How to solve arcsin(i)?

i got stuck and the correct answer is npi plus/minus i*arsinh((-1)^n)

Reply 1

arcsin i = x + iy
==> i = sin(x + iy)
==> i = sin x cos iy + cos x sin iy
==> i = sin x cosh y + i cos x sinh y

Compare real and imaginafsrjk r parts.

Reply 2

mahjt

I've got two simultaneous equations for arcsin(i) but I cannot find a way to solve them. How do you solve the simultaneous equations?

Reply 3

Hasufel

Original post by Evan247

i got stuck and the correct answer is npi plus/minus i*arsinh((-1)^n)

solve what?

Reply 4

joostan

Original post by mahjt

I've got two simultaneous equations for arcsin(i) but I cannot find a way to solve them. How do you solve the simultaneous equations?

What equations have you got? :smile:

Reply 5

Hasufel

it might possibly help (even if i don`t know the question) to give you the identity:

Unparseable latex formula:

\displaystyle arcsin(z)=\frac{1}{i}ln(iz+\sqrt{1-z^{2}})=\frac{1}{i}arcsinh(iz)}

which gives:

Unparseable latex formula:

arcsin(z)=-i \times arcsinh(-1)}

Reply 6

natninja

Original post by Evan247

i got stuck and the correct answer is npi plus/minus i*arsinh((-1)^n)

the answer is that because

sin(x) =(e^ix - e^-ix)/2i

then you can solve for x and it is rather neat :smile:

Reply 7

joostan

Original post by natninja

the answer is that because

sin(x) =(e^ix - e^-ix)/2i

then you can solve for x and it is rather neat :smile:

The person with the new problem is a couple of posts down :tongue:

Reply 8

natninja

Original post by mahjt

I've got two simultaneous equations for arcsin(i) but I cannot find a way to solve them. How do you solve the simultaneous equations?

Try using the definition sin(x)=(e^ix - e^-ix)/2i

Reply 9

Hasufel

Original post by joostan

What equations have you got? :smile:

Ditto

Reply 10

Abhishek.is.here

Let y=arcsin iI=sin yCos y = √2e^i.i = cos i i sin i

Reply 11

RichE

Original post by Abhishek.is.here

Let y=arcsin iI=sin yCos y = √2e^i.i = cos i i sin i

The previous thread was 4 years old.

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