Question

If cos(x+iy) cos(u+iv)=1,where x,y,u,v are real then show that tanh^2y cosh^2v=sin^2u

Answer 1

expanding and comparing real and imaginary parts led to nowhere...

Answer 2

@SithsAndGiggles

Answer 3

@hartnn That is sin^2(u) not sinh^2(u) ?

Answer 4

its sin^2 u

Answer 5

The reason I'm asking is because ...
\[\tanh^2(x) \cdot \cosh^2(x)=(\frac{e^x-e^{-x}}{e^x+e^{-x}})^2(\frac{e^x+e^{-x}}{2})^2=(\frac{e^x-e^{-x}}{2})^2=\sinh^2(x)\]

Answer 6

I do realize we do have those different variables though

Answer 7

typing mistakes in the question document is quite common, so if we can prove it to be sinh^2 u .....

Answer 8

Well I'm not really sure yet about that. I'm just thinking.

Answer 9

but i see no reason why it must not be sin^2 u

Answer 10

@satellite73 @surjithayer @mathmale
any ideas how to start...?

Answer 11

@experimentX

Answer 12

i think with usual cos(A+B) formula
just note that cos(ix) = cosh(x) and sin(ix)=-sinh(x)

Answer 13

Comparing real and imaginary parts
cos x cosh y cos u cosh v-sin x sinh y sin u sinh v=1
cos x cosh y sin u sinh v+sin x sinh y cos u cosh v=0

Answer 14

yes that is one of the tool needed to solve it.

Answer 15

ok...then how would tanh terms and squares of them woud come into picture ?

Answer 16

lol ... i haven't done it till now ... trolling on fb
lemme try this and post in 10 min

Answer 17

sure, thanks

Answer 18

\[
\cos(x+iy) = \cos(x)\cosh(y) +i \sin(x)\sinh(y) \\
\cos(u+iv) = \cos(u)\cosh(v) + i \sin(u)\sinh(v)
\]
multiplying them gives,
\[ \cos(x)\cosh(y) \cos(u)\cosh(v) - \sin(x)\sinh(y) \sin(u)\sinh(v) \\ + i (\cos(x)\cosh(y)\sin(u)\sinh(v)+\cos(u)\cosh(v)\sin(x)\sinh(y)) = 1 \]

Answer 19

thats what i got

Answer 20

the second part gives,
\[ \tan(u)\tanh(v) = - \tan(x) \tanh(y) \]
the first part gives,
\[
\cos(x)\cosh(y) \cos(u)\cosh(v) - \sin(x)\sinh(y) \sin(u)\sinh(v) = 1\]

Answer 21

right

Answer 22

isn't there x term in question, looks like we migth begin by eliminating x term.

Answer 23

no x term in the question,
so can we try dividing the 2nd equation by cos x and replacing for tan x
but there would still be sc x on right side...

Answer 24

yes ... that's creating problem.

Answer 25

any other approach.....
like cos x+iy = sec u+iv
or something...

Answer 26

looks worth try

Answer 27

but ... sec(a+ib) isn't going to be that nice

Answer 28

even tan^2 a+ib ?

cos^2 (x+iy) = 1+ tan^2 (u+iv)

how about this ?

Answer 29

still eliminating x would be a challenge in that expansion...

Answer 30

IDK ... but where did you get the question BTW? did you check for the values of x,y,u,v where this relation is true?
i couldn't guess any value other than 0's

Answer 31

university question paper
last year

Answer 32

which means it isn't supposed to be hard :P

Answer 33

haven't you already finished Masters?

Answer 34

yep, i did
solving previous year's question papers to help a friend

Answer 35

damn man ... this topic is supposed to be covered undergrad. finding it hard to solve lol.

Answer 36

exactly!
there could be some twist...

Answer 37

lemme me try this tonight. I am pretty distracted right now. If I solve it, i'll msg you on fb.

Answer 38

sure, np., thanks :)

Answer 39

i have also asked my friend to do it ... let's see what he gets.

Answer 40

@hartnn I thought about this problem a lot yesterday and it stumps me too. I'm still trying but I'm not coming to anything useful.

Answer 41

Even i tried solving it in different ways...no luck in google search too...this problem is bugging me from a week :O

Answer 42

\[\text{ Suppose } \cos(x+iy) \cdot \cos(u+iv)=1 \text{ where x,y,u,v are real .} \\ \text{ W.T.S. } \tanh^2(y) \cdot \cosh^2(v)=\sin^2(u)\]
I.
\[\cos(x+iy)=\cos(x)\cosh(y)-i \sin(x) \sinh(y) \]
II.
\[\cos(u+iv)=\cos(u)\cosh(v)-i \sin(u) \sinh(v)\]
Multiplying I and II gives:
\[\cos(x)\cos(u)\cosh(y)\cosh(v)\\ -\sin(x)\sin(u)\sinh(y)\sinh(v) \\ -i \cos(x)\sin(u)\cosh(y)\sinh(v) \\ -i \sin(x) \cos(u) \sinh(y) \cosh(v) \]
Now I times II is suppose to give 1 since we are supposing this is true for real values x,y,u, and v.
So we have the equation:
\[\cos(x)\cos(u)\cosh(y)\cosh(v)\\ -\sin(x)\sin(u)\sinh(y)\sinh(v) \\ -i \cos(x)\sin(u)\cosh(y)\sinh(v) \\ -i \sin(x) \cos(u) \sinh(y) \cosh(v) =1\]
This means by comparing real and imaginary parts we have two equations:
\[\ \text{ A. } cos(x)\cos(u)\cosh(y)\cosh(v)-\sin(x)\sin(u)\sinh(y)\sinh(v)=1 \\ \text{ and } \\ \text{ B. }\cos(x)\sin(u)\cosh(y)\sinh(v)+\sin(x)\cos(u)\sinh(y)\cosh(v)=0\]


So let's remind ourselves of what we want to show:
\[\tanh^2(y) \cosh^2(v)=\sin^2(u)\]

So we are suppose to take equations A. and B. and somehow eliminate the x part and basically try to get sin^2(u) on one side of an equation.

Since we have the square thing on both sides, I wonder if that means we need to multiply A and B...
Let's try...

A*B=1*0=0

So we have:
\[\cos^2(x)\sin(u)\cos(u)\cosh^2(y)\cosh(v)\sinh(v)\\ +\sin(x)\cos(x)\cos^2(u)\sinh(y)\cosh(y)\cosh^2(v) \\ - \sin(x)\cos(x)\sin^2(u)\sinh(y)\cosh(y)\sinh^2(v) \\ -\sin^2(x)\sin(u)\cos(u)\sinh^2(y)\sinh(v)\cosh(v)=0\]

I think we may be able to do some factoring here...
This is what happens if we group 1 and 4 together then 2 and 3. (talking about the terms)
\[\sin(u)\cos(u)\sinh(v)\cosh(v) \cdot [\cos^2(x)\cosh^2(y)-\sin^2(x)\sinh^2(y)] \\ +\sin(x)\cos(x)\sinh(y)\cosh(y) \cdot [\cos^2(u)\cosh^2(v)-\sin^2(u)\sinh^2(v)]=0\]

\[\sin(u)\cos(u)\sinh(v)\cosh(v)[\cos^2(x)\cos^2(y)-\sin^2(x)\sinh^2(y)]=\\-\sin(x)\cos(x)\sinh(y)\cosh(y)[\cos^2(u)\cosh^2(v)-\sin^2(u)\sinh^2(v)]\]

I'm going to go ahead and post this just in case my openstudy dies on me.
It is probably not helpful because I have been staring at this last thing I wrote forever.

Answer 43

I thought I was going to be able to use cosh^2(x)-sinh^2(x)=1 :(
but then those lousy factos cos^2(x) and sin^2(x) got in the way

Answer 44

thanks for the link @ganeshie8
i would post that link so that experimentX and myin would get the solution too

https://docs.google.com/file/d/0B7ss_Y8M_VEgMzRYVmdidjFnN00/edit

it was truly difficult problem....unexpected to come in university exam :O

Answer 45

Yeah, I don't think I would have came up with that solution.