Relationship between $\sin(x)$ and $\sinh(x)$

Question

Given that $\tan(y) =\sinh(x)$ show that $\sin(y) = \pm \tanh(x) $.

I know:

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

$\tanh\theta=\dfrac{\sinh\theta}{\cosh\theta}$

Also,

$\tanh\theta = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

$\sinh\theta=\dfrac{e^x-e^{-x}}{2}$

$\cosh\theta=\dfrac{e^x+e^{-x}}{2}$

I'm struggling to link the two together, please assume little to no knowledge of calculus.

Answer 1

with $$\frac{\sin(y)}{\cos(y)}$$ we get $$\frac{\sin(y)}{\pm\sqrt{1-\sin^2(y)}}=\sinh(x)$$ square this equation and solve the equation for $\sin(y)$ with Algebra we get $$\sin^2(y)=\frac{\sinh^2(x)}{1+\sinh^2(x)}$$ can you finish? after squaring the given equation we obtain $$\frac{\sin^2(y)}{1-\sin^2(y)}=\sinh^2(x)$$ and then $$\sin^2(y)=\sinh^2(x)(1-\sin^2(y))$$ expanding $$\sin^2(y)=\sinh^2(x)-\sinh^2(x)\sin^2(y)$$ this gives $$\sin^2(y)+\sin^2(y)\sinh^2(x)=\sinh^2(x)$$ or $$\sin^2(y)=\frac{\sinh^2(x)}{1+\sinh^2(x)}$$ and note that $$-\sinh^2(x)+\cosh^2(x)=1$$

Answer 2

First of all notice that to keep consistency $$\tanh\theta = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$

Should have been $$\tanh(x) = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$

Same with your other two functions.

There are many similarities and differences between hyperbolic functions and trig functions.

For example, trig functions are periodic but hyperbolic functions are not periodic.

$\sin(x)$ and $\cos(x)$ are bounded but $\sinh(x)$ and $\cosh(x)$ are not bounded.

The identities $$ \cos^2(x) + \sin ^2(x) =1$$

turn into $$ \cosh^2(x) - \sinh ^2(x) =1$$

and

$$\cosh(x)=\dfrac{e^x+e^{-x}}{2}$$

turns into

$$\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$$

You will learn more about their infinite series and derivatives in your calculus courses.

Answer 3

We know that,

$$\sinh x = \frac{e^{x} - e^{-x}}{2}$$

Going by the definition of $e^{ix}$,

$$ \begin{aligned} e^{ix} = \cos x + i\sin x \\ e^{-ix} = \cos x - i\sin x \end{aligned} $$

Putting $\pmb{ix}$ in place of $\pmb{x}$,

$$ \left. \begin{aligned} e^{i(ix)} = \cos (ix) + i\sin (ix) \\ e^{-i(ix)} = \cos (ix) - i\sin (ix) \end{aligned} \right\} \tag{1}\label{1} $$

Now we know,

$$i^2 = -1 \tag{2}\label{2}$$

$\eqref{2}\; in\; \eqref{1}$,

$$e^{-x} = \cos ix + i\sin ix \tag{3}\label{3}$$ $$e^{x} = \cos (ix) - i\sin (ix) \tag{4}\label{4}$$

Finally, subtract $\eqref{3}\; from\; \eqref{4}$ and divide by two, to get in the form of $\pmb{\sinh x}$,

$$\frac{e^{x} - e^{-x}}{2} =\sinh x=\frac{-i\sin (ix) - i\sin (ix)}{2}$$

$$\bbox[5px,border:2px solid red] { \sinh x=-i\sin(ix) } $$

Since, $\frac{1}{i}=-i$, we can also have,

$$\bbox[5px,border:2px solid red] { \sinh x=\frac{\sin(ix)}{i} } $$

I know this isn't strictly the relation between $\sinh x$ and $\pmb{\sin x}$, in the sense that this involves a complex part.

Hope this helps someone who needs this relationship between the two functions.