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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.

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    $\begingroup$ $\sin\theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$ $\endgroup$ – JohnColtraneisJC Mar 12 '18 at 18:45
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    $\begingroup$ $\sin i\theta = i \sinh \theta$ $\endgroup$ – JohnColtraneisJC Mar 12 '18 at 18:46
  • $\begingroup$ $${\sinh(i*\Theta)= i*\sin\Theta}$$ $\endgroup$ – ketankk Dec 20 '19 at 6:53
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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 $$-\sinh^2(x)+\cosh^2(x)=1$$

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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.

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