Free math worksheets

- 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) \(105^\circ\)

- 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 unit circle. - The x-coordinate is cosine and the y-coordinate is sine. - Apply the equation \(x^2+y^2=1\).

1. The axis angles are \(0^\circ\), \(90^\circ\), \(180^\circ\), \(270^\circ\), and \(360^\circ\). 2. The corresponding unit-circle coordinates \((\cos(\alpha), \sin(\alpha))\) are \((1, 0)\), \((0, 1)\), \((-1, 0)\), \((0, -1)\), and \((1, 0)\). 3. Thus the sine-cosine pairs are \((0, 1)\), \((1, 0)\), \((0, -1)\), \((-1, 0)\), and \((0, 1)\), respectively. 4. If both sine and cosine were \(1\), then \(\sin^2(\alpha)+\cos^2(\alpha)=1^2+1^2=2\), contradicting the unit-circle equation, which requires the sum to equal \(1\).

a) \(0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ\) b) \(0^\circ: (\sin(\alpha), \cos(\alpha))=(0, 1)\); \(90^\circ: (1, 0)\); \(180^\circ: (0, -1)\); \(270^\circ: (-1, 0)\); \(360^\circ: (0, 1)\) c) It is impossible because \(1^2+1^2=2 \ne 1\).

- 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. On the unit circle, cosine is the x-coordinate and sine is the y-coordinate. Therefore, \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), \(\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\), and \(\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2}\).

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.

- 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) \(\sin(\alpha)=-0.8\)

- 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 \(y=-x\).

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