- Think of a radian as measuring an arc by comparing it with the radius. - Compare the given arc length with the given radius. - The angle in radians has no length unit.

1. Radian measure compares an intercepted arc length with the circle's radius, so the angle is \(\frac{12}{8}=\frac{3}{2}\) radians. 2. The value \(\frac{3}{2}\) means the arc is \(1.5\) times as long as the radius.

\(\frac{3}{2}\) radians; the arc length is \(1.5\) radius-lengths.

- Compare the terminal side with the initial side on the positive x-axis. - What fraction of one full rotation is shown? - Recall the radian measure of one full rotation.

1. The terminal side points directly opposite the initial side, so the angle is one-half of a full rotation. 2. A full rotation measures \(2\pi\) radians, so one-half of a full rotation measures \(\pi\) radians.

\(\pi\) radians

- Choose a conversion factor that cancels the angle unit you start with. - For degrees to radians, compare the degree measure with a full or half rotation. - For radians to degrees, use the relationship between \(\pi\) radians and a straight angle.

1. Convert \(150^\circ\) to radians: \(150^\circ\cdot\frac{\pi}{180^\circ}=\frac{5\pi}{6}\). 2. Convert \(\frac{7\pi}{12}\) radians to degrees: \(\frac{7\pi}{12}\cdot\frac{180^\circ}{\pi}=105^\circ\).

a) \(\frac{5\pi}{6}\) b) \(105^\circ\)

- Read the marked angle on the circle. - How do the two unit-circle coordinates relate to sine and cosine? - Recall the exact ratios for a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.

1. The x-coordinate of a unit-circle point is the cosine of its angle. 2. The marked angle is \(60^\circ\), so the x-coordinate is \(\cos(60^\circ)=\frac{1}{2}\).

a) Cosine b) \(\frac{1}{2}\)

- Recall what makes an angle exactly one radian. - Compare the length of the radius with the intercepted arc. - The answer should have a unit of length, not an angle unit.

1. One radian is the central angle that cuts off an arc the same length as the circle's radius. 2. Therefore the arc is \(6\,\text{cm}\) long.

\(6\,\text{cm}\), because one radian cuts off one radius-length of arc.

- Coterminal angles finish on the same ray. - How many radians are in one full counterclockwise turn? - Add that amount without changing the final position.

1. Coterminal angles differ by whole turns of \(2\pi\). 2. Add one turn: \(-\frac{\pi}{2}+2\pi=\frac{3\pi}{2}\).

\(\frac{3\pi}{2}\)

- How does a one-radian angle relate arc length to the radius? - What happens to an arc's length when its central angle changes but its radius stays fixed? - Your result is a length even though the angle is in radians.

1. When the angle is measured in radians, arc length is \(s=r\theta\). 2. Substitute \(r=9\) and \(\theta=\frac{\pi}{3}\): \(s=9\cdot\frac{\pi}{3}=3\pi\,\text{in.}\).

\(3\pi\,\text{in.}\)

- Find the vertical position of \(P\) relative to the origin. - What is the lowest possible y-coordinate on the unit circle? - Recall whether sine is the x-coordinate or the y-coordinate.

1. Point \(P\) is at the bottom of the unit circle, so its coordinates are \((0,-1)\). 2. Sine is the y-coordinate of a unit-circle point. Therefore, \(\sin(\theta)=-1\).

\(-1\)

- On the unit circle, \(x=\cos(\alpha)\) and \(y=\sin(\alpha)\). - Use \(x^2+y^2=1\). - Determine the signs of both coordinates in Quadrant III.

1. Since the x-coordinate on the unit circle equals cosine, \(\cos(\alpha)=-0.6\). 2. Use \(x^2+y^2=1\): \((-0.6)^2+y^2=1\), so \(0.36+y^2=1\). 3. Therefore, \(y^2=0.64\), so \(y=\pm0.8\). 4. In Quadrant III, the y-coordinate is negative, so \(y=-0.8\). 5. Since the y-coordinate equals sine, \(\sin(\alpha)=-0.8\).

a) \(\cos(\alpha)=-0.6\) b) \(y=-0.8\) c) \(\sin(\alpha)=-0.8\)

- Interpret cosine as the x-coordinate and sine as the y-coordinate of a unit-circle point. - A fixed negative x-coordinate corresponds to two points on the circle. - Those two points have opposite y-coordinate signs.

1. On the unit circle, \(\sin^2(\beta)+\cos^2(\beta)=1\). 2. Substitute the x-coordinate: \(\sin^2(\beta)+\frac{144}{169}=1\). 3. Thus, \(\sin^2(\beta)=\frac{25}{169}\), so \(\sin(\beta)=\pm\frac{5}{13}\). 4. A negative x-coordinate occurs in Quadrants II and III, so both y-coordinate signs occur on the unit circle.

\(\sin(\beta)=\frac{5}{13}\) or \(\sin(\beta)=-\frac{5}{13}\)

- List the four axis-intersection points of the circle. - Which coordinate represents cosine, and which represents sine? - Check the two proposed values against the unit-circle equation.

1. The axis angles are \(0^\circ\), \(90^\circ\), \(180^\circ\), \(270^\circ\), and \(360^\circ\). 2. Their unit-circle points \((\cos(\alpha),\sin(\alpha))\) are \((1,0)\), \((0,1)\), \((-1,0)\), \((0,-1)\), and \((1,0)\), respectively. 3. Thus the x-coordinate gives cosine and the y-coordinate gives sine at each angle. 4. Both values cannot equal \(1\), because \(1^2+1^2=2\), whereas a unit-circle point must satisfy \(x^2+y^2=1\).

a) \(0^\circ,90^\circ,180^\circ,270^\circ,360^\circ\) b) Unit-circle coordinates \((\cos(\alpha),\sin(\alpha))\): \(0^\circ:(1,0)\); \(90^\circ:(0,1)\); \(180^\circ:(-1,0)\); \(270^\circ:(0,-1)\); \(360^\circ:(1,0)\). c) Impossible: \(1^2+1^2=2\neq1\).

- On the unit circle, each point has the form \((\cos\theta,\sin\theta)\). - Use the exact side ratios of a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle for the first-quadrant point. - For the second-quadrant point, keep the same reference-angle magnitudes and check the signs of the coordinates.

1. Point \(P\) is at \(\frac{\pi}{3}\). Its unit-circle coordinates are \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\). 2. Point \(Q\) is at \(\frac{2\pi}{3}\), in Quadrant II with reference angle \(\frac{\pi}{3}\). Its coordinates are \(\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\). 3. Cosine is the x-coordinate and sine is the y-coordinate, giving the values listed below.

a) \(P=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\); \(Q=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\) b) \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\); \(\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\), \(\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2}\)

- On the unit circle, identify which coordinate represents cosine and which represents sine. - Once sine and cosine are known, compare their ratio to the tangent value. - The remaining three functions are reciprocals of sine, cosine, or tangent.

1. On the unit circle, \(\cos(\theta)\) is the x-coordinate and \(\sin(\theta)\) is the y-coordinate. Thus, \(\cos(\theta)=-\frac{12}{13}\) and \(\sin(\theta)=\frac{5}{13}\). 2. Then \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{5/13}{-12/13}=-\frac{5}{12}\). 3. The reciprocal values are \(\csc(\theta)=\frac{13}{5}\), \(\sec(\theta)=-\frac{13}{12}\), and \(\cot(\theta)=-\frac{12}{5}\). 4. The coordinates give sine and cosine directly; tangent and cotangent come from their ratios, and secant and cosecant come from reciprocals.

a) \(\sin(\theta)=\frac{5}{13}\), \(\cos(\theta)=-\frac{12}{13}\), \(\tan(\theta)=-\frac{5}{12}\) b) \(\csc(\theta)=\frac{13}{5}\), \(\sec(\theta)=-\frac{13}{12}\), \(\cot(\theta)=-\frac{12}{5}\) c) The coordinates give sine and cosine; ratios give tangent and cotangent; reciprocals give secant and cosecant.

- Locate the two rays relative to the x-axis. - The coordinates at a \(45^\circ\) reference angle have equal magnitudes. - Reflection across the x-axis changes which coordinate's sign?

1. At \(45^\circ\), both coordinates are \(\frac{\sqrt{2}}{2}\), so \(P=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\). 2. The point at \(315^\circ\) is the reflection across the x-axis, so \(Q=\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\). 3. The sine values have the same absolute value and opposite signs because the y-coordinates are reflected.

a) \(P=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\); \(Q=\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\) b) Their sine values have equal absolute values but opposite signs.

- A half turn is \(180^\circ\) and also \(\pi\) radians. - What fraction of a half turn is \(120^\circ\)? - Compare your result with \(\pi\) before accepting it.

1. Sam multiplied by \(\pi\) but forgot that \(180^\circ=\pi\) radians. The conversion factor is \(\frac{\pi}{180^\circ}\). 2. Therefore, \(120^\circ\cdot\frac{\pi}{180^\circ}=\frac{2\pi}{3}\) radians. 3. Since \(120^\circ<180^\circ\), its radian measure must be less than \(\pi\). The corrected \(\frac{2\pi}{3}\) is, but \(120\pi\) is not.

Sam forgot to divide by \(180\). The correct angle is \(\frac{2\pi}{3}\) radians, which is less than \(\pi\).

- Count how many radians make one complete turn. - Remove a full turn to identify the remaining part of the rotation. - Recall which axis direction corresponds to a quarter turn counterclockwise.

1. Subtract one full turn: \(\frac{5\pi}{2}-2\pi=\frac{\pi}{2}\). 2. An angle of \(\frac{\pi}{2}\) points straight up, along the positive y-axis.

a) \(\frac{\pi}{2}\) b) Positive y-axis

- What equation describes every point on the unit circle? - A negative x-coordinate places the point on which side of the y-axis? - Recall the special sine and cosine values associated with a \(45^\circ\) reference angle. - Use the reference angle to identify angles in other quadrants.

1. Use the unit-circle equation: \(x^2+y^2=1\). 2. Substitute \(x=-\frac{\sqrt{2}}{2}\): \(\left(-\frac{\sqrt{2}}{2}\right)^2+y^2=1\), so \(\frac{1}{2}+y^2=1\). 3. Thus, \(y^2=\frac{1}{2}\), giving \(y=\pm\frac{\sqrt{2}}{2}\). 4. The point \(\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) lies in Quadrant II and corresponds to \(135^\circ\). 5. The point \(\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\) lies in Quadrant III and corresponds to \(225^\circ\).

1. \(y=\frac{\sqrt{2}}{2}\) or \(y=-\frac{\sqrt{2}}{2}\) 2. The corresponding angles are \(135^\circ\) and \(225^\circ\), respectively. 3. Quadrants II and III

- A full rotation is \(360^\circ\) or \(2\pi\) radians. - Use \(\frac{\pi}{180^\circ}\) to convert degrees to radians. - Recall exact sine values for special angles. - Check whether your calculator is in degree or radian mode.

1. For \(60^\circ\), \(x=60^\circ\cdot\frac{\pi}{180^\circ}=\frac{\pi}{3}\approx1.05\), and \(\sin(x)=\frac{\sqrt{3}}{2}\approx0.87\). 2. For \(x=\frac{3\pi}{4}\), the degree measure is \(135^\circ\), and \(\sin(x)=\frac{\sqrt{2}}{2}\approx0.71\). 3. On \(0\le x<2\pi\), \(\sin(x)=-1\) at \(x=\frac{3\pi}{2}\approx4.71\), which is \(270^\circ\). 4. For \(310^\circ\), \(x=310^\circ\cdot\frac{\pi}{180^\circ}=\frac{31\pi}{18}\approx5.41\), and \(\sin(x)\approx-0.77\).

<table> <tr> <td>Degree measure</td> <td>Radian measure \(x\)</td> <td>\(\sin(x)\)</td> </tr> <tr> <td>\(60^\circ\)</td> <td>\(\frac{\pi}{3}\approx1.05\)</td> <td>\(\frac{\sqrt{3}}{2}\approx0.87\)</td> </tr> <tr> <td>\(135^\circ\)</td> <td>\(\frac{3\pi}{4}\approx2.36\)</td> <td>\(\frac{\sqrt{2}}{2}\approx0.71\)</td> </tr> <tr> <td>\(270^\circ\)</td> <td>\(\frac{3\pi}{2}\approx4.71\)</td> <td>\(-1\)</td> </tr> <tr> <td>\(310^\circ\)</td> <td>\(\frac{31\pi}{18}\approx5.41\)</td> <td>\(-0.77\)</td> </tr> </table>

- Use the conversion factor between degrees and radians for part a). - Compare the sizes of \(1\) radian and \(1^\circ\) before comparing their cosine values. - For part c), add whole revolutions until the angle lies in the requested interval. - Once the coterminal angle is in \([0,2\pi)\), use its standard-angle position on the unit circle.

1. Convert \(225^\circ\): \(225^\circ\cdot\frac{\pi}{180^\circ}=\frac{5\pi}{4}\). 2. One radian is approximately \(57.3^\circ\). Cosine decreases on \([0^\circ,90^\circ]\), and \(1^\circ<57.3^\circ\), so \(\cos(1^\circ)>\cos(1)\). 3. Add one full revolution twice: \(-\frac{13\pi}{6}+4\pi=\frac{11\pi}{6}\). The point at \(\frac{11\pi}{6}\) on the unit circle is \(\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\).

a) \(\frac{5\pi}{4}\) b) \(\cos(1^\circ)>\cos(1)\) c) Coterminal angle: \(\frac{11\pi}{6}\); coordinates: \(\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\)

- Choose one sign convention for clockwise and counterclockwise rotation and use it consistently. - Combine the two rotations before finding a coterminal angle. - One full rotation changes the angle by \(2\pi\) without changing the terminal position. - Relate signed angle measure to signed distance traveled along the rim.

1. Treat counterclockwise rotation as positive and clockwise rotation as negative. The net angle is \(-\frac{11\pi}{6}+\frac{\pi}{3}=-\frac{11\pi}{6}+\frac{2\pi}{6}=-\frac{3\pi}{2}\). 2. Add one full rotation: \(-\frac{3\pi}{2}+2\pi=\frac{\pi}{2}\). Thus, \(\frac{\pi}{2}\) is the coterminal angle in \([0,2\pi)\). 3. Signed arc displacement equals radius times signed angle, so \(8\left(-\frac{3\pi}{2}\right)=-12\pi\,\text{cm}\).

a) \(-\frac{3\pi}{2}\) b) \(\frac{\pi}{2}\) c) \(-12\pi\,\text{cm}\)

- In radian measure, arc length depends on both the radius and the central angle. - For part a), decide which quantity in the arc-length relationship is unknown. - Use the radius you find in part a) for the new central angle in part b). - Check that each requested length has length units.

1. For an angle measured in radians, arc length satisfies \(s=r\theta\). Thus, \(15\pi=r\left(\frac{5\pi}{6}\right)\), so \(r=18\,\text{cm}\). 2. For the second angle, \(s=18\left(\frac{7\pi}{12}\right)=\frac{21\pi}{2}\,\text{cm}\).

a) \(18\,\text{cm}\) b) \(\frac{21\pi}{2}\,\text{cm}\)

- Compare arc length with radius to recover the radian measure in Circle A. - The same central angle has the same radian measure in both circles. - For part c), write the arc-length relationship once for each circle and compare the two expressions. - Focus on which factor is common when the central angles are equal.

1. In Circle A, \(\theta=\frac{s}{r}=\frac{4\pi}{6}=\frac{2\pi}{3}\). 2. In Circle B, \(s=r\theta=15\left(\frac{2\pi}{3}\right)=10\pi\,\text{cm}\). 3. For a fixed radian angle \(\theta\), each arc length has the form \(s=r\theta\). Therefore, for radii \(r_1\) and \(r_2\), \(\frac{s_2}{s_1}=\frac{r_2\theta}{r_1\theta}=\frac{r_2}{r_1}\). The common angle cancels, so arc length scales directly with radius.

a) \(\theta=\frac{2\pi}{3}\) b) \(10\pi\,\text{cm}\) c) For the same radian angle, \(s=r\theta\), so the common factor \(\theta\) cancels in the ratio and \(\frac{s_2}{s_1}=\frac{r_2}{r_1}\).

- How is the distance from \((x, y)\) to each coordinate axis found? - Translate “the same distance” into an equation involving absolute values. - Which lines contain all points that are equally distant from the axes? - How many times do those lines intersect the unit circle?

1. The distance from \(P\) to the x-axis is \(|\sin(\alpha)|\), and the distance to the y-axis is \(|\cos(\alpha)|\). 2. Equal distances require \(|\sin(\alpha)|=|\cos(\alpha)|\), or equivalently \(\sin^2(\alpha)=\cos^2(\alpha)\). 3. This occurs when \(\sin(\alpha)=\cos(\alpha)\) or \(\sin(\alpha)=-\cos(\alpha)\). 4. The first equation gives \(45^\circ\) and \(225^\circ\). The second gives \(135^\circ\) and \(315^\circ\). 5. Geometrically, these are the four intersections of the unit circle with the lines \(y=x\) and \(y=-x\).

\(\alpha\in\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}\)