Free math worksheets

- Think about what an even function does when its input changes from \(x\) to \(-x\). - The rewritten expression should use the positive input \(x\).

1. Cosine is even, so \(\cos(-x)=\cos(x)\).

\(\cos(x)\)

- The two acute angles in a right triangle are complementary. - A cofunction identity changes the trigonometric function while replacing an angle by its complement.

1. By the cofunction identity, \(\sin(90^\circ-\theta)=\cos(\theta)\).

\(\cos(\theta)\)

- Think about how the y-coordinate of a unit-circle point changes when the angle is reflected from \(x\) to \(-x\). - Sine records the y-coordinate, so decide whether the value stays the same or changes sign.

1. Sine is odd, so \(\sin(-x)=-\sin(x)\).

\(-\sin(x)\)

- 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. \(\sin(30^\circ)=\cos(60^\circ)=\frac{1}{2}\)

- Compare the order of the two products with a sine angle-combination pattern. - Decide whether the minus sign between the products corresponds to a sum or a difference of angles. - After combining the angles, simplify the coefficient of \(t\).

1. Use \(\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\). 2. With \(A=5t\) and \(B=2t\), the expression is \(\sin(5t-2t)\). 3. Therefore, the expression simplifies to \(\sin(3t)\).

Identity: \(\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\); result: \(\sin(3t)\).

- 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 \(\sin(\alpha)=\cos(90^\circ-\alpha)\).

- Start with the cofunction identity relating sine to cosine of a complementary expression. - If that produces a negative cosine angle, use cosine's evenness. - Check that each final cosine angle lies in the required interval.

1. Use \(\sin(\theta)=\cos(90^\circ-\theta)\), together with cosine symmetry when needed. 2. \(\sin(18^\circ)=\cos(72^\circ)\). 3. \(\sin(155^\circ)=\cos(-65^\circ)=\cos(65^\circ)\). 4. \(\sin(-40^\circ)=\cos(130^\circ)\). 5. \(\sin(245^\circ)=\cos(-155^\circ)=\cos(155^\circ)\). 6. Cosine is one-to-one on \([0^\circ,180^\circ]\), so the stated range makes each requested angle unique.

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

- Start with the cofunction identity relating cosine to sine. - Use symmetry or a full-period shift if the first sine angle is outside the required interval. - Check every final angle against the stated canonical range.

1. Use \(\cos(\theta)=\sin(90^\circ-\theta)\), together with symmetry or periodicity when needed. 2. \(\cos(77^\circ)=\sin(13^\circ)\). 3. \(\cos(112^\circ)=\sin(-22^\circ)\). 4. \(\cos(-15^\circ)=\cos(15^\circ)=\sin(75^\circ)\). 5. \(\cos(195^\circ)=\sin(-105^\circ)=\sin(-75^\circ)\). 6. Sine is one-to-one on \([-90^\circ,90^\circ]\), so the stated range makes each requested angle unique.

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

- Replace \(x\) with \(-x\). - Recall which of sine and cosine changes sign when its input changes sign. - Compare the result with \(f(x)\) and \(-f(x)\).

1. Evaluate \(f(-x)=\sin(-x)\cos(-x)\). 2. Use \(\sin(-x)=-\sin(x)\) and \(\cos(-x)=\cos(x)\). 3. Then \(f(-x)=-\sin(x)\cos(x)=-f(x)\), so \(f\) is odd.

\(f(x)=\sin(x)\cos(x)\) is odd.

- Describe each labeled side once relative to \(\alpha\) and again relative to \(\beta\). - The same leg can be opposite one acute angle and adjacent to the other. - Compare the resulting side ratios before simplifying the tangent product.

1. Relative to \(\alpha\), the opposite side is \(a\), the adjacent leg is \(b\), and the hypotenuse is \(c\). Thus, \(\sin(\alpha)=\frac{a}{c}\), \(\cos(\alpha)=\frac{b}{c}\), and \(\tan(\alpha)=\frac{a}{b}\). 2. Relative to \(\beta\), the opposite side is \(b\), the adjacent leg is \(a\), and the hypotenuse is \(c\). Thus, \(\cos(\beta)=\frac{a}{c}\), \(\sin(\beta)=\frac{b}{c}\), and \(\tan(\beta)=\frac{b}{a}\). 3. Therefore, \(\sin(\alpha)=\cos(\beta)\) and \(\cos(\alpha)=\sin(\beta)\). 4. Also, \(\tan(\alpha)\tan(\beta)=\frac{a}{b}\cdot\frac{b}{a}=1\).

\(\sin(\alpha)=\cos(\beta)\), \(\cos(\alpha)=\sin(\beta)\), and \(\tan(\alpha)\tan(\beta)=1\)

- A horizontal reflection changes only one coordinate. - Match the reflected point to the unit-circle angle below the positive horizontal radius. - Compare the coordinate descriptions of \(P\) and \(Q\) component by component.

1. Reflection across the horizontal diameter keeps the horizontal coordinate unchanged and reverses the sign of the vertical coordinate. 2. Therefore, \(Q=(\cos(\alpha),-\sin(\alpha))\). 3. Point \(Q\) represents the angle \(-\alpha\), so its coordinates are also \((\cos(-\alpha),\sin(-\alpha))\). 4. Comparing coordinates gives \(\cos(-\alpha)=\cos(\alpha)\) and \(\sin(-\alpha)=-\sin(\alpha)\).

\(Q=(\cos(\alpha),-\sin(\alpha))\) \(\cos(-\alpha)=\cos(\alpha)\) \(\sin(-\alpha)=-\sin(\alpha)\)

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

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

- 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.

- Which identity relates sine and cosine of the same angle? - How do unit-circle coordinates change under reflection across the y-axis or x-axis? - In which quadrant does \(360^\circ-\alpha\) lie when \(\alpha\) is acute?

1. Use \(\sin^2(\alpha)+\cos^2(\alpha)=1\): \(0.28^2+\cos^2(\alpha)=1\). 2. Then \(\cos^2(\alpha)=1-0.0784=0.9216\). Because \(\alpha\) is acute, cosine is positive, so \(\cos(\alpha)=\sqrt{0.9216}=0.96\). 3. Supplementary angles have equal sine values, so \(\sin(180^\circ-\alpha)=\sin(\alpha)=0.28\). 4. The angles \(\alpha\) and \(360^\circ-\alpha\) have equal cosine values, so \(\cos(360^\circ-\alpha)=\cos(\alpha)=0.96\).

a) \(\cos(\alpha)=0.96\) b) \(\sin(180^\circ-\alpha)=0.28\) and \(\cos(360^\circ-\alpha)=0.96\)

- 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 \(\gamma=350^\circ\)

- 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, \(\cos(180^\circ-\alpha)=-\cos(\alpha)\).

- 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.

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

- Expand the shifted sine using an angle-addition identity. - Identify which resulting term is even and which is odd. - Set the coefficient of the odd term equal to \(0\).

1. An even function must satisfy \(f(-x)=f(x)\) for every \(x\). 2. Use the angle-addition identity: \(f(x)=\sin(c)\cos\left(\frac{x}{2}\right)+\cos(c)\sin\left(\frac{x}{2}\right)\). 3. The cosine term is even and the sine term is odd. For the entire function to be even, the coefficient of the odd term must be \(0\): \(\cos(c)=0\). 4. On \([0, 2\pi]\), this gives \(c=\frac{\pi}{2}\) or \(c=\frac{3\pi}{2}\). 5. These choices produce \(f(x)=\cos\left(\frac{x}{2}\right)\) and \(f(x)=-\cos\left(\frac{x}{2}\right)\), respectively, and both are even.

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

- 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.

- Use the Pythagorean identity to relate sine and cosine. - Because the angle is acute, choose the positive square root. - Sine and cosine interchange for complementary angles.

1. Substitute \(\sin(\alpha)=0.8\) into the Pythagorean identity: \((0.8)^2+\cos^2(\alpha)=1\). 2. Then \(\cos^2(\alpha)=1-0.64=0.36\). 3. Because \(\alpha\) is acute, cosine is positive, so \(\cos(\alpha)=\sqrt{0.36}=0.6\). 4. Since \(\cos(\beta)=0.8\), \(\beta=\cos^{-1}(0.8)\approx36.87^\circ\). 5. Also, \(\sin(\alpha)=\cos(90^\circ-\alpha)\), so \(\beta=90^\circ-\alpha\). The two angles are complementary.

a) \(\cos(\alpha)=0.6\) b) \(\beta\approx36.87^\circ\), and \(\alpha+\beta=90^\circ\).

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

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

- What exact input change appears in the even-function identity for cosine? - Compare \(-x\) with \(\pi-x\). - You can check the sign by thinking about the unit-circle reflection across the y-axis.

1. The even-function identity applies to \(\cos(-x)\), not to \(\cos(\pi-x)\). 2. Use the subtraction identity: \(\cos(\pi-x)=\cos(\pi)\cos(x)+\sin(\pi)\sin(x)\). 3. Since \(\cos(\pi)=-1\) and \(\sin(\pi)=0\), \(\cos(\pi-x)=-\cos(x)\).

Mateo applied evenness to the wrong transformed angle. The correct identity is \(\cos(\pi-x)=-\cos(x)\).

- Expand \(\cos(x-b)\) and \(\cos(-x-b)\). - For an identity to hold for every \(x\), the coefficients of the variable trigonometric terms must satisfy the required conditions. - Test odd symmetry by adding \(f(-x)\) and \(f(x)\).

1. Even symmetry requires \(f(-x)=f(x)\). Using angle identities, \(f(-x)-f(x)=-2\sin(b)\sin(x)\). 2. This expression is \(0\) for every \(x\) exactly when \(\sin(b)=0\), so \(b=k\pi\) for some integer \(k\). The vertical shift \(a\) may be any nonzero real number. 3. Odd symmetry would require \(f(-x)+f(x)=0\) for every \(x\). But \(f(-x)+f(x)=2a+2\cos(b)\cos(x)\). 4. For this expression to vanish for every \(x\), both \(a=0\) and \(\cos(b)=0\) would be necessary. Since \(a\ne0\), \(f\) cannot be odd.

Even: \((a, k\pi)\), where \(a\in\mathbb{R}\setminus\{0\}\) and \(k\in\mathbb{Z}\) Odd: impossible when \(a\ne0\)

- An even graph has a maximum or minimum on the y-axis. - Identify whether the graph matches a cosine function or its reflection. - Read the value of the function at \(x=0\). - For part b, use an angle-subtraction formula and separate the even and odd components.

1. The graph has a minimum of \(-1\) at \(x=0\), so it is \(-\cos(2x)\). 2. Since \(\sin\left(2x-\frac{\pi}{2}\right)=-\cos(2x)\), set \(2(x-c)=2x-\frac{\pi}{2}\). This gives \(c=\frac{\pi}{4}\). 3. In general, use the subtraction formula: \(\sin(b(x-c))=\sin(bx)\cos(bc)-\cos(bx)\sin(bc)\). 4. The function is even exactly when the coefficient of the odd term \(\sin(bx)\) is zero. Thus, \(\cos(bc)=0\), so \(bc=\frac{\pi}{2}+k\pi\), where \(k\) is an integer. 5. Therefore, \(c=\frac{\pi/2+k\pi}{b}\).

a) \(c=\frac{\pi}{4}\) b) \(bc=\frac{\pi}{2}+k\pi\), so \(c=\frac{\pi/2+k\pi}{b}\), where \(k \in \mathbb{Z}\)

- Use the first condition to determine the parity of \(f\). - Think about which linear combination of an odd function and an even function can itself be odd for every \(x\). - After using the first condition, apply the cofunction identity to the second condition.

1. The condition \(f(-x)=-f(x)\) says that \(f\) is odd. 2. Since \(\sin(x)\) is odd and \(\cos(x)\) is even, the cosine term must vanish, so \(b=0\). 3. Then \(f\left(\frac{\pi}{2}-x\right)=a\sin\left(\frac{\pi}{2}-x\right)=a\cos(x)\). 4. Because this equals \(2\cos(x)\) for every \(x\), \(a=2\).

\(a=2\) and \(b=0\)