Chapter 2.7 Rotations

The goal of this section to continue our discussion of rotation by defining angle, angle measure, functions defined on the unit circle and their properties.

Fractions of a Circle and Measurement of Angles

To understand what is an angle. Imagine the following:

Roll to the right a circular wheel of radius \(1\) along a line so that the circle does not slip while rolling—The wheel traces out a length of \(2\pi\) on one complete rotation.

If \(p\) and \(q\) are two points on the wheel, then the arc \(\mathcal A\) from \(p\) to \(q\) will remain in contact with the line on a given interval.

This line that is drawn out has a certain length. We use that define what we call a “fraction of a circle”.

Fraction of a Circle

If the interval has length \(\ell(\mathcal A)\), then the length of \(\mathcal A\) is \(\ell(\mathcal A)\), and the fraction of a circle given by \(\mathcal A\), \({\rm Frac}(\mathcal A)\), is given to be \[{\rm Frac}(\mathcal A) = \frac{\ell(\mathcal A)}{2\pi}.\]

This definition makes sense if we have the length. But, how do we make sense of it if we have just a point?

To make sense of it, take \(p\) to be on the unit circle, \(\mathcal C\) and take \(\mathcal A\) to be the arc from \((1,0)\) to \(p\). Write \(p^n\) to mean \[p^n = \underbrace{p\star \cdots \star p}_{n-{\rm times}}.\]

Visualize this as applying rotation by the point \(p\) \(n\) times.

Suppose that \(p\) is not \((1,0)\), take \(n\) to be a natural number that is at least 2, and suppose that

1. \(n\) is the smallest natural number with \(p^n = (1,0)\);

2. for any \(q\) on \(\mathcal A\) with \(q^n = (1,0)\), \(q\) is either \((1,0)\) or \(p\).

Take \(\mathcal B\) to be the arc from \((1,0)\) to \(p^m\).

From an algebraic perspective, \(\mathcal B\) should be \(m\) \(n^{\rm ths}\) of \(\mathcal C\). This algebraic notion of a fraction of a circle agrees with the notion based on length, that is, \[{\rm Frac}(\mathcal B) = \frac{m}{n}.\]

Here is an example.

Example 1

Compare \(\frac{5}{12}^{\rm ths}\) of \(\mathcal C\) with \(\frac{1}{12}^{\rm th}\) of \(\mathcal C\).

Graph and notice this.

In this example, understand how to compare other fractions of a circle relative to two points.

Example 2

Suppose that the arc from \((1,0)\) to \(p\) is three sevenths of \(\mathcal C\). What fraction of \(\mathcal C\) is the arc from \(p\) to \((1,0)\)? What fraction of \(\mathcal C\) is the arc from \(p\) to \((-1,0)\)? What fraction of \(\mathcal C\) is the arc from \((0,1)\) to \(p\)?

Consider the following image:

The fraction of the circle \(C\) of the arc from \(p\) to \((1,0)\) is \(\frac{4}{7}\).

The fraction of the circle \(C\) of the arc from \(p\) to \((-1,0)\) is \(\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\).

The fraction of the circle \(C\) of the arc from \((0,1)\) to \(p\) is \(\frac{3}{7}-\frac{1}{4}=\frac{5}{28}\).

Now that we understand fraction of a circle, let us define what we mean by an angle.

Angle

Take \(O\) to be \((0,0)\). For any \(p\) and \(q\) on \(\mathcal C\), define the angle \(\angle pOq\) from \(p\) to \(q\) to be a point on \(\mathcal C\) that has the property that \[p\star \angle pOq = q,\quad\text{and so}\quad \angle pOq = p^{-1}\star q.\]

In this way, an angle is a point, but also can be thought of as the fraction of the circle from which the point \(\angle pOq\) makes.

This can be used to describle angles formed by using rays.

Specifically, take \(R_p\) to be the ray with tail at \(O\) that intersects \(p\), take \(R_q\) to be the ray with tail at \(O\) that intersects \(q\), where \(p\) and \(q\) are in \(\mathcal C\).

The angle between \(R_p\) and \(R_q\), \(\angle R_pR_q\), is defined to be the angle \(\angle pOq\).

Calculate the angle in the next example.

Example 3

Calculate \(\angle pOq\) where \[p = \left(\tfrac{1}{4}, \tfrac{\sqrt{15}}{4}\right) \quad\text{and} \quad q = \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right).\]

Calculate to get

\[ \begin{align*} \langle pOq\rangle&=p^{-1}*q\\ &=\left(\frac{1}{4},-\frac{\sqrt{15}}{4}\right)*\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\\ &=\left(\frac{1}{4}\cdot \frac{1}{2}+\frac{\sqrt{15}}{4}\cdot \frac{\sqrt{3}}{2},\frac{1}{4}\cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{15}}{4}\cdot\frac{1}{2}\right) \end{align*} \] and \[ \begin{align*} \langle pOq\rangle&=p^{-1}*q\\ &=\left(\frac{1}{4},-\frac{\sqrt{15}}{4}\right)*\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\\ &=\left(\frac{1}{4}\cdot \frac{1}{2}+\frac{\sqrt{15}}{4}\cdot \frac{\sqrt{3}}{2},\frac{1}{4}\cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{15}}{4}\cdot\frac{1}{2}\right) \end{align*} \]

In addition to thinking of \(\angle pOq\) as point, think of it as a measurement in the following way.

The angle \(\angle pOq\) is measured in degrees as 360 times the fraction of \(\mathcal C\) given by the arc from \((1,0)\) to \(\angle pOq\).

Additionally, the angle \(\angle pOq\) is measured in radians as the length of the arc from \(p\) to \(q\), that is, \(2\pi\) times the fraction of \(\mathcal C\) given by the arc from \((1,0)\) to \(\angle pOq\).

If an angle measure of \(\theta\) corresponds to a point \(p\), then define \(-\theta\) to correspond to the point \(p^{-1}\). In this sense, think of negative angle measures as corresponding to a measurement in a clockwise rather than a counterclockwise direction.

Identify the angle with measure 0 radians or 0 degrees to be the point \((1,0)\).

With all this in mind, determine the following measures.

Example 4

Find the degree and radian measures of the angle \(\angle pOq\) where:

1. \(p = (1,0)\) and \(q = (-1, 0)\);

2. \(p = (1,0)\) and \(q = (0, 1)\);

3. \(p = (1,0)\) and \(q = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\);

4. \(p = (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) and \(q = (-1,0)\).

Calculate to get

1. \(180^{\circ}\), \(\pi\)

2. \(90^{\circ}\), \(\frac{\pi}{2}\)

3. \(45^{\circ}\), \(\frac{\pi}{4}\)

4. \(45^{\circ}\), \(\frac{\pi}{4}\)

Now, practice converting from degrees to radians.

Example 5

What is the radian measure of the angle given by \(\theta\) where:

1. \(\theta = 25^\circ\);

2. \(\theta = 210^\circ\).

1. What fraction of a circle? It is \(\frac{25}{360}\), so the angle is \(\frac{25}{360}\cdot 2\pi\)

2. \(\frac{210}{360}\cdot 2\pi\)

In this example, practice converting from degrees to radians.

Example 6

What is the degree measure of the angle given by \(\theta\) radians, where:

1. \(\theta = \frac{\pi}{3}\);

2. \(\theta = \frac{7\pi}{12}\);

3. \(\theta = 30\).

1. \(\frac{\frac{\pi}{3}}{2\pi}\cdot 360=60^{\circ}\)

2. \(\frac{\frac{7\pi}{12}}{2\pi}\cdot 360=105^{\circ}\)

3. \(\frac{30}{2\pi}\cdot 360^{\circ}\)

If \(k\) is an integer, then an angle with measure \((\theta + k360)^\circ\) (in degrees) and an angle with measure \(\theta^\circ\) (in degrees) are the same point on the unit circle.

If \(k\) is an integer, then an angle with measures \(\theta + 2\pi k\) (in radians) and an angle with measure \(\theta\) (in radians) are the same point on the unit circle.

Because of this, we define principle angle measure.

Principle Angle Measure

We will define the principle angle measure respectively in degrees and radians to be an angle measure in \([0, 360)\) and in \([0, 2\pi).\)

In this example, find angles that are “equivalent” to the given principle angle measure in degrees.

Example 7

Find three degree measures that correspond to the point \(p\) on the unit circle with principle angle measure equal to \(30^\circ\).

Many answers; here are some

• \(30^{\circ}+7\cdot 360^{\circ}\),
• \(30^{\circ}-360^{\circ}\),
• \(30^{\circ}+2\cdot 360^{\circ}\),

In this example, find angles that are “equivalent” to the given principle angle measure in radian measures.

Example 8

Find three radian measures that correspond to the point \(p\) on the unit circle with principle angle measure equal to \(\frac{5\pi}{8}\).

Many answers; here are some

• \(\frac{5\pi}{8}\),
• \(\frac{5\pi}{8}+2\pi\),
• \(\frac{5\pi}{8}-2\pi\),
• \(\frac{5\pi}{8}+175\cdot 2\pi\)

Use the addition property of angles and the properties of points on a unit circle to deduce relationships between points and angles.

It can help to draw pictures rather than work symbolically with formulas.

Example 9

Suppose that the angle measure \(\theta\) corresponds to a point in the first quadrant with \(x\) coordinate equal to \(\frac{1}{5}\). What is the angle measure of a point in the fourth quadrant with \(x\)-coordinate equal to \(\frac{1}{5}\)?

Its measure is \(-\theta\) or \(2\cdot pi -\theta\).

The Sine, Cosine, and Tangent Functions

Because our goal is to be able to define rotation using function notation, we will define the following functions.

Sine and Cosine

Suppose angle measure \(\theta\) corresponds to a point \(p\) on the unit circle.

The vertical line intersecting \(p\) intersects the \(x\)-axis at \((x(p), 0)\).

The horizontal line intersecting \(p\) intersects the \(y\)-axis at a point \((0, y(p))\).

Define \[\cos(\theta) = x(p)\quad {\rm and}\quad \sin(\theta) = y(p).\]

Use this picture to visualize these functions.

Practice understanding these functions using the next example.

Example 10

Compute \(\cos(\theta)\) and \(\sin(\theta)\) when \(\theta\) varies through ``standard angles’’ in quadrants I, II, III, IV. Look at the cases of \(30^\circ\) and \(45^\circ\) separately.

Unit Circle

• \(\sin(30^{\circ})=\frac{1}{2}\)

• \(\sin(45^{\circ})=\frac{1}{\sqrt{2}}\)

• \(\cos(30^{\circ})=\frac{1}{2}\)

• \(\cos(45^{\circ})=\frac{1}{\sqrt{2}}\)

The other function we have is the tangent functon.

Tangent

Define the function \[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.\]

What does \(\tan(\theta)\) geometrically represent?

It represents the slope of this line.

Understand this function with this example.

Example 11

Evaluate \(\tan(0)\), \(\tan\left(\frac{\pi}{6}\right)\), \(\tan\left(\frac{\pi}{4}\right)\), and \(\tan\left(\frac{\pi}{3}\right)\)

• \(\tan(0)=0\), \(\tan\left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}\),
• \(\tan\left(\frac{\pi}{4}\right)=1\)
• \(\tan\left(\frac{\pi}{3}\right)=\sqrt{3}\)

In this example, evaluate tangent on various angles.

Example 12

Evaluate \(\tan{(\theta)}\) when \(\theta\) varies through standard angles in quadrants II, III, and IV

Use GeoGebra to verify some of this work.

One the important things in viewing these functions as functions on the unit circle is that a lot of properties come out easily and in a memorable way.

Pythagorean Identities

For any point \((x, y)\) on the unit circle, \[x^2 + y^2 = 1.\]

Therefore, \[\cos^2(\theta) + \sin^2(\theta) = 1,\] and so \[\cos^2(\theta) = 1 - \sin^2(\theta) \quad {\rm and}\quad \sin^2(\theta) = 1 - \cos^2(\theta).\]

These are known as Pythagorean Identities.

Use these identities to solve the following problem.

Example 13

The sine of an angle is \(\frac{1}{2}\). The angle is in quadrant II. What is the cosine of the angle?

Since \(\sin(\theta)=\frac{1}{2}\) \(\theta\) corresponds to \(\left(x,\frac{1}{2}\right)\). To get cosine, find \(x\). Use \(x^2+y^2=1\) to solve for \(x\) to get \(x=\frac{\sqrt{3}}{2}\) or \(x=-\frac{\sqrt{3}}{2}\). Since the angle is in quadrant II, \(x=-\frac{\sqrt{3}}{2}\) so \(\cos(\theta)=-\frac{\sqrt{3}}{2}\).

Angle Addition Formulae for Trigonometric Functions

We will now derive some important identities for the trigonometric functions.

Suppose that \(p\) and \(q\) are points on the unit circle.

If \(A\) is the measure of \(p\) and \(B\) is the measure \(q\), then \[p = (\cos(A), \sin(A)) \quad {\rm and} \quad q = (\cos(B), \sin(B)).\]

Since the measure of \(p\star q\) is \(A+B\), \[\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)\] and \[\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B).\]

These are known as the angle addition formulas

We also have the following \[\cos(-\theta) = \cos(\theta) \quad {\rm and}\quad \sin(-\theta) = -\sin(\theta).\]

These come from the following visual representation.

Use the angle addition formula to compute the following.

Example 14

Calculate

1. \(\sin(75^\circ)\).

2. \(\cos(15^\circ)\).

Calculate to get

1. \[ \begin{align*} \sin(75^{\circ})&=\sin(30^{\circ}+45^{\circ})\\ &=\sin(30^{\circ})\cos(45^{\circ})+\cos(30^{\circ})\sin(45^{\circ})\\ &=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\\ &=\frac{\sqrt{2}+\sqrt{2}\cdot\sqrt{3}}{4} \end{align*} \]

2. \[ \begin{align*} \cos(15^{\circ})&=\cos(45^{\circ}-30^{\circ})\\ &=\cos(45^{\circ})\cos(30^{\circ})+\sin(45^{\circ})\sin(30^{\circ})\\ &=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}\\ &=\frac{\sqrt{2}+\sqrt{2}\cdot\sqrt{3}}{4} \end{align*} \]

The angle summation formulas imply that \[\cos(2A) = \cos^2(A) - \sin^2(A)\quad {\rm and}\quad \sin(2A) = 2\sin(A)\cos(A).\]

These formulas are the double angle formulas.

We will use these formulas calculate \(\cos\left(\frac{\theta}{2}\right)\) and \(\sin\left(\frac{\theta}{2}\right)\).

The formulas we obtain are the half angle formulas

Example 15

Calculate

1. \(\cos\left(\frac{\theta}{2}\right) = ?\)

2. \(\sin\left(\frac{\theta}{2}\right) = ?\)

3. \(\cos(15^\circ)\) by using the half angle formula

1. \[ \begin{align*}\cos(\theta)&=\cos\left(\frac{\theta}{2}+\frac{\theta}{2}\right)\\ &=\cos^2\left(\frac{\theta}{2}\right)-\sin^2\left(\frac{\theta}{2}\right)\\ &=\cos^2\left(\frac{\theta}{2}\right)-\left(1-\cos^2\left(\frac{\theta}{2}\right)\right)\\ &=2\cos^2\left(\frac{\theta}{2}\right)-1 \end{align*}\] which means that \(\cos^2\left(\frac{\theta}{2}\right)=\frac{\cos(\theta)+1}{2}\). Take the square root we get \(\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{\cos(\theta)+1}{2}}\) if \(\frac{\theta}{2}\) in quadrant I or IV and \(\cos\left(\frac{\theta}{2}\right)=-\sqrt{\frac{\cos(\theta)+1}{2}}\) if \(\frac{\theta}{2}\) in quadrant II or III.

2. By the pythagorean theorem, \(\sin^2(X)+\cos^2(X)=1\), so \(\sin^2\left(\frac{\theta}{2}\right)+\cos^2(\frac{\theta}{2})=1\). This implies that \(\sin^2\left(\frac{\theta}{2}\right)=1-\cos^2(\frac{\theta}{2})\) or \(\sin^2\left(\frac{\theta}{2}\right)=1-\frac{\cos(\theta)+1}{2}\). By combining the expression we get \(\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos(\theta)}{2}\). Hence \(\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos(\theta)}{2}}\) if \(\frac{\theta}{2}\) in quadrant I or II and \(\sin\left(\frac{\theta}{2}\right)=-\sqrt{\frac{1-\cos(\theta)}{2}}\) if \(\frac{\theta}{2}\) in quadrant III or IV.

3. \[ \begin{align*} \cos(15^{\circ})&=\cos\left(\frac{30^{\circ}}{2}\right)\\ &=\sqrt{\frac{1+\cos(30^{\circ})}{2}}\\ &=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{2+\sqrt{3}}{4}}\\ \end{align*} \]

Parameterizing Rotations by Arclength

Now we bring it back to modeling motion.

The point on the unit circle with angle measure \(\theta\) is the point \((\cos(\theta), \sin(\theta))\).

To compress language, refer henceforth to the angle measure \(\theta\) as just the angle \(\theta\).

Rotation

The rotation \(R_\theta\) of \((a,b)\) around \((0,0)\) by angle \(\theta\) is given by \[R_\theta(a,b) = (\cos(\theta), \sin(\theta))\star (a,b).\]

Expanding out, this means

\[R_\theta(a,b) = (a\cos(\theta)-b\sin(\theta),b\cos(\theta)+a\sin(\theta)).\]

Write out the following.

Example 16

What is \(R_\theta(1, 2)\)? Explore the effect of \(\theta\) on this expression.

\(R_\theta(1,2)=(\cos(\theta),\sin(\theta))\star(1,2)=(1\cdot\cos(\theta)-2\sin(\theta),2\cos(\theta)+\sin(\theta))\)

So,

\[R_\theta(a,b) = (a\cos(\theta)-b\sin(\theta),b\cos(\theta)+a\sin(\theta)).\]

This will rotate the point \((1,2)\) around \((0,0)\) by the angle \(\theta\).

The following paramterizes the circle.

Parameterization of circle with constant speed

The path \(\gamma\) given by \[\gamma(\theta) = R_\theta(a,b)\] parameterizes the position of a point that is at \((a,b)\) initially and moves with constant speed on the circle of radius \(r\), where \[r = \sqrt{a^2 + b^2}.\]

This path will have speed equal to \(r\) if \(\theta\) is given in radians.

Try to describe the following path.

Example 17

Graph \[\gamma(t) = (\cos(2\pi t), \sin(2\pi t))\star(50,0).\]

Describe its motion.

Use GeoGebra to graph it. You will see that it rotates the point \((50,0)\). It takes \(t=1\) to go around once.

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