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Alternative ways to show that cos(ix)=cosh(x)?

  • Thread starter Nathanael
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  • #1
Nathanael
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Homework Statement


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There is an exercise which says "Show that cos(ix)=cosh(x)"

2. The attempt at a solution

The only thing I can think of is that cos(ix) is a solution to y''=y with initial conditions y(0)=1 and y'(0)=0 therefore cos(ix)=cosh(x)

There's no solution given, but it is from a chapter on complex numbers from a calculus book, so I don't think knowledge of differential equations is expected, so I doubt this is the solution they were going for.

Does anyone have another method? Perhaps you can some how use [itex]e^{i\theta}= \cos \theta+i \sin \theta[/itex] ? I can't think of a way.
 

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  • #2
SteamKing
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Homework Statement


[/B]
There is an exercise which says "Show that cos(ix)=cosh(x)"

2. The attempt at a solution

The only thing I can think of is that cos(ix) is a solution to y''=y with initial conditions y(0)=1 and y'(0)=0 therefore cos(ix)=cosh(x)

There's no solution given, but it is from a chapter on complex numbers from a calculus book, so I don't think knowledge of differential equations is expected, so I doubt this is the solution they were going for.

Does anyone have another method? Perhaps you can some how use [itex]e^{i\theta}= \cos \theta+i \sin \theta[/itex] ? I can't think of a way.
Hint: Use the definition of the hyperbolic functions sinh(x) and cosh(x) in terms of the exponential function. Then use x = iy and Euler's formula to establish the connection between the hyperbolic functions and the trig functions. It's basically an exercise in algebra and identities.
 
  • #3
BvU
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Perhaps you can some how use [itex]e^{i\theta}= \cos \theta+i \sin \theta[/itex] ? I can't think of a way.
Yes you can use that. And I can think of a way (what is ##e^{-i\theta}## ? So what is ##\cos \theta## ? So what is ##\cos {ix}## ?)

But that's exactly SteamKing's hint, so I bow to his majesty ;)
 
  • #4
Nathanael
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[itex]\cosh(x)=\frac{1}{2}(e^{x}+e^{-x})=\frac{1}{2}(e^{-i(ix)}+e^{i(ix)})=\frac{1}{2}(\cos (-ix) + i\sin (-ix) + \cos(ix) + i\sin (ix))=\cos(ix)[/itex]

Thanks guys, I think that is the solution the author was expecting. I still prefer the y''=y method but this method is also fairly straightforward.

Did you guys already know that cos(ix)=cosh(x)? It might not be very useful (the chapter didn't even mention it) but I think it's quite interesting!
 
  • #5
SteamKing
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[itex]\cosh(x)=\frac{1}{2}(e^{x}+e^{-x})=\frac{1}{2}(e^{-i(ix)}+e^{i(ix)})=\frac{1}{2}(\cos (-ix) + i\sin (-ix) + \cos(ix) + i\sin (ix))=\cos(ix)[/itex]

Thanks guys, I think that is the solution the author was expecting. I still prefer the y''=y method but this method is also fairly straightforward.
It's like you said in an earlier post: you've got to know how to solve ODEs for this to happen.

Did you guys already know that cos(ix)=cosh(x)? It might not be very useful (the chapter didn't even mention it) but I think it's quite interesting!
Yep. It's one of those surprising side effects of Euler's formula.

The link between hyperbolic functions and trig functions is well known and comes in handy at times.

This link should come up in any well-written article on trig functions, however, which is why it often pays to do a little digging on the internet.
 
  • #6
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[itex]e^{i\theta}= \cos \theta+i \sin \theta[/itex] ? I can't think of a way.
Just substitute θ=ix. Then substitute θ=-ix. Then add.

Chet
 

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