Free math worksheets

- Identify which quadrant contains the terminal side of \(\theta\). - Recall how sine and cosine correspond to the two coordinates of a unit-circle point. - Use the location of the point relative to the coordinate axes to determine the two signs.

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 \(180^\circ\): \(230^\circ-180^\circ=50^\circ\).

\(50^\circ\)

- First decide which quadrant contains the given radian angle. - A reference angle is the acute angle between the terminal side and the x-axis.

1. The angle \(\frac{5\pi}{6}\) lies in Quadrant II. 2. Its reference angle is \(\pi-\frac{5\pi}{6}=\frac{\pi}{6}\).

\(\frac{\pi}{6}\)

- Start with the signs of the unit-circle coordinates in Quadrant IV. - Relate tangent and secant to sine and cosine without finding any numerical values.

1. In Quadrant IV, sine is negative and cosine is positive. 2. Since \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\), tangent is negative. 3. Since \(\sec(\theta)=\frac{1}{\cos(\theta)}\), secant is positive.

\(\tan(\theta)<0\) and \(\sec(\theta)>0\)

- 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) \(\sin(\theta)=-\frac{1}{2}\), \(\cos(\theta)=-\frac{\sqrt{3}}{2}\)

- Add or subtract whole rotations of \(2\pi\) until the terminal angle lies in the required interval. - The reference angle is the acute angle between the terminal side and the x-axis. - Determine the sign of tangent from the signs of sine and cosine in the final quadrant.

1. For \(-\frac{19\pi}{6}\), add \(4\pi=\frac{24\pi}{6}\) to obtain the coterminal angle \(\frac{5\pi}{6}\). Its reference angle is \(\frac{\pi}{6}\). This terminal side is in Quadrant II, where sine and cosine have opposite signs, so tangent is negative. 2. For \(\frac{25\pi}{4}\), subtract \(6\pi=\frac{24\pi}{4}\) to obtain the coterminal angle \(\frac{\pi}{4}\). Its reference angle is \(\frac{\pi}{4}\). This terminal side is in Quadrant I, where sine and cosine are both positive, so tangent is positive.

a) Coterminal angle: \(\frac{5\pi}{6}\); reference angle: \(\frac{\pi}{6}\); \(\tan(\theta)<0\) b) Coterminal angle: \(\frac{\pi}{4}\); reference angle: \(\frac{\pi}{4}\); \(\tan(\theta)>0\)

- Identify the reference angle and quadrant before assigning signs. - Start with the special-angle sine and cosine values. - Use a quotient for tangent and reciprocals for secant, cosecant, and cotangent.

1. The reference angle is \(\frac{\pi}{3}\), and \(\theta\) lies in Quadrant II. Therefore, sine is positive and cosine is negative. 2. Using the special-angle values gives \(\sin(\theta)=\frac{\sqrt{3}}{2}\) and \(\cos(\theta)=-\frac{1}{2}\). 3. Then \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=-\sqrt{3}\). 4. Taking reciprocals gives \(\csc(\theta)=\frac{2\sqrt{3}}{3}\), \(\sec(\theta)=-2\), and \(\cot(\theta)=-\frac{\sqrt{3}}{3}\).

\(\sin(\theta)=\frac{\sqrt{3}}{2}\), \(\cos(\theta)=-\frac{1}{2}\), \(\tan(\theta)=-\sqrt{3}\), \(\csc(\theta)=\frac{2\sqrt{3}}{3}\), \(\sec(\theta)=-2\), \(\cot(\theta)=-\frac{\sqrt{3}}{3}\)

- Read the terminal point's coordinates from its position on the unit circle. - Use those coordinates first for sine and cosine. - For quotient and reciprocal functions, check whether a denominator becomes zero before simplifying.

1. The terminal point is the top of the unit circle, so its coordinates are \((0,1)\). Therefore, \(\cos(\theta)=0\) and \(\sin(\theta)=1\). 2. Since \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\), tangent is undefined because its denominator is zero. 3. \(\csc(\theta)=\frac{1}{\sin(\theta)}=1\). 4. \(\sec(\theta)=\frac{1}{\cos(\theta)}\) is undefined because its denominator is zero. 5. \(\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}=0\).

\(\sin(\theta)=1\), \(\cos(\theta)=0\), \(\tan(\theta)\) undefined, \(\csc(\theta)=1\), \(\sec(\theta)\) undefined, \(\cot(\theta)=0\)

- 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. - Separate the angles according to whether cosine is negative, zero, or positive. - Use standard-angle cosine values where they apply. - For \(20^\circ\), compare its unit-circle position with a nearby standard angle instead of relying on a decimal approximation.

1. Use unit-circle positions and standard values: \(\cos(180^\circ)=-1\), \(\cos(210^\circ)=-\frac{\sqrt{3}}{2}\), \(\cos(120^\circ)=-\frac{1}{2}\), \(\cos(270^\circ)=0\), and \(\cos(300^\circ)=\frac{1}{2}\). 2. Since \(20^\circ<60^\circ\) and cosine decreases on \([0^\circ,180^\circ]\), \(\cos(20^\circ)>\cos(60^\circ)=\frac{1}{2}\). 3. Therefore, the increasing order is \(180^\circ,\ 210^\circ,\ 120^\circ,\ 270^\circ,\ 300^\circ,\ 20^\circ\).

\(180^\circ,\ 210^\circ,\ 120^\circ,\ 270^\circ,\ 300^\circ,\ 20^\circ\)

- 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 for each angle before comparing magnitudes. - Use exact special-angle values rather than decimal approximations.

1. Evaluate the standard-angle sine values: \(\sin\left(\frac{\pi}{2}\right)=1\), \(\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\), \(\sin(0)=0\), \(\sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2}\), and \(\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}\). 2. Since \(1>\frac{\sqrt{2}}{2}>0>-\frac{1}{2}>-\frac{\sqrt{3}}{2}\), the decreasing order is \(\frac{\pi}{2},\ \frac{\pi}{4},\ 0,\ \frac{7\pi}{6},\ \frac{4\pi}{3}\).

\(\frac{\pi}{2},\ \frac{\pi}{4},\ 0,\ \frac{7\pi}{6},\ \frac{4\pi}{3}\)

- 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) \(\alpha=40^\circ\)

- A reciprocal function has the same sign as the function it reciprocates. - Translate the secant condition into information about cosine. - Translate the cotangent condition into information about the relative signs of sine and cosine. - Intersect the quadrant possibilities from the two conditions.

1. Since secant has the same sign as cosine, \(\sec(\theta)<0\) places \(\theta\) in Quadrant II or III. 2. Cotangent has the same sign as tangent. Since \(\cot(\theta)<0\), sine and cosine have opposite signs, which occurs in Quadrant II or IV. 3. The only quadrant common to those two possibilities is Quadrant II. 4. In Quadrant II, sine and cosecant are positive, while cosine and secant are negative. Tangent and cotangent are negative. 5. The secant condition alone permits Quadrants II and III, and the cotangent condition alone permits Quadrants II and IV, so each condition by itself is insufficient.

a) Quadrant II b) \(\sin(\theta)>0\), \(\cos(\theta)<0\), \(\tan(\theta)<0\), \(\csc(\theta)>0\) c) \(\sec(\theta)<0\) allows Quadrants II or III, while \(\cot(\theta)<0\) allows Quadrants II or IV; only their intersection fixes Quadrant II.

- Begin by undoing the reciprocal in the given cosecant value. - Use the quadrant to choose the correct sign when recovering cosine. - Once sine and cosine are known, use quotients and reciprocals for the remaining functions.

1. Since \(\csc(\theta)=-\frac{5}{3}\), \(\sin(\theta)=-\frac{3}{5}\). 2. In Quadrant IV, cosine is positive. Using \(\sin^2(\theta)+\cos^2(\theta)=1\) gives \(\cos(\theta)=\frac{4}{5}\). 3. Then \(\tan(\theta)=\frac{-3/5}{4/5}=-\frac{3}{4}\). 4. Taking reciprocals gives \(\sec(\theta)=\frac{5}{4}\), \(\csc(\theta)=-\frac{5}{3}\), and \(\cot(\theta)=-\frac{4}{3}\).

\(\sin(\theta)=-\frac{3}{5}\), \(\cos(\theta)=\frac{4}{5}\), \(\tan(\theta)=-\frac{3}{4}\), \(\csc(\theta)=-\frac{5}{3}\), \(\sec(\theta)=\frac{5}{4}\), \(\cot(\theta)=-\frac{4}{3}\)