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.)


Let AA and BB be points in \(\A,\) and let θ\theta be an angle. We write \(\RotP\theta AB\) for the point obtained by rotating BB by θ\theta around A.A.

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


Let vv be a vector in \(\V,\) and let θ\theta be an angle. We write \(\RotVparen\theta v\) for vv rotated (counterclockwise) by the angle θ.\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
\[\xymatrix@=3em{ \A \ar[r]^-{\RotP\theta A{\mathord?}} \ar[d]_-{\mathord? - A} & \A\\ \V \ar[r]_-{\RotVparen\theta{\mathord?}} & \V \ar[u]_-{A + \mathord?} }\]


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, vu,\ v and any scalar c.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 [10]\vec10 and [01]\vec01 by θ,\theta, we know how to rotate any angle at all by θ!\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 θ.\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 \[\begin{align*} \Rot\theta\vec xy &= x\vec{\cos\theta}{\sin\theta} + y\vec{-\sin\theta}{\cos\theta}\\ &= \vec{x\cos\theta - y\sin\theta}{x\sin\theta + y\cos\theta} \end{align*}\]


In this problem set, we will use angle brackets to denote points, e.g. \(\pt10,\) and square brackets to denote vectors, e.g. [32].\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 \[ \begin{equation}\label{rot} \RotV\theta{\vec xy} = x\vec{\cos\theta}{\sin\theta} + y\vec{-\sin\theta}{\phantom{\mathop-}\cos\theta} = \vec{x\cos\theta - y\sin\theta}{x\sin\theta + y\cos\theta} \end{equation} \] for rotating a vector [xy]\vec xy by an angle θ\theta about the origin. Use this to derive the "angle-sum identities" \[ \begin{equation}\label{angle-sum} \vec {\cos(\alpha+\beta)} {\sin(\alpha+\beta)} = \vec{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)} {\cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta)}. \end{equation} \]

  3. Establish the "double-angle" identities cos(2θ)=cos2(θ)sin2(θ)=2cos2(θ)1=12sin2(θ)sin(2θ)=2cos(θ)sin(θ) \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,\cos15^\circ, sin15,\sin15^\circ, cos75,\cos75^\circ, sin75.\sin75^\circ.