Rotations

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).

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. $
Answer Solution

$ \RotP\theta AB = \pt{a_1 + (b_1 - a_1)\cos\theta - (b_2 - a_2)\sin\theta}{a_2 + (b_1 - a_1)\sin\theta + (b_2 - a_2)\cos\theta} $

We determined that \[\begin{align*} \RotP\theta AB &= A + \RotVparen\theta{B-A}\\ \RotV\theta{\vec xy} &= x\vec{\cos\theta}{\sin\theta} + y\vec{-\sin\theta}{\phantom{\mathop-}\cos\theta} \end{align*}\] We have \[ B - A = \pt{b_1}{b_2} - \pt{a_1}{a_2} = \vec{b_1 - a_1}{b_2 - a_2} \] Thus, \[\begin{align*} \RotP\theta AB &= A + \RotVparen\theta{B-A}\\ &= \pt{a_1}{a_2} + \RotV\theta{\vec{b_1 - a_1}{b_2 - a_2}}\\ &= \pt{a_1}{a_2} + (b_1 - a_1)\vec{\cos\theta}{\sin\theta} + (b_2 - a_2)\vec{-\sin\theta}{\phantom{\mathop-}\cos\theta}\\ &= \pt{a_1 + (b_1 - a_1)\cos\theta - (b_2 - a_2)\sin\theta}{a_2 + (b_1 - a_1)\sin\theta + (b_2 - a_2)\cos\theta} \end{align*}\]
In the video, we established the formula \[ \begin{equation}\label{rot} \RotV\theta{\vec xy} = x\vec{\cos\theta}{\sin\theta} + y\vec{-\sin\theta}{\phantom{\mathop-}\cos\theta} = \vec{x\cos\theta - y\sin\theta}{x\sin\theta + y\cos\theta} \end{equation} \] for rotating a vector $\vec xy$ by an angle $\theta$ about the origin. Use this to derive the "angle-sum identities" \[ \begin{equation}\label{angle-sum} \vec {\cos(\alpha+\beta)} {\sin(\alpha+\beta)} = \vec{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)} {\cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta)}. \end{equation} \]

Hint Solution

Rotating by angle $\alpha+\beta$ is the same as first doing a rotation by $\alpha,$ followed by a rotation by $\beta.$ This is all that \eqref{angle-sum} is expressing, just written out in gruesome coordinates.

We know that $\vec {\cos(\alpha+\beta)}{\sin(\alpha+\beta)}$ is the result of rotating the unit vector $\vec 10$ about the origin by an angle $\alpha + \beta.$ However, doing a rotation of $\alpha+\beta$ is the same as first doing a rotation by $\alpha,$ followed by a rotation by $\beta$ —that is, \[ \RotVparen{\alpha+\beta}v = \Rot\beta(\Rot\alpha(v)) \] for any vector $v.$ Thus, we just apply the rotation formula \eqref{rot} twice to obtain the result. \[ \begin{align*} \vec {\cos(\alpha+\beta)}{\sin(\alpha+\beta)} &= \Rot{\alpha+\beta}\vec 10\\ &= \Rot\beta \Rot\alpha \vec 10\\ &= \Rot\beta \vec{\cos\alpha}{\sin\alpha}\\ &= \vec{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\cos\alpha\sin\beta + \sin\alpha\cos\beta} \end{align*} \]
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 \[ \begin{align} \cos(\theta/2) &= \pm\sqrt{\frac{1+\cos(\theta)}2} \label{half-cos}\\ \sin(\theta/2) &= \pm\sqrt{\frac{1-\cos(\theta)}2}. \label{half-sin} \end{align} \]

Solution

Double-angle identities

Taking $\alpha=\beta=\theta$ in \eqref{angle-sum} yields $\begin{aligned} \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta)\\ \sin(2\theta) &= 2\cos(\theta)\sin(\theta) \end{aligned}$ Using the Pythagorean theorem, $\begin{aligned} \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta) & \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta)\\ % &= \cos^2(\theta) - (1 - \cos^2(\theta)) & &= (1-\sin^2(\theta)) - \sin^2(\theta)\\ % &= 2\cos^2(\theta) - 1 & &= 1 - 2\sin^2(\theta) \end{aligned}$

Half-angle identities

Writing $\theta = 2(\theta/2),$ the double-angle identities give $\begin{aligned} \cos\theta &= 2\cos^2(\theta/2) - 1 & \cos\theta &= 1 - 2\sin^2(\theta/2)\\ % \cos^2(\theta/2) &= \frac{1+\cos\theta}2 & \sin^2(\theta/2) &= \frac{1-\cos\theta}2\\ % \cos(\theta/2) &= \pm\sqrt{\frac{1+\cos(\theta)}2} & \sin(\theta/2) &= \pm\sqrt{\frac{1-\cos(\theta)}2} \end{aligned}$
Find $\cos15^\circ,$ $\sin15^\circ,$ $\cos75^\circ,$ $\sin75^\circ.$

Solution

Recall that $\cos30^\circ = \sqrt3/2.$ From the half-angle formulas \eqref{half-cos} and \eqref{half-sin}, we get $\begin{aligned} \cos15^\circ &= \sqrt{\frac{1+\cos30^\circ}2} & \sin15^\circ &= \sqrt{\frac{1-\cos30^\circ}2}\\ &= \sqrt{\frac{1+\sqrt3/2}2} & &= \sqrt{\frac{1-\sqrt3/2}2}\\ &= \frac12\sqrt{2+\sqrt3} & &= \frac12\sqrt{2-\sqrt3} \end{aligned}$ Since $\cos(90^\circ-\theta) = \sin\theta$ and $\sin(90^\circ - \theta) = \cos\theta,$ this also gives $\begin{aligned} \cos75^\circ &= \frac12\sqrt{2-\sqrt3}\\[1em] \sin75^\circ &= \frac12\sqrt{2+\sqrt3} \end{aligned}$

Video Problems