## Notes

We write $$\A$$ for the set of points, and $$\V$$ for the corresponding set of vectors.

(Using blackboard bold for this is weird, normally you'd write them $$A$$ and $$V,$$ but I already used $$A$$ as the name of a point.)

### Rotating

Let $A$ and $B$ be points in $$\A,$$ and let $\theta$ be an angle. We write $$\RotP\theta AB$$ for the point obtained by rotating $B$ by $\theta$ around $A.$

We write $$\RotP\theta A{\mathord?}$$ to speak of the process.

### Simplification

Let $v$ be a vector in $$\V,$$ and let $\theta$ be an angle. We write $$\RotVparen\theta v$$ for $v$ rotated (counterclockwise) by the angle $\theta.$

We write $$\Rot\theta$$ or $$\RotVparen\theta{\mathord?}$$ to speak of the operation itself (as opposed to applying it to a specific vector).

$\RotP\theta AB = A + \RotVparen\theta{B - A}$

This equation can also be expressed via the commutative diagram
$\xymatrix@=3em{ \A \ar[r]^-{\RotP\theta A{\mathord?}} \ar[d]_-{\mathord? - A} & \A\\ \V \ar[r]_-{\RotVparen\theta{\mathord?}} & \V \ar[u]_-{A + \mathord?} }$

### Derivation

The crucial observation was that rotation behaved well with respect to these operations, in the sense that \begin{align*} \RotVparen\theta{u + v} &= \RotVparen\theta u + \RotVparen\theta v\\ \RotVparen\theta{c\cdot v} &= c\cdot\RotVparen\theta v \end{align*} for any vectors $u,\ v$ and any scalar $c.$ You can use the widget below to convince yourself of this.

This allows us to write \begin{align*} \RotV\theta{\vec xy} &= \RotVParen\theta{x\vec10 + y\vec01}\\ &= x\,\RotV\theta{\vec10} + y\,\RotV\theta{\vec01} \end{align*} Thus, once we know how to rotate $\vec10$ and $\vec01$ by $\theta,$ we know how to rotate any angle at all by $\theta!$

The easiest nontrivial case to try this out on is $$\theta=90^\circ = \Twopi/4.$$ In this case we know that $\RotV{90^\circ}{\vec10} = \vec01, \qquad \RotV{90^\circ}{\vec01} = \vec{-1}0 \!.$ Our general formula for rotation by 90 degrees is then $\RotV{90^\circ}{\vec xy} = \vec{-y}{x}$

Now let's come back to the case of a general angle $\theta.$ Using the calculation above and the fact that $$\Rot\alpha\Rot\beta = \Rot\beta\Rot\alpha,$$ \begin{align*} \RotV\theta{\vec xy} &= x\,\RotV\theta{\vec10} + y\,\RotV\theta{\vec01}\\ &= x\,\Rot\theta\vec10 + y\,\Rot\theta\Rot{90^\circ}\vec10\\ &= x\,\Rot\theta\vec10 + y\,\Rot{90^\circ}\Rot\theta\vec10 \end{align*} Thus, to calculate $$\RotV\theta{\vec xy},$$ we only need to know $$\RotV\theta{\vec10}\!.$$ But by definition this is $\Rot\theta\vec10 = \vec{\cos\theta}{\sin\theta}\!.$ Plugging this back into our calculation, we get \begin{align*} \Rot\theta\vec xy &= x\vec{\cos\theta}{\sin\theta} + y\vec{-\sin\theta}{\cos\theta}\\ &= \vec{x\cos\theta - y\sin\theta}{x\sin\theta + y\cos\theta} \end{align*}

## Problems

In this problem set, we will use angle brackets to denote points, e.g. $$\pt10,$$ and square brackets to denote vectors, e.g. $\vec32.$ This is not standard notation. Typically both are denoted by square brackets (or parentheses), and distinguished by context (if at all).

1. We reduced the problem of computing $$\RotP\theta AB$$ to the problem of computing $$\RotVparen\theta {B-A},$$ and found a formula for that, but we never explicitly wrote down the resulting formula for $$\RotP\theta AB.$$ So suppose that $$A = \pt {a_1}{a_2}$$ and $$B = \pt {b_1}{b_2},$$ and write down an explicit formula for $$\RotP\theta AB.$$
2. In the video, we established the formula $$$\label{rot} \RotV\theta{\vec xy} = x\vec{\cos\theta}{\sin\theta} + y\vec{-\sin\theta}{\phantom{\mathop-}\cos\theta} = \vec{x\cos\theta - y\sin\theta}{x\sin\theta + y\cos\theta}$$$ for rotating a vector $\vec xy$ by an angle $\theta$ about the origin. Use this to derive the "angle-sum identities" $$$\label{angle-sum} \vec {\cos(\alpha+\beta)} {\sin(\alpha+\beta)} = \vec{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)} {\cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta)}.$$$

3. Establish the "double-angle" identities \begin{aligned} \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta)\\ &= 2\cos^2(\theta) - 1\\ & = 1 - 2\sin^2(\theta)\\ \sin(2\theta) &= 2\cos(\theta)\sin(\theta) \end{aligned} and the "half-angle" identities \begin{align} \cos(\theta/2) &= \pm\sqrt{\frac{1+\cos(\theta)}2} \label{half-cos}\\ \sin(\theta/2) &= \pm\sqrt{\frac{1-\cos(\theta)}2}. \label{half-sin} \end{align}

4. Find $\cos15^\circ,$ $\sin15^\circ,$ $\cos75^\circ,$ $\sin75^\circ.$