This is my ~~de~~reconstruction of the Khan Academy videos [here](https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/trig-values-special-angles/v/trigonometry-unit-circle-symmetry) and [here](https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/trig-values-special-angles/v/tan-symmetries-unit-circle) in an attempt to understand and learn all the results in them. It provides a geometric construction of the following sets of identities: $ \begin{align*} \cos(-\theta) &= \cos \theta & \sin(-\theta) &= -\sin \theta \tag{1} \\ \cos(\pi - \theta) &= -\cos \theta & \sin(\pi - \theta) &= \sin \theta \tag{2}\\ \cos(\pi + \theta) &= -\cos \theta & \sin(\pi + \theta) &= -\sin \theta \tag{3} \\ \tan(\pi + \theta) &= \tan \theta & \tan(\pi - \theta) &= -\tan \theta = \tan(- \theta) \tag{4} \end{align*} $ As you might infer from the presence of $\pi$ these are expressed in [[radians]]. You could express the same relationships in degrees by substituting $\pi$ with $180\textdegree$. There is a helpful shortcut for understanding some of these called the "cast" or "astc" method which I will describe [[#ASTC Method|below]]. ## Setup Consider the unit circle centered on the origin, $A$, with a radius to some arbitrary point in quadrant I $B$. Let the angle that this radius forms with the x-axis be $\theta$. We know from the [[Unit Circle Definitions of Trigonometric Functions|unit circle definition]] that the coordinates of $B$ are $(x,y) = (cos \theta, sin \theta)$. ![[Unit Circle with Ray 1.png]] ## Reflection about the x-axis Now we reflect $B$ about the x-axis to point $D$. ![[Unit Circle with Ray 2.png]] $ \begin{align*} B &= (x,y) = (\cos \theta, \sin \theta) \\ D &= (x,-y) = (\cos \theta, -\sin \theta) & \text{...by reflection of B around the x-axis}\\ \end{align*} $ We say by convention $m \angle HAD = -\theta$ since it is equal in magnitude to $m \angle BAH$ but measured clockwise instead of counterclockwise starting at the x-axis, however it is clear that the absolute magnitude of this angle is $\theta$ by reflection, and that is the value we will use after this step. But this means that under our unit circle definition, $ \begin{align*} D = (\cos(-\theta), \sin(-\theta)) \\ \\ \text{Therefore } \begin{cases} \cos(-\theta) &= \cos \theta \\ \sin(-\theta) &= -\sin \theta \\ \end{cases} \tag{1} \end{align*} $ These two relationships are incredibly important. Since $\cos(-\theta) = \cos \theta$, cosine is an "[even](https://en.wikipedia.org/wiki/Even_and_odd_functions#Even_functions)" function and conversely, since $\sin(-\theta)=-\sin \theta$, sine is an "[odd](https://en.wikipedia.org/wiki/Even_and_odd_functions#Odd_functions)" function. This has tremendous implications later on for Fourier analysis and elsewhere. ## Reflection about the y-axis Now if we project $BA$ and $DA$ back through the origin until they intersect the circle on the opposite side at $E$ and $C$ we have the two remaining reflections. First let us consider $C$, which is the reflection of the original radius about the y-axis. For ease of notation, all the below will be in radians, but the same results could be obtained in degrees using the appropriate conversion of constants. ![[Unit Circle with Ray 3.png]] We can see that since $C$ is simply a reflection of $B$ about the y-axis, that $ C = (-\cos \theta, \sin \theta) $ ...by construction, but what is $m \angle CAH$? We see that $ \begin{align*} \angle GAC \cong \angle HAB &= \theta & \text{...reflection}\\ \angle CAH &= \pi - \theta & \text{...supplementary with } \angle GAC \\ \\ \text{Therefore } \begin{cases} \cos(\pi - \theta) &= -\cos \theta \\ \sin(\pi - \theta) &= \sin \theta\\ \end{cases} \tag{2} \end{align*} $ ## Reflection about both axes Finally, we consider $EA$ and its associated angles, which are the reflection of $BAH$ about both axes. ![[Unit Circle with Ray 4.png]] We know the coordinates of $E$ must be $(-\cos \theta, -\sin \theta)$, but what is the obtuse $\angle HAE > \pi$ which we can use to establish the final set of identities? $ \begin{align*} \angle EAG \cong \angle HAB &= \theta & \text{...vertically opposite angles}\\ \angle EAH &= \pi + \theta \\ \\ \text{Therefore } \begin{cases} \cos(\pi + \theta) &= -\cos \theta \\ \sin(\pi + \theta) &= -\sin \theta\\ \end{cases} \tag{3} \end{align*} $ ## Tangents Since we know that under the [[Unit Circle Definitions of Trigonometric Functions#Tangents in terms of the unit circle|unit circle definition]], $\tan \theta$ is the slope of the ray at angle $\theta$ with the positive $x$ axis, we can therefore use the diagram above to show some tangent identities geometrically. $ \begin{align*} \tan(\pi + \theta) &= \tan \theta & \text{...EA and AB are colinear and so have the same slope} \\ \tan(\pi - \theta) &= \tan(-\theta) & \text{...CA and AD are colinear and so have the same slope} \\ -\tan \theta &= \tan(-\theta) & \text{...CD and EB are reflections, therefore have opposite slope} \\ \end{align*} \tag{4} $ ### Algebraic verification using the results above We can construct a parallel algebraic proof of these three identities using the results above. $ \begin{align*} \tan(\pi + \theta) &= \frac{\sin(\pi + \theta)}{\cos(\pi + \theta)} \\ &= \frac{- \sin \theta}{- \cos \theta} & \text{....from (3)} \\ &= \frac{\sin \theta}{\cos \theta} & \text{....cancelling} \\ &= \tan \theta \\ \\ \tan(\pi - \theta) &= \frac{\sin(\pi - \theta)}{\cos(\pi - \theta)} \\ &= \frac{\sin \theta}{- \cos \theta} & \text{....from (2)} \\ &= -\tan \theta \\ \\ \tan( - \theta) &= \frac{\sin( - \theta)}{\cos(- \theta)} \\ &= \frac{-\sin \theta}{\cos \theta} & \text{....from (1)} \\ &= -\tan \theta \\ \end{align*} $ ## Summary diagram with all sets of identities ![[Unit Circle with Ray 5.png]] ## ASTC Method The "ASTC" or "CAST" method is a shortcut for remembering the summary diagram above. Firstly consider that if we have some angle $\theta$ which is _not_ on one of the axes, then the trig ratios for $\theta$ will always be related to the acute angle $\phi$ (phi, pronounced _pf-eye_) formed with the (positive or negative) x-axis in the following way. $ \begin{align*} \sin \theta &= \pm \sin \phi \\ \cos \theta &= \pm \cos \phi \\ \tan \theta &= \pm \tan \phi \\ \end{align*} $ ... and you choose which positive or negative value is correct for a given trigonometric function based on the quadrant of $\theta$. The mnemonic follows the initials "ASTC" If you start in quadrant 1 and proceed anticlockwise, or "CAST" if you start in quadrant 4 because you perhaps prefer a word. ![[Unit Circle with Ray 6.png]] Diagrams available on geogebra [1](https://www.geogebra.org/calculator/ar5rdfuk)[2](https://www.geogebra.org/calculator/sbfcuw3c)[3](https://www.geogebra.org/calculator/yegdqbez)[4](https://www.geogebra.org/calculator/rw4bma6v)[5](https://www.geogebra.org/calculator/yzngen3p)[6](https://www.geogebra.org/calculator/bgqzpaaa)