- Locate the point before it makes any turn. - Which coordinate corresponds to cosine, and which to sine?

1. At \(t=0\), the point has not moved from \((1,0)\). 2. Cosine is its x-coordinate and sine its y-coordinate, so \(\cos 0=1\) and \(\sin 0=0\).

\(\cos 0=1\) and \(\sin 0=0\)

- Focus on vertical position rather than the distance moved. - Recall which unit-circle coordinate is sine. - A quarter-turn corresponds to what familiar radian input?

1. The y-coordinate starts at \(0\) and ends at \(1\). 2. Sine is that y-coordinate, so \(\sin 0=0\) and \(\sin\left(\frac{\pi}{2}\right)=1\).

The y-coordinate changes from \(0\) to \(1\). Thus \(\sin 0=0\) and \(\sin\left(\frac{\pi}{2}\right)=1\).

- Which direction corresponds to a negative real input? - Which side of the x-axis does a small clockwise turn reach?

1. A clockwise turn of \(\frac{\pi}{6}\) ends below the x-axis in Quadrant IV. 2. Sine is the point's y-coordinate, which is negative there.

Negative; the terminal point is below the x-axis.

- Which coordinate appears in the denominator of tangent? - What happens to a quotient when its denominator is zero?

1. At this point, \(\sin t=-1\) and \(\cos t=0\). 2. Tangent would be \(\frac{-1}{0}\), which is undefined.

No. Tangent is undefined because its denominator \(\cos t\) is \(0\).

- Locate each angle relative to the nearest coordinate axis on the unit circle. - Remember that sine is the y-coordinate and cosine is the x-coordinate. - Use both the coordinate sign and its approximate size to eliminate choices.

1. For a), \(10^\circ\) is in Quadrant I near the positive x-axis, so the y-coordinate is small and positive. Thus \(\sin(10^\circ)\approx0.17\). 2. For b), \(100^\circ\) is in Quadrant II near the positive y-axis, so the x-coordinate is small and negative. Thus \(\cos(100^\circ)\approx-0.17\). 3. For c), \(260^\circ\) is in Quadrant III near the negative y-axis, so the y-coordinate is close to \(-1\). Thus \(\sin(260^\circ)\approx-0.98\). 4. For d), \(350^\circ\) is in Quadrant IV near the positive x-axis, so the x-coordinate is close to \(1\). Thus \(\cos(350^\circ)\approx0.98\); the other positive choices are too far from \(1\).

a) \(0.17\); Quadrant I, with the y-coordinate near \(0\) and positive. b) \(-0.17\); Quadrant II, with the x-coordinate near \(0\) and negative. c) \(-0.98\); Quadrant III, with the y-coordinate near \(-1\). d) \(0.98\); Quadrant IV, with the x-coordinate near \(1\).

- Add or subtract whole turns of \(2\pi\) until the input lies in the requested interval. - Keep the reduced angle as part of your answer rather than jumping directly to a trigonometric value. - Use exact unit-circle values at multiples of \(\frac{\pi}{2}\).

1. For a), \(-\pi+2\pi=\pi\), so the reduced angle is \(\pi\) and \(\sin(-\pi)=\sin(\pi)=0\). 2. For b), \(\frac{7\pi}{2}-2\pi=\frac{3\pi}{2}\), so \(\cos\left(\frac{7\pi}{2}\right)=\cos\left(\frac{3\pi}{2}\right)=0\). 3. For c), \(\frac{11\pi}{2}-4\pi=\frac{3\pi}{2}\), so \(\sin\left(\frac{11\pi}{2}\right)=\sin\left(\frac{3\pi}{2}\right)=-1\). 4. For d), \(-4\pi+4\pi=0\), so \(\cos(-4\pi)=\cos(0)=1\).

a) Reduced angle: \(\pi\); value: \(0\) b) Reduced angle: \(\frac{3\pi}{2}\); value: \(0\) c) Reduced angle: \(\frac{3\pi}{2}\); value: \(-1\) d) Reduced angle: \(0\); value: \(1\)

- Reduce each coefficient of \(\pi\) modulo a full turn before thinking about cosine. - Keep the reduced representative next to each original angle. - Use cosine values at the standard axis directions once the reductions are complete.

1. The angles \(7\pi\), \(-3\pi\), and \(15\pi\) each reduce to \(\pi\), whose cosine is \(-1\). 2. The angles \(8\pi\), \(-2\pi\), and \(20\pi\) each reduce to \(0\), whose cosine is \(1\). 3. The angles \(4.5\pi\), \(-1.5\pi\), and \(12.5\pi\) each reduce to \(\frac{\pi}{2}\), whose cosine is \(0\). 4. The groups therefore follow from their common reduced unit-circle position, not from separate decimal evaluations.

Cosine \(-1\): \(7\pi\to\pi\), \(-3\pi\to\pi\), \(15\pi\to\pi\) Cosine \(1\): \(8\pi\to0\), \(-2\pi\to0\), \(20\pi\to0\) Cosine \(0\): \(4.5\pi\to\frac{\pi}{2}\), \(-1.5\pi\to\frac{\pi}{2}\), \(12.5\pi\to\frac{\pi}{2}\)

- An angle with a degree symbol must be entered in degree mode; a unitless angle is interpreted in radians. - Use degree mode for parts a) and d). - Use radian mode for parts b) and c). - Check the fourth decimal place to round to the nearest thousandth.

1. For a), use degree mode: \(\sin(115^\circ) \approx 0.906307\), so the result is \(0.906\). 2. For b), use radian mode: \(\sin(2.8) \approx 0.334988\), so the result is \(0.335\). 3. For c), use radian mode: \(\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2} \approx -0.866025\), so the result is \(-0.866\). 4. For d), use degree mode: \(\sin(-45^\circ) = -\frac{\sqrt{2}}{2} \approx -0.707107\), so the result is \(-0.707\).

a) \(0.906\) b) \(0.335\) c) \(-0.866\) d) \(-0.707\)

- Make sure your calculator is in degree mode. - Recall the angle intervals for the four quadrants. - Use the signs of the x- and y-coordinates on the unit circle to check the signs of cosine and sine.

1. For a), \(90^\circ < 142.5^\circ < 180^\circ\), so the angle is in Quadrant II. A calculator gives \(\sin(142.5^\circ)\approx0.6088\). 2. For b), \(180^\circ < 195^\circ < 270^\circ\), so the angle is in Quadrant III. A calculator gives \(\cos(195^\circ)\approx-0.9659\). 3. For c), \(270^\circ < 280.4^\circ < 360^\circ\), so the angle is in Quadrant IV. A calculator gives \(\sin(280.4^\circ)\approx-0.9836\). 4. For d), \(270^\circ < 340^\circ < 360^\circ\), so the angle is in Quadrant IV. A calculator gives \(\cos(340^\circ)\approx0.9397\).

a) \(\sin(142.5^\circ)\approx0.6088\); Quadrant II b) \(\cos(195^\circ)\approx-0.9659\); Quadrant III c) \(\sin(280.4^\circ)\approx-0.9836\); Quadrant IV d) \(\cos(340^\circ)\approx0.9397\); Quadrant IV

- A full rotation is \(2\pi\) radians. - Add or subtract multiples of \(2\pi\). - Check that one result lies in the required negative interval.

1. Coterminal angles differ by integer multiples of \(2\pi\). 2. Subtract \(2\pi\): \(\frac{3\pi}{4}-2\pi=-\frac{5\pi}{4}\), which lies in \([-2\pi, 0)\). 3. Add \(2\pi\): \(\frac{3\pi}{4}+2\pi=\frac{11\pi}{4}\). 4. Add \(4\pi\): \(\frac{3\pi}{4}+4\pi=\frac{19\pi}{4}\).

For example, \(-\frac{5\pi}{4}\), \(\frac{11\pi}{4}\), and \(\frac{19\pi}{4}\)

- Subtract full rotations of \(360^\circ\). - Identify the quadrant of the reduced angle. - Use exact special-angle values.

1. Reduce \(480^\circ\) to \(120^\circ\). Therefore, \(\sin(480^\circ)=\sin(120^\circ)=\frac{\sqrt{3}}{2}\) and \(\cos(480^\circ)=\cos(120^\circ)=-\frac{1}{2}\). 2. Reduce \(1125^\circ\) by three full rotations: \(1125^\circ-3\cdot 360^\circ=45^\circ\). Therefore, \(\sin(1125^\circ)=\frac{\sqrt{2}}{2}\) and \(\cos(1125^\circ)=\frac{\sqrt{2}}{2}\).

a) \(\sin(480^\circ)=\frac{\sqrt{3}}{2}\); \(\cos(480^\circ)=-\frac{1}{2}\) b) \(\sin(1125^\circ)=\frac{\sqrt{2}}{2}\); \(\cos(1125^\circ)=\frac{\sqrt{2}}{2}\)

- Coterminal angles differ by an integer multiple of \(360^\circ\). - Reduce each angle to the interval \([0^\circ, 360^\circ)\). - Compare each reduced angle with \(210^\circ\). - Be careful when adding full rotations to negative angles.

1. Reduce each angle to \([0^\circ, 360^\circ)\). 2. \(570^\circ-360^\circ=210^\circ\), so it has the same terminal side. 3. \(-150^\circ+360^\circ=210^\circ\), so it has the same terminal side. 4. \(910^\circ-2\cdot 360^\circ=190^\circ\), so it has a different terminal side. 5. \(-510^\circ+2\cdot 360^\circ=210^\circ\), so it has the same terminal side.

a) Same terminal side b) Same terminal side c) Different terminal side; the coterminal angle is \(190^\circ\). d) Same terminal side

- Add or subtract multiples of \(360^\circ\) to place each angle in the standard interval. - Evaluate the trigonometric function at the reduced angle and compare it with the original expression. - A negative angle represents clockwise rotation, but coterminal angles have the same terminal side.

1. \(980^\circ-2\cdot 360^\circ=260^\circ\), so \(\cos(980^\circ)=\cos(260^\circ)\approx-0.1736\). 2. \(-440^\circ+2\cdot 360^\circ=280^\circ\), so \(\sin(-440^\circ)=\sin(280^\circ)\approx-0.9848\). 3. \(-1234^\circ+4\cdot 360^\circ=206^\circ\), so \(\cos(-1234^\circ)=\cos(206^\circ)\approx-0.8988\). 4. \(1850^\circ-5\cdot 360^\circ=50^\circ\), so \(\sin(1850^\circ)=\sin(50^\circ)\approx0.7660\).

1. \(\alpha=260^\circ\); \(\cos(980^\circ)\approx-0.1736\) 2. \(\alpha=280^\circ\); \(\sin(-440^\circ)\approx-0.9848\) 3. \(\alpha=206^\circ\); \(\cos(-1234^\circ)\approx-0.8988\) 4. \(\alpha=50^\circ\); \(\sin(1850^\circ)\approx0.7660\)

- Add or subtract multiples of \(360^\circ\) to place each angle in the standard interval. - Use the quadrant of \(\beta\) to determine the sign. - Use a reference angle of \(30^\circ\), \(45^\circ\), or \(60^\circ\) to find the exact value.

1. \(405^\circ-360^\circ=45^\circ\), so \(\sin(405^\circ)=\sin(45^\circ)=\frac{\sqrt{2}}{2}\). 2. \(-300^\circ+360^\circ=60^\circ\), so \(\cos(-300^\circ)=\cos(60^\circ)=\frac{1}{2}\). 3. \(1020^\circ-2\cdot 360^\circ=300^\circ\), so \(\sin(1020^\circ)=\sin(300^\circ)=-\frac{\sqrt{3}}{2}\). 4. \(-120^\circ+360^\circ=240^\circ\), so \(\cos(-120^\circ)=\cos(240^\circ)=-\frac{1}{2}\).

a) \(\beta=45^\circ\); \(\frac{\sqrt{2}}{2}\) b) \(\beta=60^\circ\); \(\frac{1}{2}\) c) \(\beta=300^\circ\); \(-\frac{\sqrt{3}}{2}\) d) \(\beta=240^\circ\); \(-\frac{1}{2}\)

- Check that your calculator display indicates radian mode. - Use the fifth decimal digit to decide how to round the fourth decimal digit. - Enter a negative argument with its negative sign included.

1. Set the calculator to radian mode. 2. Evaluate and round each expression: a) \(\sin(0.85)\approx0.7513\) b) \(\cos(2.12)\approx-0.5220\) c) \(\sin(-0.5)\approx-0.4794\) d) \(\cos(7.5)\approx0.3466\)

a) \(0.7513\) b) \(-0.5220\) c) \(-0.4794\) d) \(0.3466\)

- Substitute each given value directly into the function rule. - You can enter the entire expression \(\sin(x)+\cos(x)\) in one calculator line. - Include the negative sign when entering the input in part c).

1. Substitute each input into \(f(x)=\sin(x)+\cos(x)\) and evaluate in radian mode. 2. For a), \(f(1.4)=\sin(1.4)+\cos(1.4)\approx1.1554169\), so \(f(1.4)\approx1.155\). 3. For b), \(f(3.5)=\sin(3.5)+\cos(3.5)\approx-1.2872399\), so \(f(3.5)\approx-1.287\). 4. For c), \(f(-2.1)=\sin(-2.1)+\cos(-2.1)\approx-1.3680555\), so \(f(-2.1)\approx-1.368\).

a) \(f(1.4)\approx1.155\) b) \(f(3.5)\approx-1.287\) c) \(f(-2.1)\approx-1.368\)

- Make sure your calculator is in radian mode. - A number closer to \(0\) has a smaller absolute value. - Use the fifth decimal digit to round to four decimal places.

1. Evaluate each expression in radian mode: a) \(\sin(4.5)\approx-0.9775\) b) \(\cos(-2.1)\approx-0.5048\) c) \(\sin(-0.75)\approx-0.6816\) d) \(\cos(10.2)\approx-0.7143\) 2. Compare absolute values: \(0.5048<0.6816<0.7143<0.9775\). Therefore, \(\cos(-2.1)\) is closest to \(0\).

a) \(-0.9775\) b) \(-0.5048\) c) \(-0.6816\) d) \(-0.7143\) The value \(\cos(-2.1)\) is closest to \(0\).

- Substitute the given values before evaluating the trigonometric functions. - Pay close attention to the negative sign in part c). - Keep full calculator precision until the final rounding step.

1. For a), \(\sin(1.5)\cdot\cos(-2.8)\approx-0.9398621\), so the result is \(-0.9399\). 2. For b), \(\cos(1.5)+\sin(-2.8)\approx-0.2642509\), so the result is \(-0.2643\). 3. For c), \(x+y=1.5+(-2.8)=-1.3\), and \(\sin(-1.3)\approx-0.9635582\), so the result is \(-0.9636\).

a) \(-0.9399\) b) \(-0.2643\) c) \(-0.9636\)

- Sine and cosine repeat after a rotation of \(2\pi\). - Add \(2\pi\) to a negative angle to obtain a positive coterminal angle. - Express \(2\pi\) with the same denominator as the given angle. - Check that each result lies in the required interval.

1. Sine and cosine have period \(2\pi\), so add or subtract multiples of \(2\pi\). 2. \(\frac{17\pi}{4}-4\pi=\frac{\pi}{4}\), so \(\sin\left(\frac{17\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)\). 3. \(-\frac{2\pi}{3}+2\pi=\frac{4\pi}{3}\), so \(\cos\left(-\frac{2\pi}{3}\right)=\cos\left(\frac{4\pi}{3}\right)\). 4. \(-\frac{7\pi}{6}+2\pi=\frac{5\pi}{6}\), so \(\sin\left(-\frac{7\pi}{6}\right)=\sin\left(\frac{5\pi}{6}\right)\). 5. \(\frac{19\pi}{6}-2\pi=\frac{7\pi}{6}\), so \(\cos\left(\frac{19\pi}{6}\right)=\cos\left(\frac{7\pi}{6}\right)\).

a) \(\sin\left(\frac{\pi}{4}\right)\) b) \(\cos\left(\frac{4\pi}{3}\right)\) c) \(\sin\left(\frac{5\pi}{6}\right)\) d) \(\cos\left(\frac{7\pi}{6}\right)\)

- Reduce the real input by a full turn of \(2\pi\). - Identify the reference angle and quadrant of the reduced input. - Unit-circle coordinates are given by cosine first and sine second.

1. Subtract one full turn: \(\frac{17\pi}{6}-2\pi=\frac{5\pi}{6}\). 2. The reference angle for \(\frac{5\pi}{6}\) is \(\frac{\pi}{6}\), and the terminal side is in Quadrant II. 3. On the unit circle, the coordinates are \((\cos t,\sin t)\), so the point is \(\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\).

\(\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\)

- The reference angle gives coordinate magnitudes, but not their signs. - Read the quadrant from the displayed terminal point. - On the unit circle, the x-coordinate is cosine and the y-coordinate is sine.

1. The diagram places \(P\) in Quadrant II, so cosine is negative and sine is positive. 2. With reference angle \(\frac{\pi}{6}\), the coordinate magnitudes are \(\frac{\sqrt{3}}{2}\) horizontally and \(\frac{1}{2}\) vertically. 3. Applying the Quadrant-II signs gives \(P=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\).

a) Quadrant II; cosine is negative and sine is positive. b) \(P=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\)

- Locate the interval on the unit circle before thinking about function values. - Sine corresponds to the y-coordinate; cosine corresponds to the x-coordinate. - Determine the product's sign only after determining each factor's sign.

1. The interval \(\left(\frac{3\pi}{2},2\pi\right)\) places the terminal point in Quadrant IV. 2. In Quadrant IV, the y-coordinate is negative, so \(\sin t<0\). 3. The x-coordinate is positive, so \(\cos t>0\). 4. A negative number times a positive number is negative, so \(\sin t\cos t<0\).

a) Negative b) Positive c) Negative

- Read the quadrant and the marked acute angle from the diagram. - Build the positive input from the negative x-axis benchmark, then shift by one full turn for the negative input. - Equal terminal points have equal x- and y-coordinates.

1. The diagram shows a Quadrant-III terminal radius that is \(\frac{\pi}{4}\) past the negative x-axis direction. 2. Therefore, \(t_1=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\). 3. Subtracting one full turn gives the unique requested negative input: \(t_2=\frac{5\pi}{4}-2\pi=-\frac{3\pi}{4}\). 4. The two inputs reach the same unit-circle point, so their x-coordinates and y-coordinates are equal. Hence \(\cos t_1=\cos t_2\) and \(\sin t_1=\sin t_2\).

a) \(t_1=\frac{5\pi}{4}\), \(t_2=-\frac{3\pi}{4}\) b) \(\sin t_1=\sin t_2\) and \(\cos t_1=\cos t_2\) because the two inputs have the same terminal point.

- Add a multiple of \(2\pi\) to make the input easier to locate. - Use the reference angle and quadrant of the reduced input. - Write cosine as the x-coordinate and sine as the y-coordinate.

1. Add one full turn: \(-\frac{13\pi}{6}+2\pi=-\frac{\pi}{6}\). 2. The angle \(-\frac{\pi}{6}\) has reference angle \(\frac{\pi}{6}\) and lies in Quadrant IV. 3. Therefore, \((\cos t,\sin t)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\).

\(\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\)

- Express the full turn with the same denominator as the original input. - Check the arithmetic before using unit-circle facts. - Use the quadrant of the corrected input to determine signs.

1. Rewrite \(2\pi\) as \(\frac{12\pi}{6}\). 2. Then \(-\frac{7\pi}{6}+\frac{12\pi}{6}=\frac{5\pi}{6}\). 3. The input \(\frac{5\pi}{6}\) is in Quadrant II. 4. Therefore, sine is positive and cosine is negative.

The correct reduced input is \(\frac{5\pi}{6}\). Its sine is positive and its cosine is negative.

- Separate the magnitude of the radian measure from its sign. - Recall which rotation direction corresponds to a negative input. - To find a positive coterminal input, change the input by one full revolution.

1. The magnitude \(\frac{5\pi}{6}\) gives the size of the turn, but the negative sign determines its direction. 2. Therefore, \(-\frac{5\pi}{6}\) means a clockwise turn, not a counterclockwise turn. 3. Add one full revolution: \(-\frac{5\pi}{6}+2\pi=\frac{7\pi}{6}\). 4. Thus the positive coterminal input is \(\frac{7\pi}{6}\), not \(\frac{5\pi}{6}\).

The error is ignoring the negative sign. The rotation is clockwise, and the positive coterminal input is \(\frac{7\pi}{6}\).

- Connect the x- and y-coordinates of a unit-circle point with cosine and sine. - Tangent compares those two coordinates as a ratio. - Use the quadrant to check whether the ratio should be positive or negative.

1. On the unit circle, the x-coordinate is \(\cos t\) and the y-coordinate is \(\sin t\), so \(\cos t=-\frac{8}{17}\) and \(\sin t=\frac{15}{17}\). 2. Tangent is the coordinate ratio \(\tan t=\frac{\sin t}{\cos t}\). 3. Therefore, \(\tan t=\frac{15/17}{-8/17}=-\frac{15}{8}\). 4. The point is in Quadrant II, where sine is positive and cosine is negative, so tangent must be negative.

a) \(\cos t=-\frac{8}{17}\), \(\sin t=\frac{15}{17}\) b) \(\tan t=-\frac{15}{8}\) c) \(P(t)\) is in Quadrant II, so sine and cosine have opposite signs and tangent is negative.

- Compare both graph values at \(t=0\) with the two coordinates of the unit-circle starting point. - Use the quarter-turn position as a second check. - Connect the first unit-circle coordinate with cosine and the second with sine.

1. At \(t=0\), \(P(0)=(1,0)\). The graph has \(f(0)=1\) and \(g(0)=0\), so \(f\) records the x-coordinate and \(g\) records the y-coordinate. 2. At \(t=\frac{\pi}{2}\), \(P\left(\frac{\pi}{2}\right)=(0,1)\). The graph has \(f\left(\frac{\pi}{2}\right)=0\) and \(g\left(\frac{\pi}{2}\right)=1\), confirming the match. 3. Therefore, \(f(t)=\cos t\) and \(g(t)=\sin t\).

\(f\) is the x-coordinate function, so \(f(t)=\cos t\). \(g\) is the y-coordinate function, so \(g(t)=\sin t\).

- Which coordinate gives sine, and which gives cosine? - What relation connects tangent with sine and cosine? - Check that the signs of the coordinates agree with the sign of the result.

1. At \(P\), \(\cos t=\frac{20}{29}\) and \(\sin t=\frac{21}{29}\). 2. Use \(\tan t=\frac{\sin t}{\cos t}=\frac{21/29}{20/29}=\frac{21}{20}\).

\(\tan t=\frac{21}{20}\), since tangent is sine divided by cosine.

- Read the two curve heights at the left edge of the graph. Which coordinate does each one represent? - What do the signs of the coordinates tell you about the quadrant? - Which coordinate must be zero when a point lies on the y-axis?

1. At \(t=0\), the graph gives \(u(0)\approx0.5\) and \(v(0)\approx0.9\). Both are positive, so the point is in Quadrant I. 2. The point is on the y-axis when its horizontal coordinate \(u(t)\) is zero. The \(u\) curve first crosses zero for positive \(t\) at \(t=\frac{\pi}{6}\).

a) About \((0.5, 0.9)\); Quadrant I. b) \(t=\frac{\pi}{6}\), where the \(u\) curve crosses the horizontal axis.

- Locate \(0.6\) between the two given radian benchmarks before thinking about decimal values. - Track how the x-coordinate and y-coordinate change as a Quadrant I point moves counterclockwise. - On the unit circle, connect the first coordinate with cosine and the second with sine.

1. The input \(0.6\) lies in Quadrant I between \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\). 2. Cosine decreases on this part of the unit circle, so \(\frac{\sqrt{2}}{2}<\cos(0.6)<\frac{\sqrt{3}}{2}\). Therefore, \(\frac{\sqrt{2}}{2}<x<\frac{\sqrt{3}}{2}\). 3. Sine increases on this part of the unit circle, so \(\frac{1}{2}<\sin(0.6)<\frac{\sqrt{2}}{2}\). Therefore, \(\frac{1}{2}<y<\frac{\sqrt{2}}{2}\). 4. In radian mode, \(\cos(0.6)\approx0.825\) and \(\sin(0.6)\approx0.565\). 5. On the unit circle, \(P=(\cos t,\sin t)\), so \(P\approx(0.825,0.565)\).

a) \(\frac{\sqrt{2}}{2}<x<\frac{\sqrt{3}}{2}\) and \(\frac{1}{2}<y<\frac{\sqrt{2}}{2}\) b) \(\cos(0.6)\approx0.825\) and \(\sin(0.6)\approx0.565\), so \(P\approx(0.825,0.565)\).

- Imagine how a point moves on the unit circle as the angle increases. - What happens to the point’s height from \(0^\circ\) to \(90^\circ\)? - What happens to its horizontal coordinate? - When are the horizontal and vertical coordinates equal?

1. In Quadrant I, sine is the y-coordinate, which increases as the angle increases. Since \(75^\circ > 20^\circ\), \(\sin(75^\circ) > \sin(20^\circ)\). 2. In Quadrant I, cosine is the x-coordinate, which decreases as the angle increases. Since \(10^\circ < 80^\circ\), \(\cos(10^\circ) > \cos(80^\circ)\). 3. Sine and cosine are equal when the x- and y-coordinates are equal. In Quadrant I, this occurs on the line \(y=x\), at \(\alpha = 45^\circ\).

a) \(\sin(75^\circ)\) b) \(\cos(10^\circ)\) c) \(45^\circ\)

- What are the coordinates of a unit-circle point in terms of its angle? - How do coordinates change when a point is reflected across the y-axis? - Which quadrant contains a point with a negative x-coordinate and a positive y-coordinate?

1. A unit-circle point at angle \(\alpha\) has coordinates \((\cos(\alpha), \sin(\alpha))\). Thus, \(P=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). 2. Point \(Q=\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) is the reflection of \(P\) across the y-axis, so it lies in Quadrant II. Therefore, \(\beta=180^\circ-60^\circ=120^\circ\). 3. The coordinates of \(Q\) give \(\cos(120^\circ)=-\frac{1}{2}\) and \(\sin(120^\circ)=\frac{\sqrt{3}}{2}\).

a) \(P\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) b) \(\beta=120^\circ\) c) \(\sin(120^\circ)=\frac{\sqrt{3}}{2}\) and \(\cos(120^\circ)=-\frac{1}{2}\)

- Use the signs of \(x\) and \(y\) to identify the quadrant. - On the unit circle, how are cosine and sine related to coordinates? - Apply the Pythagorean theorem to the coordinates. - How does reflection across the y-axis change an ordered pair?

1. Since the x-coordinate is negative and the y-coordinate is positive, \(P\) lies in Quadrant II. 2. On the unit circle, \(\cos(\alpha)=x=-0.8\) and \(\sin(\alpha)=y=0.6\). 3. Then \(\sin^2(\alpha)+\cos^2(\alpha)=0.6^2+(-0.8)^2=0.36+0.64=1\). This confirms the Pythagorean identity. 4. The angle \(180^\circ-\alpha\) reflects the point across the y-axis, so \(Q=(0.8, 0.6)\).

a) Quadrant II b) \(\cos(\alpha)=-0.8\); \(\sin(\alpha)=0.6\) c) \(1\); the Pythagorean identity d) \(Q(0.8, 0.6)\)

- Track how the x- and y-coordinates change as a point moves from \(0^\circ\) to \(90^\circ\). - Use \(\sin(180^\circ-\theta)=\sin(\theta)\) and \(\cos(180^\circ-\theta)=-\cos(\theta)\). - Recall the exact sine values for \(30^\circ\) and \(60^\circ\). - Use signs first when ordering values.

1. In Quadrant I, cosine decreases from \(1\) to \(0\), while sine increases from \(0\) to \(1\). They are equal at \(45^\circ\), so \(\cos(\alpha)>\sin(\alpha)\) for \(0^\circ \le \alpha <45^\circ\). 2. The value \(\cos(160^\circ)\) is negative. Also, \(\sin(160^\circ)=\sin(20^\circ)\), and \(\cos(20^\circ)\) is positive and larger than \(\sin(20^\circ)\). Thus, \(\cos(160^\circ)<\sin(20^\circ)=\sin(160^\circ)<\cos(20^\circ)\). 3. \(\sin(30^\circ)=\frac{1}{2}\), so twice this value is \(1\). But \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\ne1\). Therefore, sine is not proportional to the angle.

a) \(0^\circ \le \alpha <45^\circ\) b) \(\cos(160^\circ)<\sin(20^\circ)=\sin(160^\circ)<\cos(20^\circ)\) c) \(2\sin(30^\circ)=1\), but \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\), so the claim is false.

- Adjust each input by whole rotations until it lies in the requested interval. - Use the coterminal angle, not the original large or negative input, when checking the quadrant and sign. - Round only after evaluating the trigonometric value.

1. For a), \(510^\circ-360^\circ=150^\circ\), so \(\sin(510^\circ)=\sin(150^\circ)=0.500\). 2. For b), \(-130.4^\circ+360^\circ=229.6^\circ\), so \(\cos(-130.4^\circ)=\cos(229.6^\circ)\approx-0.648\). 3. For c), \(1000^\circ-720^\circ=280^\circ\), so \(\sin(1000^\circ)=\sin(280^\circ)\approx-0.985\). 4. For d), \(825^\circ-720^\circ=105^\circ\), so \(\cos(825^\circ)=\cos(105^\circ)\approx-0.259\).

a) Coterminal angle: \(150^\circ\); value: \(0.500\) b) Coterminal angle: \(229.6^\circ\); value: \(-0.648\) c) Coterminal angle: \(280^\circ\); value: \(-0.985\) d) Coterminal angle: \(105^\circ\); value: \(-0.259\)

- Use degree mode on your calculator. - Locate \(310^\circ\) by dividing the unit circle into four quadrants. - Negative angles are measured clockwise. Find a positive coterminal angle for \(-215^\circ\).

1. For \(\gamma\), \(\sin(310^\circ)\approx-0.766\) and \(\cos(310^\circ)\approx0.643\). 2. The angle \(310^\circ\) lies in Quadrant IV, where x-coordinates are positive and y-coordinates are negative. Therefore, cosine is positive and sine is negative. 3. For \(\delta\), \(\sin(-215^\circ)\approx0.574\) and \(\cos(-215^\circ)\approx-0.819\).

a) \(\sin(310^\circ)\approx-0.766\); \(\cos(310^\circ)\approx0.643\). The angle is in Quadrant IV, so sine is negative and cosine is positive. b) \(\sin(-215^\circ)\approx0.574\); \(\cos(-215^\circ)\approx-0.819\)

- Add or subtract full rotations to reduce each angle. - Cosine is the x-coordinate on the unit circle. - Reflecting across the x-axis preserves the x-coordinate.

1. For \(480^\circ\), subtract one full rotation: \(480^\circ=120^\circ+1\cdot 360^\circ\). 2. The angle reflected across the x-axis is \(360^\circ-120^\circ=240^\circ\), and it has the same cosine. 3. For \(-150^\circ\), add one full rotation: \(-150^\circ=210^\circ-1\cdot 360^\circ\). 4. The reflected angle is \(360^\circ-210^\circ=150^\circ\), and it has the same cosine.

a) \(\alpha_1=120^\circ+1\cdot 360^\circ\); \(\alpha_2=210^\circ-1\cdot 360^\circ\) b) For \(\alpha_1\): \(240^\circ\); for \(\alpha_2\): \(150^\circ\)

- Add or subtract multiples of \(360^\circ\). - Coterminal angles have the same sine and cosine values. - Check the required inequalities for \(\beta_1\) and \(\beta_2\).

1. Subtract one full rotation: \(620^\circ-360^\circ=260^\circ\), so \(\alpha_0=260^\circ\). 2. A negative coterminal angle is \(260^\circ-360^\circ=-100^\circ\). 3. An angle greater than \(720^\circ\) is \(260^\circ+720^\circ=980^\circ\). 4. Both angles differ from \(260^\circ\) by integer multiples of \(360^\circ\), so they have the same sine value.

a) \(\alpha_0=260^\circ\) b) For example, \(\beta_1=-100^\circ\) and \(\beta_2=980^\circ\)

- First use whole turns of \(2\pi\) to move each input into the requested interval. - For each reduced input, identify its acute reference angle and the signs of the relevant unit-circle coordinates. - Record the symmetry relationship explicitly before using the exact special-angle value.

1. For a), \(-\frac{13\pi}{6}+4\pi=\frac{11\pi}{6}\). Its reference angle is \(\frac{\pi}{6}\), and \(\sin\left(\frac{11\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=-\frac{1}{2}\). 2. For b), \(2025\pi+\frac{\pi}{3}\) is coterminal with \(\frac{4\pi}{3}\). Its reference angle is \(\frac{\pi}{3}\), and \(\cos\left(\frac{4\pi}{3}\right)=-\cos\left(\frac{\pi}{3}\right)=-\frac{1}{2}\). 3. For c), \(\frac{11\pi}{4}\) reduces to \(\frac{3\pi}{4}\), so \(\sin\left(\frac{3\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\). Also, \(-\frac{7\pi}{3}\) reduces to \(\frac{5\pi}{3}\), so \(\cos\left(\frac{5\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\). The sum is \(\frac{\sqrt{2}+1}{2}\). 4. For d), both inputs reduce to \(\frac{7\pi}{6}\). Using the reference angle \(\frac{\pi}{6}\), \(\cos\left(\frac{7\pi}{6}\right)=-\cos\left(\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}\) and \(\sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=-\frac{1}{2}\). Therefore, the difference is \(\frac{1-\sqrt{3}}{2}\).

a) Reduced input: \(\frac{11\pi}{6}\); symmetry: \(\sin\left(\frac{11\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)\); value: \(-\frac{1}{2}\) b) Reduced input: \(\frac{4\pi}{3}\); symmetry: \(\cos\left(\frac{4\pi}{3}\right)=-\cos\left(\frac{\pi}{3}\right)\); value: \(-\frac{1}{2}\) c) Reduced inputs: \(\frac{3\pi}{4}\) and \(\frac{5\pi}{3}\); symmetries: \(\sin\left(\frac{3\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)\), \(\cos\left(\frac{5\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)\); value: \(\frac{\sqrt{2}+1}{2}\) d) Reduced inputs: \(\frac{7\pi}{6}\) and \(\frac{7\pi}{6}\); symmetries: \(\cos\left(\frac{7\pi}{6}\right)=-\cos\left(\frac{\pi}{6}\right)\), \(\sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)\); value: \(\frac{1-\sqrt{3}}{2}\)

- Use radian mode. - Reduce angles by multiples of \(2\pi\). - Use even and odd symmetry for negative angles. - Keep full precision until the final rounding step.

1. Reduce \(2.8\pi\) by \(2\pi\): \(\sin(2.8\pi)=\sin(0.8\pi)\approx0.588\). 2. A calculator in radian mode gives \(\cos(10)\approx-0.839\). 3. A calculator in radian mode gives \(\sin(-4.5)\approx0.978\). 4. Add \(2\pi\) to \(-\frac{13\pi}{6}\): \(\cos\left(-\frac{13\pi}{6}\right)=\cos\left(-\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\approx0.866\).

a) \(0.588\) b) \(-0.839\) c) \(0.978\) d) \(0.866\)

- Compare each angle with multiples of \(\frac{\pi}{2}\). - Negative angles are measured clockwise. - Reduce angles greater than \(2\pi\). - Use the quadrant to check the sign.

1. Since \(\pi < 3.5 < \frac{3\pi}{2}\), the angle is in Quadrant III. Thus \(\cos(3.5)\approx-0.936\). 2. The angle \(-2\) is coterminal with \(2\pi-2\), which lies in Quadrant III. Thus \(\sin(-2)\approx-0.909\). 3. Reduce \(7.2\pi\) by \(6\pi\) to \(1.2\pi\), which lies in Quadrant III. Thus \(\cos(7.2\pi)\approx-0.809\). 4. The angle \(-\frac{4\pi}{9}\) lies between \(-\frac{\pi}{2}\) and \(0\), so its terminal side is in Quadrant IV. Thus \(\sin\left(-\frac{4\pi}{9}\right)\approx-0.985\).

a) \(-0.936\); Quadrant III b) \(-0.909\); Quadrant III c) \(-0.809\); Quadrant III d) \(-0.985\); Quadrant IV

- Reduce inputs outside \([0,2\pi)\) by adding or subtracting \(2\pi\). - After reducing, compare positions with the reference angle \(\frac{\pi}{6}\). - Use the sign of the y-coordinate in each quadrant to separate the two values.

1. The reduced inputs are \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), \(\frac{11\pi}{6}\), \(\frac{7\pi}{6}\), \(\frac{11\pi}{6}\), and \(\frac{7\pi}{6}\), respectively. 2. The angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) have sine \(\frac{1}{2}\), so the corresponding four original expressions form one group. 3. The angles \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\) have sine \(-\frac{1}{2}\), so the remaining four expressions form the second group.

Value \(\frac{1}{2}\): \(\frac{\pi}{6}\to\frac{\pi}{6}\), \(\frac{5\pi}{6}\to\frac{5\pi}{6}\), \(\frac{13\pi}{6}\to\frac{\pi}{6}\), \(-\frac{7\pi}{6}\to\frac{5\pi}{6}\) Value \(-\frac{1}{2}\): \(-\frac{\pi}{6}\to\frac{11\pi}{6}\), \(\frac{7\pi}{6}\to\frac{7\pi}{6}\), \(\frac{11\pi}{6}\to\frac{11\pi}{6}\), \(-\frac{5\pi}{6}\to\frac{7\pi}{6}\)

- Reduce any angle outside one revolution before evaluating a trigonometric value. - Use the quadrant or axis location to determine signs. - Match the reference angle with familiar exact unit-circle coordinates.

1. For a), the coterminal angle is \(225^\circ\), in Quadrant III with reference angle \(45^\circ\). Thus \(\sin(225^\circ)=-\frac{\sqrt{2}}{2}\) and \(\cos(225^\circ)=-\frac{\sqrt{2}}{2}\). 2. For b), the coterminal angle is \(330^\circ\), in Quadrant IV with reference angle \(30^\circ\). Thus \(\sin(-30^\circ)=-\frac{1}{2}\) and \(\cos(-30^\circ)=\frac{\sqrt{3}}{2}\). 3. For c), the coterminal angle is \(60^\circ\), in Quadrant I. Thus \(\sin(420^\circ)=\frac{\sqrt{3}}{2}\) and \(\cos(420^\circ)=\frac{1}{2}\). 4. For d), the coterminal angle is \(180^\circ\), on the negative x-axis. Thus \(\sin(900^\circ)=0\) and \(\cos(900^\circ)=-1\).

a) \(225^\circ\), Quadrant III; \(\sin=-\frac{\sqrt{2}}{2}\), \(\cos=-\frac{\sqrt{2}}{2}\) b) \(330^\circ\), Quadrant IV; \(\sin=-\frac{1}{2}\), \(\cos=\frac{\sqrt{3}}{2}\) c) \(60^\circ\), Quadrant I; \(\sin=\frac{\sqrt{3}}{2}\), \(\cos=\frac{1}{2}\) d) \(180^\circ\), negative x-axis; \(\sin=0\), \(\cos=-1\)

- Distinguish between the unitless angle \(2\) and the angle \(2^\circ\). - Which calculator mode is required for each term? - Evaluate the two sine values separately, but keep full precision until the final addition.

1. Interpret the unitless angle in radians: \(\sin(2) \approx 0.9093\). 2. Interpret the angle with the degree symbol in degrees: \(\sin(2^\circ) \approx 0.0349\). 3. Add using full calculator precision, then round: \(T = \sin(2) + \sin(2^\circ) \approx 0.9442\).

\(\sin(2) \approx 0.9093\) \(\sin(2^\circ) \approx 0.0349\) \(T \approx 0.9442\)

- Distinguish degree measures from radian measures. - Use the correct calculator mode for each expression. - Compare \(4\) with \(\pi\) and \(\frac{3\pi}{2}\). - Use quadrant signs to check the coordinates.

1. Evaluate the four expressions, using degrees for (1) and (2) and radians for (3) and (4): (1) \(\cos(110^\circ)\approx-0.342\) (2) \(\sin(260^\circ)\approx-0.985\) (3) \(\cos(4)\approx-0.654\) (4) \(\sin(-0.8)\approx-0.717\) 2. Since \(\pi < 4 < \frac{3\pi}{2}\), an angle of \(4\) radians has its terminal side in Quadrant III. Therefore, the point \(P\) lies in Quadrant III.

a) (1) \(-0.342\); (2) \(-0.985\); (3) \(-0.654\); (4) \(-0.717\) b) Quadrant III, because \(\pi < 4 < \frac{3\pi}{2}\)

- At which inputs does the denominator become zero, and how can the unit circle help? - Think about how a full turn affects the coordinates used in the formula. - A counterexample needs just one input where two proposed values differ.

1. The denominator is zero when \(\cos x=\frac{1}{2}\). The unit circle gives \(x=2\pi k\pm\frac{\pi}{3}\), where \(k\in\mathbb{Z}\). 2. Since \(\cos(x+2\pi)=\cos x\), the numerator and denominator of \(g\) do not change after a full turn. Thus \(g(x+2\pi)=g(x)\) whenever defined. 3. At \(x=0\), \(g(0)=\frac{2}{2-1}=2\). At \(x=\pi\), \(g(\pi)=\frac{2}{-2-1}=-\frac{2}{3}\). Therefore adding \(\pi\) does not always preserve \(g\).

a) \(x=2\pi k\pm\frac{\pi}{3}\), where \(k\in\mathbb{Z}\) b) \(g(x+2\pi)=g(x)\) since cosine repeats after one full turn; \(g(0)=2\ne-\frac{2}{3}=g(\pi)\).

- A full rotation is \(360^\circ\). - Write a formula for all angles coterminal with \(225^\circ\). - Test both positive and negative integer multiples of a full rotation. - Keep only angles in the given interval, and exclude the original angle.

1. Coterminal angles have the form \(\beta=225^\circ+360^\circ k\), where \(k\) is an integer. 2. Test integer values of \(k\) that keep \(\beta\) in the given interval. 3. For \(k=-3, -2, -1, 1\), the angles are \(-855^\circ\), \(-495^\circ\), \(-135^\circ\), and \(585^\circ\). The values for \(k=-4\) and \(k=2\) lie outside the interval, and \(k=0\) gives the original angle.

\(\beta \in \{-855^\circ, -495^\circ, -135^\circ, 585^\circ\}\)

- A full rotation is \(360^\circ\), or \(2\pi\) radians. - Convert radians to degrees by multiplying by \(\frac{180^\circ}{\pi}\). - Add or subtract full rotations to compare terminal sides. - Reduce each angle to the interval \([0^\circ, 360^\circ)\).

1. Multiply each radian measure by \(\frac{180^\circ}{\pi}\): \(x_1=60^\circ\), \(x_2=420^\circ\), \(x_3=-300^\circ\), and \(x_4=240^\circ\). 2. Coterminal angles differ by a multiple of \(360^\circ\), or equivalently by a multiple of \(2\pi\). 3. Since \(420^\circ-360^\circ=60^\circ\) and \(-300^\circ+360^\circ=60^\circ\), the angles \(x_1\), \(x_2\), and \(x_3\) have the same terminal side. The angle \(x_4=240^\circ\) has a different terminal side.

a) \(x_1=60^\circ\), \(x_2=420^\circ\), \(x_3=-300^\circ\), \(x_4=240^\circ\) b) \(x_1\), \(x_2\), and \(x_3\) have the same terminal side because their measures differ by integer multiples of \(360^\circ\). The angle \(x_4\) does not.

- A point returns to the same location after a rotation of \(2\pi\). - Write a formula using integer multiples of \(2\pi\). - Test both positive and negative integer values. - Check every result against the interval endpoints.

1. All coterminal angles have the form \(\alpha=\frac{2\pi}{3}+2\pi k\), where \(k\) is an integer. 2. For \(k=-1\), \(\alpha=-\frac{4\pi}{3}\), which lies in the interval. 3. For \(k=0\), \(\alpha=\frac{2\pi}{3}\), which lies in the interval. 4. For \(k=1\), \(\alpha=\frac{8\pi}{3}\), which lies in the interval. 5. The next values, obtained with \(k=-2\) and \(k=2\), lie outside the interval.

\(\alpha \in \left\{-\frac{4\pi}{3}, \frac{2\pi}{3}, \frac{8\pi}{3}\right\}\)

- Use the signs of the coordinates to identify the quadrant. - Equal coordinate magnitudes point to a familiar reference angle. - A negative coterminal input differs by one full turn of \(2\pi\).

1. Both coordinates are negative and have magnitude \(\frac{\sqrt{2}}{2}\), so the point is in Quadrant III with reference angle \(\frac{\pi}{4}\). 2. The input in \([0,2\pi)\) is \(\pi+\frac{\pi}{4}=\frac{5\pi}{4}\). 3. Subtract \(2\pi\): \(\frac{5\pi}{4}-2\pi=-\frac{3\pi}{4}\).

a) \(\frac{5\pi}{4}\) b) \(-\frac{3\pi}{4}\)

- Interpret \(t+2\pi\) as movement on the unit circle rather than as an algebraic expression only. - Ask where the terminal point is after one additional complete turn. - Connect the point's coordinates to sine and cosine.

1. Increasing an input by \(2\pi\) adds exactly one full counterclockwise rotation. 2. After one full rotation, the terminal point on the unit circle is unchanged. 3. Therefore, both coordinates are unchanged: \(\cos(t+2\pi)=\cos t\) and \(\sin(t+2\pi)=\sin t\).

Adding \(2\pi\) makes one full rotation and returns to the same unit-circle point. Thus \(\cos(t+2\pi)=\cos t\) and \(\sin(t+2\pi)=\sin t\).

- Compare \(2.6\) with familiar radian benchmarks. - Use the quadrant to determine signs before comparing magnitudes. - Think about which coordinate axis the terminal point is nearer.

1. Since \(\frac{\pi}{2}\approx1.57<2.6<\pi\approx3.14\), the terminal point is in Quadrant II. 2. In Quadrant II, sine is positive and cosine is negative. 3. The input \(2.6\) is closer to \(\pi\) than to \(\frac{\pi}{2}\), so the point is closer to the negative x-axis than to the positive y-axis. 4. Therefore, the x-coordinate has greater magnitude, so \(|\cos t|>|\sin t|\).

a) Quadrant II b) \(\sin t>0\) and \(\cos t<0\) c) \(|\cos t|>|\sin t|\)

- Interpret \(\cos t=0\) as a condition on the unit-circle x-coordinate. - Identify the two locations in one complete turn where the x-coordinate is zero. - Include every coterminal input that lies inside the given negative-to-positive interval.

1. Cosine is the x-coordinate on the unit circle, so \(\cos t=0\) at the top and bottom points. 2. In one turn, those inputs are \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). 3. Including negative coterminal inputs in the stated interval gives \(-\frac{3\pi}{2}\) and \(-\frac{\pi}{2}\). 4. Thus the complete set in \([-2\pi,2\pi]\) is \(-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\).

\(t=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\)

- Use the quarter-turn benchmark positions shown on the unit circle. - Relate each point's x-coordinate to the denominator in the tangent ratio. - Look at the spacing between successive y-axis terminal points.

1. The upper y-axis point is reached at \(t=\frac{\pi}{2}\), and the lower y-axis point is reached at \(t=\frac{3\pi}{2}\). 2. At both points the x-coordinate, and therefore \(\cos t\), is \(0\). 3. Since \(\tan t=\frac{\sin t}{\cos t}\), tangent is undefined at both inputs. 4. These inputs repeat every \(\pi\), so tangent is undefined when \(t=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb{Z}\).

a) \(P\): \(\frac{\pi}{2}\); \(Q\): \(\frac{3\pi}{2}\) b) At both points, \(\cos t=0\), so \(\tan t=\frac{\sin t}{\cos t}\) is undefined. c) \(t=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb{Z}\)

- Write tangent as a ratio of the y-coordinate to the x-coordinate at each point. - Compare what a half-turn does to both coordinate signs. - Consider when the coordinate ratio itself is not defined.

1. Since \(P(t)=(a,b)\), \(\tan t=\frac{b}{a}\). 2. At \(t+\pi\), both coordinates change sign, so \(\tan(t+\pi)=\frac{-b}{-a}=\frac{b}{a}=\tan t\). 3. Therefore, tangent repeats after \(\pi\) radians, so its period is \(\pi\). 4. The condition \(a\ne0\) is necessary because tangent uses the ratio \(\frac{y}{x}\); if \(a=0\), the ratio is undefined.

a) \(\tan(t+\pi)=\frac{-b}{-a}=\frac{b}{a}=\tan t\) b) The period of tangent is \(\pi\). c) If \(a=0\), the coordinate ratio \(\frac{b}{a}\) is undefined.

- How does the height of the curve relate to the point's position above or below the x-axis? - Which graph value gives the point's vertical coordinate at the named input? - How does a point on the unit circle relate its two coordinates to the radius? - Check whether a point exactly on the axis counts as below it.

1. The curve is below zero for \(-\pi<t<-\frac{\pi}{6}\) and for \(\frac{5\pi}{6}<t<\pi\). Those are the times the point lies below the x-axis. 2. The graph reaches \(s(t)=-1\) at \(t=-\frac{2\pi}{3}\). On the unit circle, \(x^2+(-1)^2=1\) gives \(x=0\), so the point is \((0,-1)\).

a) \(-\pi<t<-\frac{\pi}{6}\) or \(\frac{5\pi}{6}<t<\pi\) b) \((0,-1)\); the graph gives y-coordinate \(-1\), so the unit-circle x-coordinate must be \(0\).

- Where does a half-turn place the point relative to the center? - What ratio of coordinates gives tangent? - Write both tangent fractions before deciding whether the values match.

1. A half-turn reverses both coordinates, so \(Q=\left(\frac{7}{25},-\frac{24}{25}\right)\). 2. At \(P\), \(\tan t=\frac{24/25}{-7/25}=-\frac{24}{7}\). 3. At \(Q\), \(\tan(t+\pi)=\frac{-24/25}{7/25}=-\frac{24}{7}\). Reversing both signs leaves the ratio unchanged.

\(Q=\left(\frac{7}{25},-\frac{24}{25}\right)\); \(\tan t=\tan(t+\pi)=-\frac{24}{7}\). Both coordinate signs change, so their ratio stays the same.

- Which quadrants could contain a point with this positive sine value? - What are the possible x-coordinates if the y-coordinate is fixed on the unit circle? - Compare the complete ordered pairs, not just their y-coordinates.

1. Sine is positive in Quadrants I and II. Its reference angle for \(\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{3}\). 2. The inputs are \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\). 3. Their points are \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\) and \(\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\), so they are not coterminal.

\(t=\frac{\pi}{3}\) gives \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\); \(t=\frac{2\pi}{3}\) gives \(\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\). The points differ because their x-coordinates differ.

- How could one specific input disprove a claim about every input? - Compare where the terminal point lies after half a revolution and after a full revolution. - Check both coordinates when stating a rule that works for every input.

1. At \(t=0\), \(\cos 0=1\), but \(\cos\pi=-1\). The claim fails. 2. A half-turn sends \((\cos t,\sin t)\) to the opposite point \((-\cos t,-\sin t)\). 3. Therefore, \(\cos(t+\pi)=-\cos t\) and \(\sin(t+\pi)=-\sin t\).

\(\cos 0=1\ne-1=\cos\pi\). In general, \(\cos(t+\pi)=-\cos t\) and \(\sin(t+\pi)=-\sin t\).