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I know that the same way circular trigonometry is defined over the circle $ x^2 + y^2 = 1 $, hyperbolic trigonometry is defined over the hyperbola $ x^2 - y^2 = 1 $.

What I don't know is how deduced the formulas $$ \sinh x = \frac {e^x - e^{-x}} {2} \quad \text{and} \quad \cosh x = \frac {e^x + e^{-x}} {2} $$ are deduced.

My question is: How are the formulas for $ \sinh x $ and $ \cosh x $ deduced from the equation $ x^2 - y^2 = 1 $ of the unit hyperbola?

Travis Willse
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francolino
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2 Answers2

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I'll prove $\sinh(x)$ = $\frac {e^x - e^{-x}} {2}$ and leave the proof of $\cosh(x)$ as an exercise for you. So, $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$

And we know that, $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$

And we also know that $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...$$

Therefore, $$e^x - e^{-x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...)$$

Now, after we open the brackets, the negative terms will become positive and positive terms will become negative, so let's just do that

$$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - 1 + x - \frac{x^2}{2!} + \frac{x^3}{3!} - ...$$

And we are left with : $$2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!} + ...$$ and we'll take the $2$ common so we get,

$$2(x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...)$$

And, we saw earlier that $$\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$$.

So we get that $$e^x - e^{-x} = 2 \times \sinh(x)$$ and hence we get that $$\sinh(x) = \frac {e^x - e^{-x}}{2}$$

$Q.E.D$

Hope it helps

p_square
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The parametric equation $$\left\{\begin{array}{rcl}x(t) &=&\cos t \\y(t) &=&\sin t \\ \end{array}\right.$$ parameterizes the unit circle $x^2 + y^2 = 1$, and we can check that the image of this curve lies inside the unit circle by checking that $$x(t)^2 + y(t)^2 = 1,$$ which upon substitution reduces to the Pythagorean identity $$\cos^2 t + \sin^2 t = 1.$$

Similarly, we can check that the image of the parametric equation $$\left\{\begin{array}{rcl}x(t) &=&\cosh t \\y(t) &=&\sinh t \\ \end{array}\right.$$ is contained inside the unit hyperbola $x^2 - y^2 = 1$ by substituting in that equation and checking that both sides are always equal. This in turn follows from the given definitions of $\cosh$ and $\sinh$.

Travis Willse
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    I still can't see how is deduced that $ \sinh x = \frac {e^x - e^{-x}} {2} \quad \text{and} \quad \cosh x = \frac {e^x + e^{-x}} {2} $. What you showed is that both formulas have to verify the identity (hyperbola equation), but it doesn't explain how the exponential appears. – francolino Jul 20 '15 at 04:43