Free Math Worksheets

- What is the sum of the two acute angles in a right triangle? - Compare the side used as the opposite leg for one angle with the side used as the adjacent leg for the other. - Which identity relates tangent to sine and cosine?

1. The other acute angle is \(90^\circ-25^\circ=65^\circ\). The side opposite the \(25^\circ\) angle is the side adjacent to the \(65^\circ\) angle, and both ratios use the same hypotenuse. Therefore, \(\sin(25^\circ)=\cos(65^\circ)\). 2. Use \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\). 3. Then \(\tan(25^\circ)\approx\frac{0.4226}{0.9063}\approx 0.4663\).

1. The side opposite \(25^\circ\) is the side adjacent to \(65^\circ\), and both ratios use the same hypotenuse. 2. \(\tan(25^\circ)\approx 0.4663\)

- Label the sides of one right triangle relative to both acute angles. - How many degrees do the two acute angles total? - Use the exact sine and cosine values for \(30^\circ\) and \(60^\circ\).

1. The leg opposite \(\alpha\) is the same leg that is adjacent to \(\beta\). Both sine and cosine ratios use the same hypotenuse, so \(\sin(\alpha)=\cos(\beta)\). 2. The two acute angles of a right triangle are complementary, so \(\alpha+\beta=90^\circ\) and \(\beta=90^\circ-\alpha\). 3. For \(\alpha=30^\circ\), \(\sin(30^\circ)=\frac{1}{2}\) and \(\cos(60^\circ)=\frac{1}{2}\). Therefore, the identity is true in this example.

1. The leg opposite \(\alpha\) is adjacent to \(\beta\), so the two ratios are equal. 2. \(\beta=90^\circ-\alpha\) 3. \(\sin(30^\circ)=\cos(60^\circ)=\frac{1}{2}\)

- Express each angle using \(40^\circ\) and a benchmark angle such as \(180^\circ\) or \(360^\circ\). - Identify the reflection or rotation relating each point to the point for \(40^\circ\). - Determine whether the y-coordinate changes sign.

1. Since \(140^\circ=180^\circ-40^\circ\), \(\sin(140^\circ)=\sin(40^\circ) \approx 0.643\). 2. Since \(220^\circ=180^\circ+40^\circ\), \(\sin(220^\circ)=-\sin(40^\circ) \approx -0.643\). 3. Since \(320^\circ=360^\circ-40^\circ\), \(\sin(320^\circ)=-\sin(40^\circ) \approx -0.643\).

a) \(0.643\) b) \(-0.643\) c) \(-0.643\)

- Use unit-circle symmetries to compare the values. - Identify each angle’s quadrant and the signs of sine and cosine there. - Consider the cofunction identity and reflection symmetry. - Use the positions of the corresponding unit-circle points to check your reasoning.

1. For a), \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\) and \(\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\). The equation is true. It also follows from the cofunction identity \(\sin(x)=\cos\left(\frac{\pi}{2}-x\right)\). 2. For b), \(\frac{3\pi}{4}\) is in Quadrant II, where sine is positive and cosine is negative. In fact, \(\sin\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2}\) while \(\cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}\). The equation is false. 3. For c), the angles \(\frac{5\pi}{3}\) and \(\frac{\pi}{3}\) are reflections across the x-axis on the unit circle, so they have the same x-coordinate. Thus both cosine values equal \(\frac{1}{2}\), and the equation is true.

a) True; both sides equal \(\frac{\sqrt{3}}{2}\). b) False; \(\frac{\sqrt{2}}{2}\ne-\frac{\sqrt{2}}{2}\). c) True; both sides equal \(\frac{1}{2}\).

- Use degree mode when evaluating the functions. - Compare pairs of angles that add to \(90^\circ\). - Look for repeated decimal values in opposite rows.

1. The missing values are \(\sin(12^\circ)\approx0.21\), \(\cos(28^\circ)\approx0.88\), \(\sin(45^\circ)\approx0.71\), \(\cos(62^\circ)\approx0.47\), and \(\sin(78^\circ)\approx0.98\). 2. The completed sine row is \(0.21, 0.47, 0.71, 0.88, 0.98\), while the completed cosine row is \(0.98, 0.88, 0.71, 0.47, 0.21\). 3. The rows reverse because \(\sin(\alpha)=\cos(90^\circ-\alpha)\). For example, \(\sin(12^\circ)=\cos(78^\circ)\), and \(\sin(28^\circ)=\cos(62^\circ)\). At \(45^\circ\), sine and cosine are equal.

Missing values: \(\sin(12^\circ)\approx0.21\), \(\cos(28^\circ)\approx0.88\), \(\sin(45^\circ)\approx0.71\), \(\cos(62^\circ)\approx0.47\), and \(\sin(78^\circ)\approx0.98\) Pattern: The values reverse because \(\sin(\alpha)=\cos(90^\circ-\alpha)\).

- Use the cofunction identity relating sine and cosine. - Simplify negative cosine angles using the even symmetry of cosine. - Several equivalent cosine expressions may be possible.

1. Use \(\sin(\theta)=\cos(90^\circ-\theta)\). 2. For a), \(\sin(18^\circ)=\cos(72^\circ)\). 3. For b), \(\sin(155^\circ)=\cos(-65^\circ)=\cos(65^\circ)\). 4. For c), \(\sin(-40^\circ)=\cos(130^\circ)\). 5. For d), \(\sin(245^\circ)=\cos(-155^\circ)=\cos(155^\circ)\).

a) \(\cos(72^\circ)\) b) \(\cos(65^\circ)\) c) \(\cos(130^\circ)\) d) \(\cos(155^\circ)\)

- Use the cofunction identity relating cosine and sine. - Simplify negative cosine angles using the even symmetry of cosine. - Check the sign of each value from its quadrant.

1. Use \(\cos(\theta)=\sin(90^\circ-\theta)\). 2. For a), \(\cos(77^\circ)=\sin(13^\circ)\). 3. For b), \(\cos(112^\circ)=\sin(-22^\circ)\). 4. For c), \(\cos(-15^\circ)=\cos(15^\circ)=\sin(75^\circ)\). 5. For d), \(\cos(195^\circ)=\sin(-105^\circ)\).

a) \(\sin(13^\circ)\) b) \(\sin(-22^\circ)\) c) \(\sin(75^\circ)\) d) \(\sin(-105^\circ)\)

- Use the period \(2\pi\) of sine and cosine. - Recall the phase-shift relationships between sine and cosine. - Check the signs in the identities involving shifts by \(\frac{\pi}{2}\).

1. For part b, \(\sin(2(x+\pi))=\sin(2x+2\pi)=\sin(2x)\) by periodicity. 2. For part c, use \(\cos\left(\alpha-\frac{\pi}{2}\right)=\sin(\alpha)\). Thus, \(\cos\left(2x-\frac{\pi}{2}\right)=\sin(2x)\). 3. For part d, \(\cos\left(\alpha+\frac{\pi}{2}\right)=-\sin(\alpha)\). Thus, \(\cos\left(2x+\frac{\pi}{2}\right)=-\sin(2x)\), which is different.

d) \(y=\cos\left(2x+\frac{\pi}{2}\right)\)

- Interpret tangent as a ratio of the legs. - Use the Pythagorean theorem to find the hypotenuse. - Recall the cofunction relationships for complementary angles.

1. Use proportional leg lengths \(15\) and \(8\). The hypotenuse is \(\sqrt{15^2+8^2}=17\). 2. Therefore, \(\sin(\beta)=\frac{15}{17}\) and \(\cos(\beta)=\frac{8}{17}\). 3. By the cofunction identities, \(\sin(90^\circ-\beta)=\cos(\beta)=\frac{8}{17}\) and \(\cos(90^\circ-\beta)=\sin(\beta)=\frac{15}{17}\).

\(\sin(\beta)=\frac{15}{17}\) \(\cos(\beta)=\frac{8}{17}\) \(\sin(90^\circ-\beta)=\frac{8}{17}\) \(\cos(90^\circ-\beta)=\frac{15}{17}\)

- Replace the trigonometric functions of the complementary angles. - Look for the Pythagorean identity after substitution. - The cofunction relationships come from the complementary acute angles in a right triangle.

1. Use the cofunction identities: \(\sin(90^\circ-\beta)=\cos(\beta)\) and \(\cos(90^\circ-\beta)=\sin(\beta)\). 2. The expression becomes \(\cos^2(\beta)+\sin^2(\beta)\). 3. By the Pythagorean identity, the result is \(1\).

\(1\)

- View the two legs from each acute angle. - What happens when a nonzero fraction is multiplied by its reciprocal? - How are \(22.5^\circ\) and \(67.5^\circ\) related?

1. If the legs are \(a\) and \(b\), then \(\tan(\alpha)=\frac{a}{b}\) and \(\tan(\beta)=\frac{b}{a}\). 2. Their product is \(\frac{a}{b}\cdot\frac{b}{a}=1\). 3. Since \(22.5^\circ\) and \(67.5^\circ\) are complementary, their tangent values are reciprocals. 4. Thus, \(\tan(67.5^\circ)=\frac{1}{0.4142}\approx 2.4143\).

a) \(\tan(\alpha)\tan(\beta)=\frac{a}{b}\cdot\frac{b}{a}=1\) b) \(\tan(67.5^\circ)\approx 2.4143\)

- Picture one angle measured counterclockwise from the positive x-axis and the other just below that axis. - Which axis is the line of symmetry between the two points? - How do the coordinates of reflected points compare?

1. The point for \(\alpha\) has coordinates \((\cos(\alpha), \sin(\alpha))\). 2. The angle \(360^\circ-\alpha\) produces the reflection of this point across the x-axis. 3. Reflection across the x-axis keeps the x-coordinate and reverses the sign of the y-coordinate. 4. Therefore, \(\cos(360^\circ-\alpha)=\cos(\alpha)\). 5. Also, \(\sin(360^\circ-\alpha)=-\sin(\alpha)\).

a) \(\sin(360^\circ-\alpha)=-\sin(\alpha)\) b) \(\cos(360^\circ-\alpha)=\cos(\alpha)\) c) The two unit-circle points are reflections across the x-axis, so their x-coordinates are equal and their y-coordinates are opposites.

- Use \(\sin(180^\circ-\theta)=\sin(\theta)\). - Recall the exact values for \(30^\circ\), \(45^\circ\), and \(60^\circ\). - Find the Quadrant III angle with the same x-coordinate as the point for \(120^\circ\). - When does a point have equal x- and y-coordinates?

1. \(\sin(150^\circ)=\sin(30^\circ)=\frac{1}{2}\), and \(\cos(60^\circ)=\frac{1}{2}\). Therefore, \(x=1\). 2. The cosine of \(120^\circ\) is \(-\frac{1}{2}\). In Quadrant III, the angle with reference angle \(60^\circ\) is \(180^\circ+60^\circ=240^\circ\), so \(\beta=240^\circ\). 3. On the unit circle, sine and cosine are the y- and x-coordinates. Equality requires \(y=x\). In Quadrant I, the line \(y=x\) meets the unit circle at the point corresponding to \(45^\circ\).

a) \(x=1\) b) \(\beta=240^\circ\) c) Equality requires the unit-circle point to lie on \(y=x\), which occurs in Quadrant I only at \(45^\circ\).

- Use the cofunction relationship between sine and cosine of complementary angles. - Express \(170^\circ\) and \(190^\circ\) in terms of \(10^\circ\) and \(180^\circ\). - Simplify \(\cos(90^\circ-\gamma)\) before solving.

1. By the cofunction identity, \(\cos(80^\circ)=\sin(10^\circ) \approx 0.1736\). 2. Since \(170^\circ=180^\circ-10^\circ\), \(\sin(170^\circ)=\sin(10^\circ) \approx 0.1736\). 3. Since \(190^\circ=180^\circ+10^\circ\), \(\sin(190^\circ)=-\sin(10^\circ) \approx -0.1736\). 4. Rewrite \(\cos(90^\circ-\gamma)\) as \(\sin(\gamma)\). Thus, solve \(\sin(\gamma)=-\sin(10^\circ)\). 5. Using the reference angle \(10^\circ\), the solutions are \(180^\circ+10^\circ=190^\circ\) and \(360^\circ-10^\circ=350^\circ\).

a) \(\cos(80^\circ) \approx 0.1736\) b) \(\sin(170^\circ) \approx 0.1736\); \(\sin(190^\circ) \approx -0.1736\) c) \(\gamma=190^\circ\) or \(\gamma=350^\circ\)

- Use the cofunction relationships for complementary angles. - Rewrite each tangent as a quotient of sine and cosine.

1. By the cofunction identity, \(\cos(90^\circ-\delta)=\sin(\delta)\). Therefore, the expression in part a equals \(1\). 2. Also, \(\tan(\delta)=\frac{\sin(\delta)}{\cos(\delta)}\) and \(\tan(90^\circ-\delta)=\frac{\cos(\delta)}{\sin(\delta)}\). 3. Their product is \(\frac{\sin(\delta)}{\cos(\delta)}\cdot\frac{\cos(\delta)}{\sin(\delta)}=1\).

a) \(1\) b) The tangent values are reciprocals, so their product is \(1\).

- Use the value an odd function must have at \(x=0\). - Solve \(\cos(c)=0\) on the given interval. - Verify each candidate by rewriting the shifted cosine as a sine function.

1. An odd function that is defined at \(0\) must satisfy \(f(0)=0\). 2. Here, \(f(0)=\cos(c)\), so \(\cos(c)=0\). 3. On \([0, 2\pi]\), the solutions are \(c=\frac{\pi}{2}\) and \(c=\frac{3\pi}{2}\). 4. Verify: when \(c=\frac{\pi}{2}\), \(f(x)=-\sin(x)\), which is odd. When \(c=\frac{3\pi}{2}\), \(f(x)=\sin(x)\), which is also odd.

\(c\in\left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\}\)

- Consider the point for \(35^\circ\) on the unit circle. - Express each angle in terms of \(35^\circ\), \(180^\circ\), or \(360^\circ\). - Which unit-circle coordinate represents cosine? - Recall the period and even symmetry of cosine.

1. Since \(325^\circ=360^\circ-35^\circ\), \(\cos(325^\circ)=\cos(35^\circ) \approx 0.819\). 2. Since \(145^\circ=180^\circ-35^\circ\), \(\cos(145^\circ)=-\cos(35^\circ) \approx -0.819\). 3. Cosine is even, so \(\cos(-35^\circ)=\cos(35^\circ) \approx 0.819\). 4. Cosine has period \(360^\circ\), so \(\cos(395^\circ)=\cos(35^\circ) \approx 0.819\).

a) \(0.819\) b) \(-0.819\) c) \(0.819\) d) \(0.819\)

- Use unit-circle identities for shifts by \(90^\circ\), \(180^\circ\), and \(270^\circ\). - Track whether sine and cosine switch roles and whether the sign changes. - Testing a simple angle such as \(10^\circ\) can help you check a proposed grouping.

1. Rewrite each expression using unit-circle identities: - A: \(\sin(180^\circ+\alpha)=-\sin(\alpha)\) - B: \(\cos(270^\circ+\alpha)=\sin(\alpha)\) - C: \(\sin(360^\circ-\alpha)=-\sin(\alpha)\) - D: \(\cos(90^\circ-\alpha)=\sin(\alpha)\) - E: \(\sin(180^\circ-\alpha)=\sin(\alpha)\) - F: \(\cos(270^\circ-\alpha)=-\sin(\alpha)\) 2. Therefore, B, D, and E all equal \(\sin(\alpha)\), while A, C, and F all equal \(-\sin(\alpha)\).

Group 1: B, D, E Group 2: A, C, F

- Use symmetry identities for sine and cosine. - Recall how shifts by \(90^\circ\) and \(270^\circ\) relate sine and cosine. - A single counterexample, such as \(\alpha=0^\circ\), is enough to disprove an identity.

1. Both sides of equation 1 equal \(-\cos(\alpha)\), so it is an identity. 2. Both sides of equation 2 equal \(\cos(\alpha)\), so it is an identity. 3. Both sides of equation 3 equal \(-\sin(\alpha)\), so it is an identity. 4. Both sides of equation 4 equal \(-\cos(\alpha)\), so it is an identity. 5. The left side of equation 5 is \(\cos(\alpha)\), not \(-\cos(\alpha)\) for every angle, so equation 5 is not an identity.

Equations 1, 2, 3, and 4 are identities. Equation 5 is not.

- Reflecting a unit-circle point across the x-axis keeps its x-coordinate unchanged. - Use \(\cos(\theta)=\cos(-\theta)\). - Add or subtract \(360^\circ\) to place an angle in the required interval.

1. Cosine is even and has period \(360^\circ\), so reflecting an angle across the x-axis preserves its cosine value. 2. For a), \(\beta=360^\circ-130^\circ=230^\circ\). 3. For b), \(\beta=360^\circ-310^\circ=50^\circ\). 4. For c), \(\cos(-45^\circ)=\cos(45^\circ)\), so \(\beta=45^\circ\) works. The coterminal angle \(315^\circ\) also works.

a) \(\beta=230^\circ\) b) \(\beta=50^\circ\) c) \(\beta=45^\circ\) or \(\beta=315^\circ\)

- The amplitude and vertical shift do not change when converting between sine and cosine forms. - Recall the phase-shift identity relating cosine to sine. - Rewrite an addition inside the argument as subtraction of a negative value.

1. The amplitude and vertical shift remain \(a=2.5\) and \(d=-1\). 2. Use \(\cos(\alpha)=\sin\left(\alpha+\frac{\pi}{2}\right)\), with \(\alpha=x-\frac{\pi}{4}\). 3. Then \(f(x)=2.5\sin\left(x-\frac{\pi}{4}+\frac{\pi}{2}\right)-1=2.5\sin\left(x+\frac{\pi}{4}\right)-1\). 4. Since \(x+\frac{\pi}{4}=x-\left(-\frac{\pi}{4}\right)\), \(c=-\frac{\pi}{4}\).

\(g(x)=2.5\sin\left(x+\frac{\pi}{4}\right)-1\); \(a=2.5\), \(c=-\frac{\pi}{4}\), \(d=-1\)

- Replace the cosine expression with an equivalent shifted sine expression. - For the functions to match for every \(x\), their sine inputs can be matched modulo \(2\pi\). - Use a common denominator when subtracting fractions involving \(\pi\).

1. Apply the identity to the full argument: \(\cos(x-c)=\sin\left(x-c+\frac{\pi}{2}\right)\). 2. Match the sine arguments: \(x-c_2=x-c+\frac{\pi}{2}\). Therefore, \(c_2=c-\frac{\pi}{2}\). Equivalent values differ by integer multiples of \(2\pi\). 3. When \(c=\frac{3\pi}{4}\), \(c_2=\frac{3\pi}{4}-\frac{\pi}{2}=\frac{\pi}{4}\).

1. \(c_2=c-\frac{\pi}{2}\), up to integer multiples of \(2\pi\) 2. \(c_2=\frac{\pi}{4}\)

- How do the coordinates of a unit-circle point relate to sine and cosine? - Which reflection preserves the y-coordinate? - Which reflection preserves the x-coordinate?

1. A unit-circle point has coordinates \((\cos(\alpha), \sin(\alpha))\). Since \(\cos(115^\circ)\approx-0.42\) and \(\sin(115^\circ)\approx0.91\), \(P\approx(-0.42, 0.91)\). 2. Angles reflected across the y-axis have the same sine. Thus, \(\beta=180^\circ-115^\circ=65^\circ\). 3. Angles reflected across the x-axis have the same cosine. Thus, \(\gamma=360^\circ-115^\circ=245^\circ\).

a) \(P(-0.42, 0.91)\) b) \(\beta=65^\circ\) c) \(\gamma=245^\circ\)

- Which coordinate represents sine, and which represents cosine? - Express \(200^\circ\) and \(160^\circ\) using \(180^\circ\pm20^\circ\). - Track how reflections across an axis or through the origin change coordinates.

1. The angles \(200^\circ=180^\circ+20^\circ\) and \(160^\circ=180^\circ-20^\circ\) correspond to points reflected across the x-axis. Their x-coordinates are equal, so \(\cos(200^\circ)=\cos(160^\circ)\). 2. The point for \(200^\circ\) is the reflection of the point for \(20^\circ\) through the origin. This changes the sign of the y-coordinate, so \(\sin(200^\circ)=-\sin(20^\circ)\).

a) The unit-circle points are reflections across the x-axis, so their x-coordinates, and therefore their cosine values, are equal. b) The unit-circle points are opposites through the origin, so their y-coordinates, and therefore their sine values, have opposite signs.

- You may keep the outside coefficient negative. - Recall the phase shift that converts cosine to sine. - Factor the coefficient of \(x\) from the entire input.

1. Use \(\cos(u)=\sin\left(u+\frac{\pi}{2}\right)\): \(y=-2\sin\left(0.5x+\frac{3\pi}{2}\right)\). 2. Factor out \(0.5\): \(y=-2\sin\left(0.5(x+3\pi)\right)\). 3. By periodicity, an equivalent form is \(y=-2\sin\left(0.5(x-\pi)\right)\).

\(y=-2\sin\left(0.5(x+3\pi)\right)\); equivalently, \(y=-2\sin\left(0.5(x-\pi)\right)\)

- Identify each angle’s quadrant or complementary relationship. - Use identities involving angles that sum to \(90^\circ\) or \(180^\circ\). - Recall the even symmetry of cosine.

1. Since \(146^\circ=180^\circ-34^\circ\), \(\sin(146^\circ)=\sin(34^\circ)\approx0.56\). 2. Since \(159^\circ=180^\circ-21^\circ\), \(\cos(159^\circ)=-\cos(21^\circ)\approx-0.93\). 3. Since \(69^\circ=90^\circ-21^\circ\), \(\sin(69^\circ)=\cos(21^\circ)\approx0.93\). 4. Cosine is even, so \(\cos(-21^\circ)=\cos(21^\circ)\approx0.93\).

a) \(0.56\) b) \(-0.93\) c) \(0.93\) d) \(0.93\)

- Compare corresponding zeros or maximum points of the two graphs. - A shift right appears as subtraction inside the function input. - Shift the location of the parent sine function's first maximum by the same amount.

1. The graph of \(g\) crosses the x-axis with positive slope at \(x=\frac{\pi}{2}\), so the parent sine graph has been shifted right by \(\frac{\pi}{2}\). 2. Therefore, \(g(x)=\sin\left(x-\frac{\pi}{2}\right)\), which is also \(-\cos(x)\). 3. The first maximum of \(\sin(x)\) occurs at \(x=\frac{\pi}{2}\). After the shift, it occurs at \(x=\pi\), with y-coordinate \(1\).

a) \(g(x)=\sin\left(x-\frac{\pi}{2}\right)\), equivalently \(g(x)=-\cos(x)\) b) \((\pi, 1)\)

- Find each graph's midline, amplitude, and period. - Check for horizontal shifts and reflections. - Use identities that relate sine and cosine after a phase shift.

1. Graph (1) has midline \(y=1\), amplitude \(2\), period \(4\), and an increasing midline crossing at \(x=0\). Thus, equation A matches. 2. In equation B, \(0.5\pi(x-1)=0.5\pi x-\frac{\pi}{2}\), and \(\cos\left(\theta-\frac{\pi}{2}\right)=\sin(\theta)\). Therefore, B is equivalent to A and also matches graph (1). 3. Graph (2) has the same midline, amplitude, and period but is shifted right \(1\) unit. Thus, equation C matches. 4. Since \(-\cos(\theta)=\sin\left(\theta-\frac{\pi}{2}\right)\), equation D is equivalent to C and also matches graph (2).

(1) A and B (2) C and D

- An odd function must pass through the origin. - Compare the graph with the parent sine function and its reflection. - Recall the phase-shift relationship between sine and cosine. - Express the minimum location near \(1.57\) as a multiple of \(\pi\).

1. A graph symmetric about the origin must be an odd function. The graph passes through the origin and decreases for small positive \(x\), so it matches \(-\sin(x)\). 2. Use \(\cos\left(x+\frac{\pi}{2}\right)=-\sin(x)\). 3. The first minimum of \(-\sin(x)\) occurs at \(x=\frac{\pi}{2}\), matching the graph. 4. Therefore, the least positive value is \(c=\frac{\pi}{2}\).

\(c=\frac{\pi}{2}\)

- Calculate the first two values and compare them. - Picture the two unit-circle points and identify their line of symmetry. - Which coordinate represents sine, and which represents cosine? - How does reflection across the y-axis change an ordered pair?

1. \(\sin(70^\circ) \approx 0.9397\) and \(\sin(110^\circ) \approx 0.9397\), so the values are equal. 2. The point for \(180^\circ-\alpha\) is the reflection of the point for \(\alpha\) across the y-axis. The y-coordinate does not change, so \(\sin(180^\circ-\alpha)=\sin(\alpha)\). 3. The x-coordinate changes sign under this reflection, so \(\cos(180^\circ-\alpha)=-\cos(\alpha)\). 4. For example, \(\cos(70^\circ) \approx 0.3420\), while \(\cos(110^\circ) \approx -0.3420\). Therefore, the proposed cosine equation is not true in general.

a) \(\sin(70^\circ)=\sin(110^\circ) \approx 0.9397\) b) \(\sin(180^\circ-\alpha)=\sin(\alpha)\) c) No. In general, \(\cos(180^\circ-\alpha)=-\cos(\alpha)\).