## Summary ### Sides | Angles (degrees) | Angles (radians) | Side 1 | Side 2 | Hypotenuse | | -------- | ------ | ------ | ---------- | - | | 30:60:90$\textdegree$ | $\frac{\pi}{6}:\frac{\pi}{3}:\frac{\pi}{2}$ | $x$ | $x \sqrt{3}$ | $2x$ | | 45:45:90$\textdegree$ | $\frac{\pi}{4}:\frac{\pi}{4}:\frac{\pi}{2}$ | $x$ | $x$ | $x\sqrt{2}$ | ### Resulting Special Ratios | Function | 30$\textdegree$/$\frac{\pi}{6}$ radians | 45$\textdegree$/$\frac{\pi}{4}$ radians | 60$\textdegree$/$\frac{\pi}{3}$ radians | | -------- |: -- :|: --- :|: --- :| | $\sin$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | | $\cos$ | $\frac{\sqrt{3}}{2}$ |$\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | | $\tan$ | $\frac{\sqrt{3}}{3}$ | $1$ | $\sqrt{3}$ | ### Hand rule There is a mnemonic trick called the "hand rule" for memorizing these, explained [here](https://www.youtube.com/watch?v=jI81WXyFrL0). If you take the angles in order $0\textdegree, 30\textdegree, 45\textdegree, 60\textdegree$ and $90\textdegree$ and assign each to a finger starting with your thumb, then: - $\sin$ of that angle is the square root of number of fingers on the "thumb-side" of the finger corresponding with the angle divided by two - $\cos$ is the same, except using the number of fingers on the "pinky-side" instead ##### Worked example of the hand rule Say we want the $\sin 60\textdegree$ and $\cos 60\textdegree$. $60\textdegree$ corresponds with our ring finger and numbering from our thumb, we have 3 fingers on the same side as the thumb and 1 on the other side. So $ \begin{align*} \sin 60\textdegree &= \frac{\sqrt{3}}{2} \\ \cos 60\textdegree &= \frac{\sqrt{1}}{2} = \frac{1}{2}\\ \end{align*} $ ### Special triangles in radians | Degrees | Radians | | ------- | ------- | | 30:60:90$\textdegree$ | $\frac{\pi}{6}:\frac{\pi}{3}:\frac{\pi}{2}$ | | 45:45:90$\textdegree$ | $\frac{\pi}{4}:\frac{\pi}{4}:\frac{\pi}{2}$ | ## Derivation: 30-60-90/$\frac{\pi}{6}$-$\frac{\pi}{3}$-$\frac{\pi}{2}$triangle Consider an equilateral triangle with sides $a$,$b$ and $c$: We drop an altitude from $C$ and call this length $x$ ![[Equilateral Triangle 2.png]] Now we know that $a=b=c$. For the purposes of this construction, let this length be $2$. It's easy to see that $c$ is bisected by our altitude given that the two outer triangles are clearly congruent having a common side, an equal side and 3 equal angles. Therefore their base must also be equal. So the length of this bottom/shortest side of each of the two triangles is $\frac{c}{2}=1$. But what is the length of $x$? Well, by Pythagoras: $ \begin{align*} x &= \sqrt{b^2-c^2}\\ &=\sqrt{2^2 - 1^2} \\ &= \sqrt{3} \end{align*} $ So the sides of this triangle are $1$, $2$ and $\sqrt{3}$. Having derived this result, let us remove the half of the triangle on the right and relabel everything to make things clearer. ![[Equilateral Triangle 3.png]] If we consider these angle sizes in degrees, we have $ \begin{cases} \angle A = 60 \textdegree ~ \text{(angles of an equilateral triangle)}\\ \angle B = 90 \textdegree ~ \text{(by construction)}\\ \angle C = 180 \textdegree - 90 \textdegree - 60 \textdegree = 30 \textdegree \\ \end{cases} $ In [[Radians]] this is $ \begin{cases} \angle A = \frac{\pi}{3}\\ \angle B = \frac{\pi}{2}\\ \angle C = \pi - \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \\ \end{cases} $ Having all the sides and angles, it is simple to derive the trig ratios of this triangle. We will do it in both degrees and radians. $ \begin{align*} \sin 30\degree &= \frac{1}{2} & \sin \frac{\pi}{6} &= \frac{1}{2} \\ \cos 30\degree &= \frac{\sqrt{3}}{2} & \cos \frac{\pi}{6} &= \frac{\sqrt{3}}{2} \\ \tan 30\degree &= \frac{\sqrt{3}}{3} & \tan \frac{\pi}{6} &= \frac{\sqrt{3}}{3} \\ \sin 60\degree &= \frac{\sqrt{3}}{2} & \sin \frac{\pi}{3} &= \frac{\sqrt{3}}{2} \\ \cos 60\degree &= \frac{1}{2} & \cos \frac{\pi}{3} &= \frac{1}{2} \\ \tan 60\degree &= \sqrt{3} & \tan \frac{\pi}{3} &= \sqrt{3} & \end{align*} $ ## Derivation: Isosceles right triangle (45-45-90/$\frac{\pi}{4}-\frac{\pi}{4}-\frac{\pi}{2}$ triangle) Consider $\Delta ABC$ with the following properties: $ \begin{cases} m \angle CAB = 45 \textdegree = \frac{\pi}{4}\\ m \angle ACB = 90 \textdegree = \frac{\pi}{2}\\ m \angle ABC = 45 \textdegree = \frac{\pi}{4}\\ \end{cases} $ ![[45-45-90 Triangle.png]] We can see that $AC \cong BC$ because it is an isosceles triangle, but what is $AB$? Let $AC = BC = x$. Then $ \begin{align*} AB^2 &= AC^2 + CB^2 \; \text{...from Pythagoras} \\ &= 2x^2 \\ \therefore AB &= x\sqrt{2} \\ \\ \begin{cases} AB &= x\sqrt{2} \\ AC &= x \\ BC &= x \\ \end{cases} \\ \end{align*} $ ![[45-45-90 Triangle with labels.png]] From this we can deduce the following ratios: $ \begin{align*} \sin 45 \degree = \cos 45 \degree &= \frac{x}{x\sqrt{2}} & \sin \frac{\pi}{4} = \cos \frac{\pi}{4} &= \frac{x}{x\sqrt{2}} \\ &= \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} & &= \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\\ \tan 45 \degree &= \frac{x}{x} & \tan \frac{\pi}{4} &= \frac{x}{x} \\ &= 1 & &= 1 \\ \end{align*} $ Diagrams available on geogebra [1](https://www.geogebra.org/geometry/zad6y4ke) [2](https://www.geogebra.org/calculator/rqpsxnkk) [3](https://www.geogebra.org/geometry/usph9zuz) [4](https://www.geogebra.org/geometry/tschmfhu)