# Why does SOH CAH TOA hold?

Even looking on the Wiki pages I have a hard time figuring this one out.

But why does SOH CAH TOA hold? As in, why is $\sin(x)$ the same as the opposite over hypotenuse of a right triangle? Why is $\cos(x)$ the adjacent over hypotenuse? Why is $\tan(x)$ the opposite over the adjacent? And why don't any of these work for right angles as the reference theta?

I understand that these facts are easily used but if I were trying to invent this for the first time I'd be totally lost. I don't understand where these trig functions come from, what their format definitions are, why they're defined this way, where they come from, why they're true, how we know they're true, etc. To me they are mysterious functions that everyone just uses and takes for granted but I have no idea how they work.

• These are definitions of sine, cosine, and tangent. This is a way of remembering definitions, not a way of remembering theorems. – Thomas Andrews Jan 30 '18 at 22:51
• Is the "intro definition" the actual definition or an oversimplified one from a much more general definition? What's the non-intro definition? – user525966 Jan 30 '18 at 22:54
• The non-intro definition could be in terms of a power series (see Taylor polynomials) or a functional equation (see differential equations). – Michael Burr Jan 30 '18 at 22:57
• There are a lot of advanced definitions, from "the real part of $e^{ix}$" to similarly. But all require calculus or some such. – Thomas Andrews Jan 30 '18 at 23:00
• One of the semi-advanced definitions that I prefer, and which is more tightly related to the triangle definition, is that $\sin(\theta)$, for $\theta > 0$, is equal to "the $y$-coordinate at the end of a path that travels counterclockwise around the unit circle in the Cartesian coordinate plane for a total length of $\theta$". But even that requires some analytical thinking in order to define lengths of paths along the circle. – Lee Mosher Jan 30 '18 at 23:02

If two right triangles share an angle $\theta\in(0,\pi/2)$, then they are similar. In particular the ratio of corresponding side lengths are equal. So the quantities $$\frac{\text{adjacent}}{\text{hypoteneuse}};\quad \frac{\text{opposite}}{\text{hypoteneuse}}$$ are the same for every right triangle with angle $\theta$. Hence these quantities are functions of $\theta$ and we may define $$\cos\theta=\frac{\text{adjacent}}{\text{hypoteneuse}};\quad \sin\theta=\frac{\text{opposite}}{\text{hypoteneuse}}$$

• And, if you have the usual picture in mind with the adjacent along the $x$-axis: $\cos\theta$ is the central displacement and $\sin\theta$ is the sideways displacement. – Rob Arthan Jan 30 '18 at 23:14