# Intro - Trig Functions for Dummies Now, if you've done anything with parametric functions, you'll know that the *sine* and *cos* functions are the result of parametrising the unit circle ($x^2 + y^2 = 1$). So... can you guess what hyperbolic trig is...? OF COURSE YOU CAN! Wait. If you read ahead, CURSES! But anyhow, yes, **hyperbolic trigonometric** functions arise from the parametrisation of a hyperbola's ($x^2-y^2 = 1$) function. We call these '*sinh*' and '*cosh*', or whatever trig function you had at the beginning, but with a 'h' put right behind it. Intuitive, huh? Now let's get into everything about these damned functions! ![[hyperbolictrigsurfing.png]] *Sign up now at a travel agency near you. We are NOT liable for any injuries you sustain surfing a cosh graph.* # Hyperbolic Function Identities ## Function Identities Now, somehow, someway, there's a way to express these functions with linear variables. For the proof, skip ahead to the fifth subheader - but just bear with me, and take these on the chin for now. $\sinh x = \frac{1}{2}(e^x - e^{-x}) \tag{1}$ $\cosh x = \frac{1}{2}(e^x + e^{-x}) \tag{2}$ $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\tag{3}$ FYI, equation (1) is also known as "Osborne's Rule!" ## Trig Operations As high school might've taught you, these functions also have their own identities - as you'll see here! Spot the similarities? $\cosh^2x - \sinh^2x = 1 \tag{4}$ $2\sinh x\cosh x = \sinh{2}x \tag{5}$ $\cosh^2x + \sinh^2x = \cosh{2}x \tag{6}$ $1 - \tanh^2x = \text{sech}^2x \tag{7}$ $1 - \coth^2 x = \csch^2x \tag{8}$ # Differentiating Hyperbolic Functions Once again, these aren't really much different from the trigonometric ones, but do note signs and stuff when you copy these down on a cheat sheet somewhere. **AND REMEMBER THAT COSH X DIFFERENTIATED IS STILL POSITIVE! PLEASE!** $\frac{d}{dx} (\sinh x) = \cosh x \tag{9}$ $\frac{d}{dx} (\cosh x) = \sinh x \tag{10}$ $\frac{d}{dx} (\tanh x) = \text{{sech}}^2x \tag{11}$ $\frac{d}{dx} (\csch x) = -\coth x \csch x \tag{12}$ $\frac{d}{dx} (\coth x) = - \csch^2x \tag{13}$ $\frac{d}{dx} (\text{{sech}} \ x) = - \tanh x \ \text{{sech}} \ x \tag{14}$ # Integrating Hyperbolic Functions NOW for the interesting part. Hyperbolic trig helps you skirt around A LOT of functions that would otherwise be unintegrable, as you'll see in the example question I've linked! Also, no, I don't know why some of the functions are capitalised. WILL FIND A FIX! $\int \frac{1}{\sqrt{ a^2 + x^2}} \, dx = \frac{1}{a} \arcsinh x \tag{15}$ $\int \frac{1}{\sqrt{ x^2 - a^2}} \, dx = \frac{1}{a} \arccosh x \tag{16}$ $\int \frac{1}{ a^2 - x^2} \, dx = \frac{1}{a} \arctanh x = \frac{1}{a} \arccoth x \tag{17}$ $\int \frac{1}{u\sqrt{ a^2 + x^2}} \, dx = \frac{1}{a} \arccsch x \tag{18}$ $\int \frac{1}{u\sqrt{ a^2 - x^2}} \, dx = \frac{1}{a} \text{{arcsech}} \ x \tag{19}$ >[!Example]- Integrating Example Questions >**Q: Integrate $\int_{\frac{1}{2}}^\frac{3}{2} \frac{1}{4x^2-4x+5} \, dx$**. > >*A:* >We need to complete the square for this one. So: >$\int \frac{1}{4x^2-4x+5} \, dx = \int \frac{1}{4\left( x-\frac{1}{2} \right)^2 + 4} \, dx$ >And from here, it's just trivial. Don't be afraid of those fractions in the brackets, they'll cancel out. >$\int \frac{1}{4\left( x-\frac{1}{2} \right)^2 + 4} \, dx = \left[ \frac{1}{2}\arctanh \left( x-\frac{1}{2} \right) \right]^{\frac{3}{2}}_{\frac{1}{2}}$ >$\int_{\frac{1}{2}}^\frac{3}{2} \frac{1}{4x^2-4x+5} \, dx = \frac{1}{4} \ln 2$ >And that's a pretty nice number, isn't it? # FINALLY- Proving the Function Identities And we're here! We'll try first to prove the $\sinh x$ expression. Once again, if we look back to equation (1), we'll think that they're completely unrelated, yes? That is, until we find the MacLaurin series for both of them! See: $\sinh x \approx x + \frac{x^3}{6} \dots$ Because the $\sinh x$ terms will give you nothing whereas the $\cosh x$ terms will give you 1. So, let's compare that to the $e^x$ and $e^{-x}$ expressions, as shown: $e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \dots$ $e^{-x} \approx 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \dots$ So, if we add both expressions together: $e^x + e^{-x} = 2x + \frac{x^3}{3}$ Which is double the expression for $\sinh x$, which proves that $\sinh x = \frac{1}{2}(e^x + e^{-x})$. # What's Next? To navigate back to the Maths page, click here: [[Maths Contents Page]] To navigate back to the home page, click here: [[Home]]