Rotations

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 \[ \begin{align*} \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta)\\ &= 2\cos^2(\theta) - 1\\ & = 1 - 2\sin^2(\theta)\\ \sin(2\theta) &= 2\cos(\theta)\sin(\theta) \end{align*} \] 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 \[ \begin{align*} \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta)\\ \sin(2\theta) &= 2\cos(\theta)\sin(\theta) \end{align*} \] Using the Pythagorean theorem, \[ \begin{align*} \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta) & \cos(2\theta) &= \cos^2(\theta) - \sin^2(\theta)\\ % &= \cos^2(\theta) - (1 - \cos^2(\theta)) & &= (1-\sin^2(\theta)) - \sin^2(\theta)\\ % &= 2\cos^2(\theta) - 1 & &= 1 - 2\sin^2(\theta) \end{align*} \]

Half-angle identities

Writing \(\theta = 2(\theta/2),\) the double-angle identities give \[ \begin{align*} \cos\theta &= 2\cos^2(\theta/2) - 1 & \cos\theta &= 1 - 2\sin^2(\theta/2)\\ % \cos^2(\theta/2) &= \frac{1+\cos\theta}2 & \sin^2(\theta/2) &= \frac{1-\cos\theta}2\\ % \cos(\theta/2) &= \pm\sqrt{\frac{1+\cos(\theta)}2} & \sin(\theta/2) &= \pm\sqrt{\frac{1-\cos(\theta)}2} \end{align*} \]
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 \[ \begin{align*} \cos15^\circ &= \sqrt{\frac{1+\cos30^\circ}2} & \sin15^\circ &= \sqrt{\frac{1-\cos30^\circ}2}\\ &= \sqrt{\frac{1+\sqrt3/2}2} & &= \sqrt{\frac{1-\sqrt3/2}2}\\ &= \frac12\sqrt{2+\sqrt3} & &= \frac12\sqrt{2-\sqrt3} \end{align*} \] Since \(\cos(90^\circ-\theta) = \sin\theta\) and \(\sin(90^\circ - \theta) = \cos\theta,\) this also gives \[ \begin{align*} \cos75^\circ &= \frac12\sqrt{2-\sqrt3}\\[1em] \sin75^\circ &= \frac12\sqrt{2+\sqrt3} \end{align*} \]

Video Problems