin^-1(-i.w)=sinh^-1(w)?

Question

the dots between i, sin^-1 and -i,w means multiply. Also i means imaginery number.

PLEASE SHOW WORK.

Answer 1

Let y = sinh(x) = (e^x - e^-x)/2
e^2x - 2y*e^2 - 1 = 0
solving the quadratic an d simplifying
x = sinh^-1 (y) = ln[ y + sqrt(y^2 + 1)]

Let's do the same thing for y = sinx = (e^ix - e^-ix)/2i
x = sin^-1(y) = (1/i) * ln [iy + sqrt(1-y^2)]

Now let's apply that to the problem at hand.
i*sin^-1 (-iw) = ln [w + sqrt(1 + w^2)]
And sinh^-1(w) = ln[ w + sqrt(w^2 + 1)]

Clearly they are equal, so the solution is w.
It's an identity not an equation. I knew it was something straightforward like that, I just couldn't put my finger on it.

Answer 2

Note that
sinh(t)=-i*sin(i*t) is an identity.
Therefore
w=sinh(sinh^-1(w))
w=-i*sin(i*sinh^-1(w))
-i.w=-sin(i*sinh^-1(w))
-i.w=sin(-i*sinh^-1(w))
sin(sin^-1(-i.w))
=sin(-i*sinh^-1(w))
sin^-1(-i.w)=-i*sinh^-1(w)
i*sin^-1(-i*w)=sinh^-1(w)

Note: In applying this identity, you should be careful to note that sin^-1 and sinh^-1 are multivalued mappings, so this "identity" really expresses the equality of two sets of values, not two numbers.

Answer 3

we know that

sin (w) = ( e^ iw - e^ -iw ) / 2i
sinh (w) = ( e^ w - e^ -w ) / 2

using the above definition for sinh & sin
let us prove the following property

for any angle A

sinh ( iA) = (e^iA - e^ -iA)/2
= i { (e^iA - e^ -iA)/2i}
= i sin A



let B = sin^ -1 ( -iw)

thus ur question is reduced as

i.B = sinh^ -1 w

or w = sinh (i.B)
using the property we had proved

w = i sin B
= i sin sin^-1 ( -iw)
= i (-iw )
= -i^2 w
= w
hence proved