By utilizing rotation, we can derive further identities, a daunting list of which can be found on [wikipedia](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Reflections,_shifts,_and_periodicity). This is a distillation of two Khan academy videos, one covering [sine and cosine rotations](https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:trig-id/v/trig-angle-rotations) and the second covering [tan](https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:trig-id/v/tan-periodicity). The identities we will derive are: $ \begin{align*} \cos \left( \frac{\pi}{2} + \theta \right) &= -\sin \theta & \quad \sin \left( \frac{\pi}{2} + \theta \right) &= \cos \theta & \tan \left( \frac{\pi}{2} + \theta \right) &= -\cot \theta \tag{1}\\[12pt] \cos \left( \frac{3\pi}{2} - \theta \right) &= -\sin \theta & \quad \sin \left( \frac{3\pi}{2} - \theta \right) &= -\cos \theta & \tan \left( \frac{3\pi}{2} - \theta \right) &= \cot \theta \tag{2}\\[12pt] \cos \left( \frac{3\pi}{2} + \theta \right) &= \sin \theta & \quad \sin \left( \frac{3\pi}{2} + \theta \right) &= -\cos \theta & \tan \left( \frac{3\pi}{2} + \theta \right) &= -\cot \theta \tag{3}\\ \end{align*} $ All angles are in [[radians]], but could be converted to use degrees in the obvious way. ## Rotation by $\frac{\pi}{2}$ Going back to our original [[Geometric Construction of Trigonometric Identities from the Unit Circle|unit circle]] with a single ray to $B$ creating the angle $\theta$, consider we construct an additional ray to point $C$ on the circumference such that $\angle BAC$ is a right angle (ie $\angle BAC = \frac{\pi}{2}$ in radians. What are the coordinates of $C$? ![[Unit Circle with Ray Rotation.png]] Call the point $(0,1)$ $I$. Consider $\angle CAI$. It is clear by rotation of $\angle BAH$ that this angle is $\theta$. Construct point $D$ such that $\angle IAD = \theta$. ![[Unit Circle with Ray Rotation 2.png]] $ \begin{align*} D &=\left(\cos\left(\frac{\pi}{2}-\theta\right), \sin\left(\frac{\pi}{2}-\theta\right)\right) & \text{...by construction} \\ \text{but } &\begin{cases} \cos(\frac{\pi}{2}-\theta) = \sin \theta \\ \sin(\frac{\pi}{2}-\theta) = \cos \theta \\ \end{cases} & \text{...complementary angles} \\[8pt] \text{so } D &= \left( \sin \theta, \cos \theta \right). \\[8pt] AC &\cong AI \cong AD = 1 \\ \angle CAI &\cong \angle IAD = \theta \\ \text{so } \Delta CAI &\cong \Delta IAD \\ \end{align*} \\ $ Therefore we can see that $\Delta CAI$ is simply a reflection of $\Delta IAD$ about the y-axis, and the coordinates of $C$ must be $(-\sin \theta, \cos \theta)$. Now considering $\angle CAH$, we notice $\angle CAH = \angle CAB + \angle BAH = \frac{\pi}{2} + \theta$ Therefore $ \begin{cases} \cos \left( \frac{\pi}{2} + \theta \right) = -\sin \theta \\ \sin \left( \frac{\pi}{2} + \theta \right) = \cos \theta \\ \tan \left( \frac{\pi}{2} + \theta \right) = -\cot \theta \\ \end{cases} \tag{1} $ ## Rotation by $\pi$ Note that rotation by $\pi$ is simply the reflection of $\angle BAH$ [[Geometric Construction of Trigonometric Identities from the Unit Circle#Reflection about both axes|about both axes]] which we have already covered. ## Rotation by $\frac{3\pi}{2}$ Continuing $AC$ to intersect the circle at $E$, $AI$ to $J$, and $AD$ to $F$ we can construct two further sets of related rotational identities. ![[Unit Circle with Ray Rotation 3.png]] $ \begin{align*} \angle FAJ &\cong \angle IAD = \theta & \text{...opposite angles} \\ \angle FAH &= \angle JAB - \angle FAJ & \left( \angle JAB > \frac{\pi}{2} \right)\\ &= \frac{3\pi}{2} - \theta \\[8pt] \text{so } F &= \left( \cos \left( \frac{3\pi}{2} - \theta \right), \sin \left( \frac{3\pi}{2} - \theta \right) \right) & \text{...unit circle definition} \\ \text{but } F &= \left( -\sin \theta, -\cos \theta \right) & \text{...by symmetry with D} \\[10pt] \text{Therefore } \begin{cases} \cos \left( \frac{3\pi}{2} - \theta \right) = -\sin \theta \\ \sin \left( \frac{3\pi}{2} - \theta \right) = -\cos \theta \\ \tan \left( \frac{3\pi}{2} - \theta \right) = \cot \theta \\ \end{cases} \tag{2} \end{align*} $ ...and similarly $ \begin{align*} \angle JAE &\cong \angle CAI = \theta & \text{...opposite angles}\\ \angle EAH &= 3\pi + \theta \\[8pt] \text{so } E &= \left( \cos \left( \frac{3\pi}{2} + \theta \right), \sin \left( \frac{3\pi}{2} + \theta \right) \right) & \text{...unit circle definition} \\ \text{but } E &= \left( \sin \theta, -\cos \theta \right) & \text{...by symmetry with D}\\[10pt] \text{Therefore }\begin{cases} \cos \left( \frac{3\pi}{2} + \theta \right) = \sin \theta \\ \sin \left( \frac{3\pi}{2} + \theta \right) = -\cos \theta \\ \tan \left( \frac{3\pi}{2} + \theta \right) = -\cot \theta \\ \end{cases} \tag{3} \end{align*} $ Geogebra diagrams here: [1](https://www.geogebra.org/calculator/mwxzpvmu)[2](https://www.geogebra.org/calculator/zzrej4z4)[3](https://www.geogebra.org/calculator/wyp8bjzr)