Prove the hyperbolic function corresponding to the given trigonometric function

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  • #1
chwala
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Homework Statement:
Prove the hyperbolic function corresponding to the given trigonometric function.

##8 \sin^4u = 3-4\cos 2u+\cos 4u##
Relevant Equations:
Hyperbolic trig. equations properties.
##8 \sin^4u = 3-4\cos 2u+\cos 4u##

##8 \sinh^4u = 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##

##8 \sin^4u = 3-4-8\sinh^2 u+ \cosh 2u \cosh 2u + \sinh 2u \sinh 2u##

##8 \sinh^4u = 3-4+1-8\sinh^2 u+ 4\sinh^2u +4\sinh^4 u + 4\sinh^2 u + 4\sinh^4 u##

##8 \sinh^4u = -8\sinh^2 u+ 8\sinh^2u +8\sinh^4 u##

##8 \sinh^4u=8\sinh^4 u##

Thus proved that lhs = rhs

Insight welcome...nothing much here...
 
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  • #2
robphy
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To me, the question sounds odd.

How would you answer the question if the given equation were
$$1=\cos^2 u +\sin^2 u$$
instead?
 
  • #3
chwala
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To me, the question sounds odd.

How would you answer the question if the given equation were
$$1=\cos^2 u +\sin^2 u$$
instead?
That's what the textbook suggests...

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1669379541915.png
 
  • #4
robphy
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That's what the textbook suggests...
Ok, then.
How would you use your approach to solve this version of the problem?
 
  • #5
chwala
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Ok, then.
How would you use your approach to solve this version of the problem?
I will need to check on this... Do we have a direct relationship between trig functions and hyperbolic ...i do not think so...

##\sin^2 u + \cos^2 u =1## is a trig property and its proof is easy! ...make use of the right angle triangle and pythagoras theorem that is if i understand your question.
 
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  • #6
robphy
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I will need to check on this... Do we have a direct relationship between trig functions and hyperbolic ...i do not think so...

##\sin^2 u + \cos^2 u =1## is a trig property and its proof is easy! ...make use of the right angle triangle and pythagoras theorem that is if i understand your question.
Yes, [itex] \sin^2 u + \cos^2 u =1 [/itex] is easy to prove as an example of the Pythagorean theorem.
But the question asks for the "hyperbolic function corresponding to the given trigonometric function".
So, use [itex] \sin^2 u + \cos^2 u =1 [/itex] as "the given".



  • Possibly useful:
    [itex]e^u =\cosh u +\sinh u[/itex]
    and thus [itex]e^{-u} = \cosh (-u) + \sinh (-u)=\cosh u - \sinh u[/itex]
    So, one can write [itex]\cosh u[/itex] and [itex]\sinh u[/itex] in terms of [itex]e^u[/itex] and [itex]e^{-u}[/itex].

    [itex]e^{ix} =\cos x +i\sin x[/itex]
    and thus [itex]e^{-ix} =e^{i(-x)}=\cos (-x) + i\sin (-x)=\cos x - i\sin x[/itex]
    So, one can write [itex]\cos x[/itex] and [itex]\sin x[/itex] in terms of [itex]e^{ix}[/itex] and [itex]e^{-ix}[/itex].

  • Further,
    [itex]\cosh (ix) +\sinh (ix) =e^{ix} =\cos x +i\sin x[/itex] ,
    where each side is written as an even function of x plus an odd function of x.
 
  • #7
chwala
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Yes, [itex] \sin^2 u + \cos^2 u =1 [/itex] is easy to prove as an example of the Pythagorean theorem.
But the question asks for the "hyperbolic function corresponding to the given trigonometric function".
So, use [itex] \sin^2 u + \cos^2 u =1 [/itex] as "the given".



  • Possibly useful:
    [itex]e^u =\cosh u +\sinh u[/itex]
    and thus [itex]e^{-u} = \cosh (-u) + \sinh (-u)=\cosh u - \sinh u[/itex]
    So, one can write [itex]\cosh u[/itex] and [itex]\sinh u[/itex] in terms of [itex]e^u[/itex] and [itex]e^{-u}[/itex].

    [itex]e^{ix} =\cos x +i\sin x[/itex]
    and thus [itex]e^{-ix} =e^{i(-x)}=\cos (-x) + i\sin (-x)=\cos x - i\sin x[/itex]
    So, one can write [itex]\cos x[/itex] and [itex]\sin x[/itex] in terms of [itex]e^{ix}[/itex] and [itex]e^{-ix}[/itex].

  • Further,
    [itex]\cosh (ix) +\sinh (ix) =e^{ix} =\cos x +i\sin x[/itex] ,
    where each side is written as an even function of x plus an odd function of x.
Ok...I had initially thought of Euler's formula but discounted(ignored) it because of the complex exponent...cheers learning point for me...I'll definitely look at this...
 
  • #8
PeroK
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Homework Statement:: Prove the hyperbolic function corresponding to the given trigonometric function.

##8 \sin^4u = 3-4\cos 2u+\cos 4u##
Relevant Equations:: Hyperbolic trig. equations properties.

##8 \sin^4u = 3-4\cos 2u+\cos 4u##

##8 \sinh^4u = 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##

##8 \sin^4u = 3-4-8\sinh^2 u+ \cosh 2u \cosh 2u + \sinh 2u \sinh 2u##

##8 \sinh^4u = 3-4+1-8\sinh^2 u+ 4\sinh^2u +4\sinh^4 u + 4\sinh^2 u + 4\sinh^4 u##

##8 \sinh^4u = -8\sinh^2 u+ 8\sinh^2u +8\sinh^4 u##

##8 \sinh^4u=8\sinh^4 u##

Insight welcome...nothing much here...
I think the question wants you to find an expression for ##8\sinh^4 u## in terms of ##\cosh 2u, \cosh 4u## etc.
 
  • #9
chwala
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I think the question wants you to find an expression for ##8\sinh^4 u## in terms of ##\cosh 2u, \cosh 4u## etc.
I thought proof of identities one just has to show that the lhs = rhs. I changed everything on the rhs in terms of ##\sinh u##...that should be fine.

Also consider the textbook solution remarks on post ##3##.
 
  • #10
robphy
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I thought proof of identities one just has to show that the lhs = rhs. I changed everything on the rhs in terms of ##\sinh u##...that should be fine.
Ok, then try it for [itex]\sin^2 u + \cos^2 u =1[/itex].
 
  • #11
pasmith
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I will need to check on this... Do we have a direct relationship between trig functions and hyperbolic ...i do not think so...

We have
[tex]\begin{split} \cos x &= \cosh (ix) \\
\sin x &= -i\sinh(ix) \end{split}.[/tex] Since these hold for all [itex]x \in \mathbb{C}[/itex], it is enough to replace each instance of [itex]\cos(x)[/itex] with [itex]\cosh(x)[/itex] and each instance of [itex]\sin(x)[/itex] with [itex]-i \sinh(x)[/itex]. Thus [tex]
\cos^2 x + \sin^2 x = 1 \rightarrow \cosh^2 x - \sinh^2 x = 1.[/tex]
 
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  • #12
chwala
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We have
[tex]\begin{split} \cos x &= \cosh (ix) \\
\sin x &= -i\sinh(ix) \end{split}.[/tex] Since these hold for all [itex]x \in \mathbb{C}[/itex], it is enough to replace each instance of [itex]\cos(x)[/itex] with [itex]\cosh(x)[/itex] and each instance of [itex]\sin(x)[/itex] with [itex]-i \sinh(x)[/itex]. Thus [tex]
\cos^2 x + \sin^2 x = 1 \rightarrow \cosh^2 x - \sinh^2 x = 1.[/tex]
With this said, then the approach in post ##1## is sufficient.
 
  • #13
chwala
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I think the question wants you to find an expression for ##8\sinh^4 u## in terms of ##\cosh 2u, \cosh 4u## etc.
...but isn't that what I did in post ##1##?...I made use of the hyperbolic trigonometry properties as required...unless I am missing something here...
 
  • #14
PeroK
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...but isn't that what I did in post ##1##?...I made use of the hyperbolic trigonometry properties as required...unless I am missing something here...
Sort of, but in a back-to-front way!
 
  • #15
PeroK
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What you want is something like:
$$3 - 4\cosh 2u + \cosh 4u$$$$ = 3 - 4 \big [2\sinh^2 u + 1 \big ] + 2\cosh^2 2u -1$$$$= -2 - 8\sinh^2 u + 2\big[2\sinh^2 u + 1\big]^2$$$$= -2 - 8\sinh^2 u +2 \big[4\sinh^4 u + 4\sinh^2 u +1 \big]$$$$= 8\sinh^4 u$$
 
  • #17
robphy
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...but isn't that what I did in post ##1##?...I made use of the hyperbolic trigonometry properties as required...unless I am missing something here...
Your method happens to work for that problem, in the spirit of the textbook solutions answers to (a),(d),(f),(g).
I asked you about [itex]1=\cos^2 u +\sin^2 u[/itex]
because it's not clear to me that a general method is being applied in your first post.
(It happens to be that [itex] (i^4)=1[/itex].)
So, I wanted to see how you would handle [itex]1=\cos^2 u +\sin^2 u[/itex].

My post #6 was trying to suggest the method used by @pasmith in #11.
 
  • #18
pasmith
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Why use one method for (a),(d),(f),(g) and not the others?

It happens in those examples that replacing [itex]\sin[/itex] with [itex]- i \sinh[/itex] results in every term being multiplied by the same power of [itex]-i[/itex], so the result is the same as replacing [itex]\sin[/itex] with [itex]\sinh[/itex]. This is not true of the other examples.

For the examples which involve derivatives and integrals, it is necessary to set [itex]y = ix[/itex] when applying the identities and use the chain rule/integration by substitution as relevant.
 
  • #19
robphy
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It happens in those examples that replacing [itex]\sin[/itex] with [itex]- i \sinh[/itex] results in every term being multiplied by the same power of [itex]-i[/itex], so the result is the same as replacing [itex]\sin[/itex] with [itex]\sinh[/itex]. This is not true of the other examples.

For the examples which involve derivatives and integrals, it is necessary to set [itex]y = ix[/itex] when applying the identities and use the chain rule/integration by substitution as relevant.

Yes, I know.
My point is that the “solutions” didn’t explain why naive substitution happens to work for those cases and not for others.
So, the general method isn’t revealed in the “solution”.
Those are merely “answers”… with no explanations.
 
  • #20
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chwala said:
##8 \sin^4u = 3-4\cos 2u+\cos 4u##
##8 \sinh^4u = 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##
##8 \sin^4u = 3-4-8\sinh^2 u+ \cosh 2u \cosh 2u + \sinh 2u \sinh 2u##
##8 \sinh^4u = 3-4+1-8\sinh^2 u+ 4\sinh^2u +4\sinh^4 u + 4\sinh^2 u + 4\sinh^4 u##
##8 \sinh^4u = -8\sinh^2 u+ 8\sinh^2u +8\sinh^4 u##
##8 \sinh^4u=8\sinh^4 u##
In addition to what I said in your other thread about a proof involving hyperbolic trig functions, about not repeating the LHS in every equation, your last equation here shows that ##8\sinh^4 u## is equal to itself. Surely this is not something that anyone would doubt, but on its own is not proof that the LHS
is equal to the original RHS expression.

The work done by @PeroK in this thread shows the expression on the LHS appearing exactly once, followed by a chain of equal expressions that lead ultimately to the desired RHS expression.
 
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  • #21
vela
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I thought proof of identities one just has to show that the lhs = rhs. I changed everything on the rhs in terms of ##\sinh u##...that should be fine.
The problem is your proof that L=R is likely based on faulty reasoning. You assume L=R and show that this leads to L=L. Then the problem occurs: you appear to reason that since L=L is true, your assumption that L=R is therefore true. That's a logical fallacy.

If your reasoning is that the steps to reach L=L are all reversible, so that L=L also implies L=R, then you should say that explicitly. Or better yet, just write the steps in reverse order in the first place, which is that @PeroK essentially did.
 
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  • #22
chwala
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The problem is your proof that L=R is likely based on faulty reasoning. You assume L=R and show that this leads to L=L. Then the problem occurs: you appear to reason that since L=L is true, your assumption that L=R is therefore true. That's a logical fallacy.

If your reasoning is that the steps to reach L=L are all reversible, so that L=L also implies L=R, then you should say that explicitly. Or better yet, just write the steps in reverse order in the first place, which is that @PeroK essentially did.
Noted...I'll look at this ...but in general; citing exam boards (exams and markscheme)... that's how identities are proved ie showing lhs= rhs...you're indicating this approach isn't correct?
I will check reference and share here.

I looked at @PeroK approach, and I really can't tell the difference between his approach and mine...he worked on the rhs of the equation by simplifying the identities ...which is exactly what I did. Note that I did not substitute step-step for all simplification as I assumed reader is conversant with the properties of hyperbolic function. I think I'll need to post my step by step simplification to the required identity.
 
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  • #23
chwala
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Homework Statement:: Prove the hyperbolic function corresponding to the given trigonometric function.

##8 \sin^4u = 3-4\cos 2u+\cos 4u##
Relevant Equations:: Hyperbolic trig. equations properties.

##8 \sinh^4u = 3-4\cos 2u+\cos 4u##

##8 \sinh^4u = 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##

##8 \sin^4u = 3-4-8\sinh^2 u+ \cosh 2u \cosh 2u + \sinh 2u \sinh 2u##

##8 \sinh^4u = 3-4+1-8\sinh^2 u+ 4\sinh^2u +4\sinh^4 u + 4\sinh^2 u + 4\sinh^4 u##

##8 \sinh^4u = -8\sinh^2 u+ 8\sinh^2u +8\sinh^4 u##

##8 \sinh^4u=8\sinh^4 u##

Thus proved that lhs = rhs

Insight welcome...nothing much here...
Since;

##\sin ix = i \sinh x## and ##\cos ix = i \ cosh x ##

then we shall replace ##\sin x## with ##\sinh x## and ##\cos x## with ##\cosh x##

thus;

##8 \sinh^4u ##

##= 3-4\cosh 2u+\cosh 4u##

##= 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##

##= 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##

##= 3-4-8\sinh^2 u+ (1+2\sinh^2 u) (1+2\sinh^2 u) + (2\sinh u \cosh u)(2\sinh u \cosh u)##

##= 3-4-8\sinh^2 u+ (1+4\sinh^2 u+4\sinh^4 u) + (4\sinh^2 u (1+\sinh^2 u)##

##= 3-4-8\sinh^2 u+ 1+4\sinh^2 u+4\sinh^4 u + 4\sinh^2 u +4\sinh^4 u##

##= 3-4-8\sinh^2 u+ 1+4\sinh^2 u+4\sinh^2 u + 4\sinh^4 u +4\sinh^4 u##

##= 3-4+1-8\sinh^2 u+8\sinh^2 u + 8\sinh^4 u ##

##= 0-0+ 8\sinh^4 u##

##=8\sinh^4 u##

Thus proved.
 
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  • #24
chwala
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Why use one method for (a),(d),(f),(g) and not the others?
This gives "answers" but not useful solutions.
What book is this?
Pure Maths 5 & 6 - Hugh Neill and Douglas Quadling -OCR
 
  • #25
vela
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I looked at @PeroK approach, and I really can't tell the difference between his approach and mine.
I laid the problem out pretty simply in my previous post. I'm not sure what else I could say that's going to help you see the point. But until you understand the point I was making, I doubt you'll be able to see the difference between his approach and yours.
 
  • #26
robphy
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##8 \sin^4u ##

##= 3-4\cos 2u+\cos 4u##

##= 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##
...snip...

Thus proved.
The first equality is true.
The second is not.

Maybe your first line should read
##3-4\cosh 2u+\cosh 4u ##
##=3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)##
##=...##

The problem is really about finding the hyperbolic-analogue of a trigonometric identity.
So, start with a hyperbolic analogue on one side (probably the simpler side)
and find an identity for it involving hyperbolic functions.

If you do it by substitution, you must use complex numbers (as I hinted in my early post #6).
 
  • #27
chwala
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I laid the problem out pretty simply in my previous post. I'm not sure what else I could say that's going to help you see the point. But until you understand the point I was making, I doubt you'll be able to see the difference between his approach and yours.
@vela thanks...I've seen your point. We can't start the proof by assuming that the lhs= rhs. Cheers.
 
  • #28
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We can't start the proof by assuming that the lhs= rhs.
It is possible to do things this way, provided that you have a sequence of equivalent equations. You get equivalent equations by applying reversible (i.e., invertible) operations to both sides, or by adding zero (in some form) to one side, multiplying one side by one (in some form), or replacing expressions by simplified or expanded versions of themselves. Operations that aren't invertible include raising both sides to some power, or taking roots of both sides, and other noninvertible operations. Here's a simple example.

Show that ##2\cos^2(x) - 1 = \cos^2(x) - \sin^2(x)##

Proof:
##2\cos^2(x) - 1 = \cos^2(x) - \sin^2(x)##
##\Leftrightarrow 2\cos^2(x) - 1 = \cos^2(x) - (1 - \cos^2(x))##
##\Leftrightarrow 2\cos^2(x) - 1 = \cos^2(x) - 1 + \cos^2(x))##
##\Leftrightarrow 2\cos^2(x) - 1 = 2\cos^2(x) - 1##

Each step after the first is reversible, since all I did was to replace an expression by another that was identically equal to the one I replaced. Because I ended with an identity, and each step was reversible, this means that the first equation is also an identity, and so the original equation is proven to be true.

Having done all this, my preference is to start with one of the sides of the original equation, and work with the other side until I reach the expression on the other side.
 
  • #29
chwala
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It is possible to do things this way, provided that you have a sequence of equivalent equations. You get equivalent equations by applying reversible (i.e., invertible) operations to both sides, or by adding zero (in some form) to one side, multiplying one side by one (in some form), or replacing expressions by simplified or expanded versions of themselves. Operations that aren't invertible include raising both sides to some power, or taking roots of both sides, and other noninvertible operations. Here's a simple example.

Show that ##2\cos^2(x) - 1 = \cos^2(x) - \sin^2(x)##

Proof:
##2\cos^2(x) - 1 = \cos^2(x) - \sin^2(x)##
##\Leftrightarrow 2\cos^2(x) - 1 = \cos^2(x) - (1 - \cos^2(x))##
##\Leftrightarrow 2\cos^2(x) - 1 = \cos^2(x) - 1 + \cos^2(x))##
##\Leftrightarrow 2\cos^2(x) - 1 = 2\cos^2(x) - 1##

Each step after the first is reversible, since all I did was to replace an expression by another that was identically equal to the one I replaced. Because I ended with an identity, and each step was reversible, this means that the first equation is also an identity, and so the original equation is proven to be true.

Having done all this, my preference is to start with one of the sides of the original equation, and work with the other side until I reach the expression on the other side.
Thanks @Mark44 ...nice weekend mate!
 
  • #30
robphy
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[snip]

Thus proved.
...Must improve before it can be proved.

(A variant of #6 )
Since [itex] \sin (ix) = i \sinh (x) [/itex],
then [itex] \sin (x) = i \sinh (x/i) = i \sinh (-ix)= -i\sinh(ix) = -i\sinh(u) [/itex] where [itex] u=ix [/itex].
check: [itex]\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right) \stackrel{u=ix}{=}-i \left(\frac{e^{u}-e^{-u}}{2}\right) =-i\sinh(u) [/itex]​

And since [itex] \cos ix = \cosh (x) [/itex],
then [itex] \cos x = \cosh (x/i)= \cosh(-ix) =\cosh(ix) = \cosh(u)[/itex] where [itex] u=ix [/itex].
check: [itex]\displaystyle \cos(x)=\frac{e^{ix}+e^{-ix}}{2} \stackrel{u=ix}{=} \left(\frac{e^{u}+e^{-u}}{2}\right) =\cosh(u) [/itex]​

my revisions in red
Since;
##\sin ix = i \sinh x## and ##\cos ix = i \ cosh x ##
then we shall replace ##\sin x## with ##\sinh x## and ##\cos x## with ##\cosh x##
then we shall replace ##\sin x## with ##(\color{red}{-i}\sinh \color{red}{u})## and ##\cos x## with ##(\cosh \color{red}{u})##
thus;

[snip]

So,
\begin{align*}
8 \sin^4x &= 3-4\cos 2x+\cos 4x\\
8 \color{red}{(-i \sinh u)}^4 &= 3-4(\color{red}{\cosh 2u})+(\color{red}{\cosh 4u})\\
8 (-i)^4 \sinh^4 u &= \\
8 \phantom{XX} \sinh^4 u &= \\
8 \sinh^4 u &= 3-4\cosh 2u+\cosh 4u\\
&= \ldots\\
&\ \ \vdots\\
\end{align*}
Because [itex] (-i)^4=1 [/itex] (ref: #17 ), you were lucky that you never saw the problem.

In general, one might not be so lucky. (Ref: #2 )
\begin{align*}
1 &= \cos^2 x +\sin^2x \\
&= \color{red}{(\cosh u)}^2 + \color{red}{(-i \sinh u)}^2 \\
&=\cosh^2 u +(-i)^2 \sinh^2 u \\
&=\cosh^2 u +(-1) \sinh^2 u \\
&=\cosh^2 u - \sinh^2 u \\
\end{align*}
which is not what one gets by
replacing [incorrectly] "##\sin x## with ##\sinh x## and ##\cos x## with ##\cosh x##".
 
  • #31
chwala
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...Must improve before it can be proved.

(A variant of #6 )
Since [itex] \sin (ix) = i \sinh (x) [/itex],
then [itex] \sin (x) = i \sinh (x/i) = i \sinh (-ix)= -i\sinh(ix) = -i\sinh(u) [/itex] where [itex] u=ix [/itex].
check: [itex]\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right) \stackrel{u=ix}{=}-i \left(\frac{e^{u}-e^{-u}}{2}\right) =-i\sinh(u) [/itex]​

And since [itex] \cos ix = \cosh (x) [/itex],
then [itex] \cos x = \cosh (x/i)= \cosh(-ix) =\cosh(ix) = \cosh(u)[/itex] where [itex] u=ix [/itex].
check: [itex]\displaystyle \cos(x)=\frac{e^{ix}+e^{-ix}}{2} \stackrel{u=ix}{=} \left(\frac{e^{u}+e^{-u}}{2}\right) =\cosh(u) [/itex]​

my revisions in red


So,
\begin{align*}
8 \sin^4x &= 3-4\cos 2x+\cos 4x\\
8 \color{red}{(-i \sinh u)}^4 &= 3-4(\color{red}{\cosh 2u})+(\color{red}{\cosh 4u})\\
8 (-i)^4 \sinh^4 u &= \\
8 \phantom{XX} \sinh^4 u &= \\
8 \sinh^4 u &= 3-4\cosh 2u+\cosh 4u\\
&= \ldots\\
&\ \ \vdots\\
\end{align*}
Because [itex] (-i)^4=1 [/itex] (ref: #17 ), you were lucky that you never saw the problem.

In general, one might not be so lucky. (Ref: #2 )
\begin{align*}
1 &= \cos^2 x +\sin^2x \\
&= \color{red}{(\cosh u)}^2 + \color{red}{(-i \sinh u)}^2 \\
&=\cosh^2 u +(-i)^2 \sinh^2 u \\
&=\cosh^2 u +(-1) \sinh^2 u \\
&=\cosh^2 u - \sinh^2 u \\
\end{align*}
which is not what one gets by
replacing [incorrectly] "##\sin x## with ##\sinh x## and ##\cos x## with ##\cosh x##".
check: ##\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right)##

just to be absolutely clear on the referenced...you multiplied the left hand side by ##-i## to arrive at the right hand side. Correct?
 
  • #32
robphy
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check: ##\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}=-i \left(\frac{e^{ix}-e^{-ix}}{2}\right)##

just to be absolutely clear on the referenced...you multiplied the left hand side by ##-i## to arrive at the right hand side. Correct?

I multiplied by 1 = [itex] \frac{-i}{-i}[/itex].
 
  • #33
vanhees71
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To me, the question sounds odd.

How would you answer the question if the given equation were
$$1=\cos^2 u +\sin^2 u$$
instead?
I guess the idea is to use
$$\cosh u=\cos(\mathrm{i} u), \quad \sinh u=-\mathrm{i} \sin(\mathrm{i} u).$$
This gives
$$1=\cos^2(\mathrm{i} u) + \sin^2(\mathrm{i} u)=\cosh^2 u-\sinh^2 u.$$
;-)).
 

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