MHB Are Equal: $$\sin \alpha_nx - \sinh \alpha_nx$$

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The discussion centers on the equality of the expressions involving sine and hyperbolic sine functions, specifically $\sin \alpha_nx - \sinh \alpha_nx$. It is noted that both expressions converge at every eigenvalue defined by the equation $\sin x \cosh x = \cos x \sinh x$ within the range [0,1]. A plot confirms their equivalence over this finite range, which is relevant for applications like beam length analysis. The equality holds under the condition that $\sin \alpha_n \cosh \alpha_n = \cos \alpha_n \sinh \alpha_n$. This establishes a clear relationship between the trigonometric and hyperbolic functions in the context discussed.
Dustinsfl
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$$
\sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n + \sinh \alpha_n}{\cos\alpha_n + \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx) = \sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n - \sinh \alpha_n}{\cos\alpha_n - \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx)
$$

When I plotted them, they were the same but I was only looking at a finite range since I am considering a beam of length 1.

In the range [0,1], they converge at every eigenvalue which are defined by $\sin x\cosh x = \cos x\sinh x$
 
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dwsmith said:
$$
\sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n + \sinh \alpha_n}{\cos\alpha_n + \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx) = \sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n - \sinh \alpha_n}{\cos\alpha_n - \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx)
$$

When I plotted them, they were the same but I was only looking at a finite range since I am considering a beam of length 1.

In the range [0,1], they converge at every eigenvalue which are defined by $\sin x\cosh x = \cos x\sinh x$

Hi dwsmith, :)

You can show that,

\[\sin \alpha_n\cosh \alpha_n = \cos \alpha_n\sinh \alpha_n\]

\[\Rightarrow\frac{\sin \alpha_n - \sinh \alpha_n}{\cos\alpha_n - \cosh\alpha_n}=\frac{\sin \alpha_n + \sinh \alpha_n}{\cos\alpha_n + \cosh\alpha_n}\]

So both sides are equal provided, \(\sin \alpha_n\cosh \alpha_n = \cos \alpha_n\sinh \alpha_n\).

Kind Regards,
Sudharaka.
 
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