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I am confused as to why you only change the sign for powers of sine that are 4n+2.

As I understand,

$sin(i\theta)=isinh(\theta)$

$sin^2(i\theta)=-sinh^2(\theta)$

$sin^3(i\theta)=-isinh^3(\theta)$

$sin^4(i\theta)=sinh^4(\theta)$

So I would think that the third power would also have the sign changed, although I am also slightly confused about what difference it makes whether there is an i present or not...

So I have 2 questions:

  1. What is the significance of the i when you have odd powers of sinh? How does it affect the relationship between $sin(i\theta)$ and $sinh(\theta)$?

  2. Why is it that the sign change only applies for powers of sine that are 4n+2?

I have read this post but it really has not clarified very much for me. Explanations that use fewer technical terms would be much appreciated!

Thank you :)

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Meep
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  • Since $$\sin(i\theta)= i;\sinh\theta \quad\to\quad \sin^n(i\theta)=i^n;\sinh^n\theta$$ this really isn't a question about hyperbolic functions so much as it is about the powers of $i$. Are you not familiar with the pattern $$i^1 = i \qquad i^2 = -1 \qquad i^3 = -i \qquad i^4 = 1 \qquad \cdots$$? – Blue Mar 20 '15 at 16:37
  • I am familiar with this, however there is a minus sign both when i is raised to second and the third power. Why then do you only switch the sign when sine is raised to the 4n+2 power? Why not also for the 4n+3 power? And how do the coefficient i's affect the relationship between sine and sinh? – Meep Mar 20 '15 at 16:51
  • Because -i multiplied by i results $(-1)$ multiplied by $i^2$ =-1 and the result is +1 – Moti Mar 20 '15 at 19:47
  • But what about for the identity $sin(3x)=3sin(x)-4sin^3(x)$? The hyperbolic equivalent is $sinh(3x)=3sinh(x)+4sinh^3(x)$ so there has been a change in sine when cubing it... – Meep Mar 20 '15 at 21:11
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    Think less in terms of "signs" (and their changes), and more in terms of "powers of $i$". For instance, $$\begin{align} \sin(3x) &= 3 \sin(x) - 4 \sin^3(x) \ \implies \quad \sin(3ix) &= 3\sin(ix) - 4\sin^3(ix) \ \implies \quad i\sinh(3x) &= 3\cdot i\sinh(x) - 4\cdot(i\sinh(x))^3 \ \implies \quad i\sinh(3x) &= 3\cdot i \sinh(x) - 4\cdot i^3\sinh^3(x) \ \implies \quad \sinh(3x) &= 3\sinh(x) - 4\cdot i^2\sinh^3(x) \quad\text{(dividing-through by $i$)} \ \implies \quad \sinh(3x) &= 3\sinh(x) - 4\cdot (-1)\sinh^3(x) \ \implies \quad \sinh(3x) &= 3\sinh(x) + 4\cdot \sinh^3(x) \end{align}$$ – Blue Mar 21 '15 at 02:47
  • Related: "How do you prove Osborn's rule?". In my answer, I quote Osborn's Rule as originally published, and suggest that the Rule is more of a "rule of thumb", intended as a teaching tool for memorizing the simplest cases of hyperbolic identities. – Blue Aug 26 '21 at 03:28

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You have to also look at when 4n+1 powers of sinh occurs, for example if it only occurs when expressing sinh(a+b) then you have to divide by i once therefore leaving the identity unchanged. For 4n+2 if it only occurs with identities involving cosh(a+b) you don't divide because cos(ix) = cosh(x). for similar reasons with 4n+3 you divide by -i. This rule can be proven with individual cases but imo they require far too much unnecessary thinking to describe using some sort of intuition

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