Free math worksheets

- Identify which quadrant contains the terminal side of \(\theta\). - On the unit circle, \(\cos(\theta)\) is the x-coordinate and \(\sin(\theta)\) is the y-coordinate. - In Quadrant II, x-values are negative and y-values are positive.

1. The marked angle \(\theta\) lies in Quadrant II. 2. In Quadrant II, points on the unit circle have a positive y-coordinate and a negative x-coordinate. 3. Since \(\sin(\theta)\) is the y-coordinate and \(\cos(\theta)\) is the x-coordinate, \(\sin(\theta)>0\) and \(\cos(\theta)<0\).

\(\sin(\theta)>0\) and \(\cos(\theta)<0\)

- First identify the quadrant containing \(230^\circ\). - A reference angle is the acute angle between the terminal side and the x-axis. - In Quadrant III, compare the angle with the straight angle on the negative x-axis.

1. The angle \(230^\circ\) lies in Quadrant III, so its reference angle is the distance from \(180^\circ\): \(230^\circ-180^\circ=50^\circ\).

\(50^\circ\)

- Sine is the y-coordinate and cosine is the x-coordinate. - Determine the sign of each coordinate in each quadrant. - Locate every given angle before comparing values.

1. The angle \(130^\circ\) lies in Quadrant II, where y-coordinates are positive. Therefore, a) is true. 2. The angle \(210^\circ\) lies in Quadrant III, where x-coordinates are negative. Therefore, b) is false. 3. The angle \(315^\circ\) lies in Quadrant IV, where y-coordinates are negative. Therefore, c) is true. 4. The angle \(100^\circ\) is in Quadrant II, so its cosine is negative, while \(80^\circ\) is in Quadrant I, so its cosine is positive. Therefore, d) is false.

a) True b) False c) True d) False

- Recall the range of sine and cosine. - Determine the quadrant of each angle. - Use the reference angle and the sign of cosine.

1. For a), the range of sine is \([-1, 1]\), so a sine value of \(1.5\) is impossible. The statement is true. 2. For b), \(150^\circ\) lies in Quadrant II, where cosine is negative. 3. Since \(150^\circ=180^\circ-30^\circ\), \(\cos(150^\circ)=-\cos(30^\circ)\), so the statement is false.

a) True; the range of sine is \([-1, 1]\). b) False; \(\cos(150^\circ)=-\cos(30^\circ)\).

- Identify where x- and y-coordinates are both positive. - Identify where x- and y-coordinates are both negative. - Recall that cosine is x and sine is y.

1. In Quadrant I, both the x-coordinate and y-coordinate are positive, so cosine and sine have the same sign. 2. In Quadrant II, cosine is negative and sine is positive, so the signs differ. 3. In Quadrant III, both coordinates are negative, so the signs are the same. 4. In Quadrant IV, cosine is positive and sine is negative, so the signs differ. 5. Therefore, \(\alpha\) can lie in Quadrant I or Quadrant III.

Quadrant I or Quadrant III

- Sine is the y-coordinate and cosine is the x-coordinate on the unit circle. - Identify the sign of each function in the angle’s quadrant. - Find each reference angle relative to \(180^\circ\) or \(360^\circ\).

1. The angle \(160^\circ\) is in Quadrant II, where cosine is negative. Since its reference angle is \(20^\circ\), the correct equation is \(\cos(160^\circ)=-\cos(20^\circ)\). 2. The angle \(215^\circ\) is in Quadrant III, where sine is negative. Its reference angle is \(35^\circ\), so \(\sin(215^\circ)=-\sin(35^\circ)\). This equation is correct. 3. The angle \(290^\circ\) is in Quadrant IV, where cosine is positive. Its reference angle is \(70^\circ\), so \(\cos(290^\circ)=\cos(70^\circ)\). This equation is correct.

1. False; \(\cos(160^\circ)=-\cos(20^\circ)\) 2. True 3. True

- Sine is the y-coordinate and cosine is the x-coordinate. - Determine where x- and y-coordinates are positive or negative. - Axis angles are excluded because one coordinate is \(0\).

1. Sine is the y-coordinate, which is negative in Quadrants III and IV. Thus \(180^\circ < \alpha < 360^\circ\). 2. Cosine is the x-coordinate, which is negative in Quadrants II and III. Thus \(90^\circ < \alpha < 270^\circ\). 3. Both coordinates are negative only in Quadrant III. Thus \(180^\circ < \alpha < 270^\circ\). 4. Sine is positive and cosine is negative only in Quadrant II. Thus \(90^\circ < \alpha < 180^\circ\).

a) \(180^\circ < \alpha < 360^\circ\) b) \(90^\circ < \alpha < 270^\circ\) c) \(180^\circ < \alpha < 270^\circ\) d) \(90^\circ < \alpha < 180^\circ\)

- Identify the quadrant of each angle. - Determine the sign of sine or cosine in that quadrant. - Find the acute reference angle by measuring to the nearest x-axis.

1. In Quadrant II, sine is positive and cosine is negative. Therefore, \(\sin(105^\circ)=\sin(75^\circ)\) and \(\cos(167^\circ)=-\cos(13^\circ)\). 2. In Quadrant III, both sine and cosine are negative. Therefore, \(\sin(212^\circ)=-\sin(32^\circ)\) and \(\cos(195^\circ)=-\cos(15^\circ)\). 3. In Quadrant IV, sine is negative and cosine is positive. Therefore, \(\sin(290^\circ)=-\sin(70^\circ)\) and \(\cos(348^\circ)=\cos(12^\circ)\).

a) \(\sin(75^\circ)\); \(-\cos(13^\circ)\) b) \(-\sin(32^\circ)\); \(-\cos(15^\circ)\) c) \(-\sin(70^\circ)\); \(\cos(12^\circ)\)

- Identify the quadrant containing each angle. - Find the acute reference angle. - Determine the signs of sine and cosine in that quadrant. - Recall the exact values for \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\).

1. The angle \(\frac{5\pi}{6}\) is in Quadrant II with reference angle \(\frac{\pi}{6}\). Therefore, \(\sin\left(\frac{5\pi}{6}\right)=\frac{1}{2}\) and \(\cos\left(\frac{5\pi}{6}\right)=-\frac{\sqrt{3}}{2}\). 2. The angle \(\frac{4\pi}{3}\) is in Quadrant III with reference angle \(\frac{\pi}{3}\). Therefore, \(\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) and \(\cos\left(\frac{4\pi}{3}\right)=-\frac{1}{2}\). 3. The angle \(\frac{7\pi}{4}\) is in Quadrant IV with reference angle \(\frac{\pi}{4}\). Therefore, \(\sin\left(\frac{7\pi}{4}\right)=-\frac{\sqrt{2}}{2}\) and \(\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}\).

a) \(\sin\left(\frac{5\pi}{6}\right)=\frac{1}{2}\); \(\cos\left(\frac{5\pi}{6}\right)=-\frac{\sqrt{3}}{2}\) b) \(\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}\); \(\cos\left(\frac{4\pi}{3}\right)=-\frac{1}{2}\) c) \(\sin\left(\frac{7\pi}{4}\right)=-\frac{\sqrt{2}}{2}\); \(\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

- Identify the quadrant of the terminal side before using special-angle values. - A reference angle is the acute angle between the terminal side and the x-axis. - Use the reference-angle magnitudes first, then apply the quadrant signs.

1. The diagram shows \(\theta=\frac{7\pi}{6}\), which is \(\frac{\pi}{6}\) past \(\pi\). Therefore, the reference angle is \(\beta=\frac{\pi}{6}\). 2. The terminal side lies in Quadrant III, where both sine and cosine are negative. 3. The reference-angle values are \(\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\) and \(\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\). Applying the Quadrant III signs gives \(\sin(\theta)=-\frac{1}{2}\) and \(\cos(\theta)=-\frac{\sqrt{3}}{2}\).

a) \(\beta=\frac{\pi}{6}\) b) \(\sin(\theta)<0\) and \(\cos(\theta)<0\) c) \(\sin(\theta)=-\frac{1}{2}\), \(\cos(\theta)=-\frac{\sqrt{3}}{2}\)

- 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

- Cosine is the x-coordinate on the unit circle. - Determine where cosine is negative, zero, or positive. - Recall the special values \(-1\), \(0\), and \(1\). - Points farther left on the unit circle have smaller cosine values.

1. Determine the cosine values: \(\cos(180^\circ)=-1\) \(\cos(210^\circ)=-\frac{\sqrt{3}}{2}\approx-0.866\) \(\cos(120^\circ)=-\frac{1}{2}\) \(\cos(270^\circ)=0\) \(\cos(300^\circ)=\frac{1}{2}\) \(\cos(20^\circ)\approx0.940\) 2. In increasing order, \(-1<-0.866<-\frac{1}{2}<0<\frac{1}{2}<0.940\), so the angle order is \(180^\circ, 210^\circ, 120^\circ, 270^\circ, 300^\circ, 20^\circ\). 3. The corresponding letters are C, O, S, I, N, E, which spell COSINE.

Increasing cosine order: \(180^\circ, 210^\circ, 120^\circ, 270^\circ, 300^\circ, 20^\circ\) Word: COSINE

- Use reference angles and unit-circle symmetry. - Determine the sign of cosine in each quadrant. - Angles with the same reference angle have cosine values with the same magnitude. - An angle closer to the positive or negative x-axis has a cosine value with greater magnitude.

1. The angle \(40^\circ\) is in Quadrant I and \(220^\circ=180^\circ+40^\circ\) is in Quadrant III. Their cosine values have the same magnitude and opposite signs. Since \(\cos(40^\circ)\approx0.77\), \(\cos(220^\circ)\approx-0.77\). 2. The angles \(130^\circ=180^\circ-50^\circ\) and \(310^\circ=360^\circ-50^\circ\) have reference angle \(50^\circ\). Cosine is negative in Quadrant II and positive in Quadrant IV. Since \(\cos(50^\circ)\approx0.64\), \(\cos(130^\circ)\approx-0.64\) and \(\cos(310^\circ)\approx0.64\).

\(\cos(40^\circ)\approx0.77\) \(\cos(130^\circ)\approx-0.64\) \(\cos(220^\circ)\approx-0.77\) \(\cos(310^\circ)\approx0.64\)

- Sine is the y-coordinate on the unit circle. - Identify the sign of sine in each quadrant. - Use exact special-angle values to compare the entries.

1. Evaluate the sine values from the unit circle: \(\sin\left(\frac{\pi}{2}\right)=1\) \(\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\) \(\sin(0)=0\) \(\sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2}\) \(\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) 2. Therefore, \(1>\frac{\sqrt{2}}{2}>0>-\frac{1}{2}>-\frac{\sqrt{3}}{2}\), so the angle order is \(\frac{\pi}{2}, \frac{\pi}{4}, 0, \frac{7\pi}{6}, \frac{4\pi}{3}\). 3. The corresponding letters spell POLAR.

Decreasing sine order: \(\frac{\pi}{2}, \frac{\pi}{4}, 0, \frac{7\pi}{6}, \frac{4\pi}{3}\) Word: POLAR

- Use unit-circle symmetry near \(180^\circ\). - Identify the quadrants and signs of the two angles. - Both angles are the same distance from \(180^\circ\).

1. The angle \(175^\circ\) is in Quadrant II, where sine is positive. The angle \(185^\circ\) is in Quadrant III, where sine is negative. 2. Both angles have reference angle \(5^\circ\). Therefore, \(\sin(175^\circ)=\sin(5^\circ)\) and \(\sin(185^\circ)=-\sin(5^\circ)\). 3. In Quadrants II and III, cosine is negative. Thus, \(\cos(175^\circ)=-\cos(5^\circ)\) and \(\cos(185^\circ)=-\cos(5^\circ)\), proving the values are equal.

a) \(175^\circ\): Quadrant II, positive sine; \(185^\circ\): Quadrant III, negative sine b) \(\sin(175^\circ)=\sin(5^\circ)\); \(\sin(185^\circ)=-\sin(5^\circ)\) c) Both cosine values equal \(-\cos(5^\circ)\).

- Find where sine is negative. - Find where cosine is negative or zero. - Intersect the two angle sets. - Check boundary angles carefully.

1. Sine is negative in Quadrants III and IV, so \(180^\circ < \alpha < 360^\circ\). 2. Cosine is nonpositive in Quadrants II and III, including the y-axis boundaries, so \(90^\circ \le \alpha \le 270^\circ\). 3. The intersection is \(180^\circ < \alpha \le 270^\circ\). The endpoint \(180^\circ\) is excluded because sine is \(0\), while \(270^\circ\) is included because sine is \(-1\) and cosine is \(0\).

\(180^\circ < \alpha \le 270^\circ\)

- Use reference angles in Quadrants III and IV. - Reduce angles greater than \(360^\circ\) to coterminal angles in one full rotation. - For a negative angle, first find a positive coterminal angle between \(0^\circ\) and \(360^\circ\). - Check that each answer is acute or a right angle.

1. The angle \(192^\circ\) is \(12^\circ\) past \(180^\circ\), so its reference angle is \(12^\circ\). Sine is negative in Quadrant III, so \(\sin(192^\circ)=-\sin(12^\circ)\). Thus, \(\alpha=12^\circ\). 2. The angle \(285^\circ\) has reference angle \(360^\circ-285^\circ=75^\circ\), and cosine is positive in Quadrant IV. Thus, \(\alpha=75^\circ\). 3. Reduce \(600^\circ\) to the coterminal angle \(240^\circ\). Its reference angle is \(60^\circ\), and sine is negative in Quadrant III. Thus, \(\alpha=60^\circ\). 4. The angle \(-40^\circ\) is coterminal with \(320^\circ\). Its reference angle is \(40^\circ\), and cosine is positive in Quadrant IV. Thus, \(\cos(-40^\circ)=\cos(40^\circ)\), so \(\alpha=40^\circ\).

a) \(\alpha=12^\circ\) b) \(\alpha=75^\circ\) c) \(\alpha=60^\circ\) d) \(\alpha=40^\circ\)