Question

Prove hyperbolic functions

Answer 1

\[\sin(x+iy)= sinx*coshy+i cosx*sinhy\]

Answer 2

RHS is \[\frac{ e^{ix+y}-e^{-ix-y} }{ 2i }\]

Answer 3

Don't know where I'm wrong

Answer 4

\[\sin(x+iy)= \frac{ e^{ix-y}-e^{-ix+y} }{ 2i }\]

Answer 5

forgotten this stuff i'm afraid

Answer 6

Wish I remembered it! Check through

http://math2.org/math/trig/hyperbolics.htm

for the identities you need; try a new 'Net search if need be.

Answer 7

I have never studied this at all. If I'm not wrong, this is how they're defined:\[\sinh x = \frac{e^x - e^{-x}}{2}\]\[\cosh x = \frac{e^x + e^{-x}}2 \]I guess we'll have to use these two, then?

Answer 8

Yup, I used these to solve RHS

Answer 9

for sinx denominator should be 2i not 2

Answer 10

Oops, correct. Typo.

Answer 11

Oh, and it may be very critical to know that\[i\sin x = \frac{e^{ix}-e^{-ix}}{2}\]\[\cos x = \frac{e^{ix} + e^{-ix}}{2}\]

Answer 12

But that's not the point. I guess you can solve the problem now.

Answer 13

I tried to solve, but the thing is I get different signs

Answer 14

for the exponents, I thought it might be a silly mistake and now I can't figure out where is the mistake

Answer 15

\[\sin(x+iy)= \sin x \cos iy + \cos x\sin iy\]Now \(\cos iy\) is just \(\cosh y\) and \(\sin iy\) is just \(i \sinh y\)

Answer 16

Okay tell me why is this true \[\frac{ e^{-y}-e^y }{2i }= i \frac{ e^y-e^{-y} }{ 2 }\]

Answer 17

Probably because \(1/i = -i\).

Answer 18

Found it! It was because of my stupidity! 1/i =-i didnt strike me

Answer 19

Thanks

Answer 20

No problem. =)