AP Precalculus Quiz: Equivalent Representations Of Trigonometric Functions
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Equivalent Representations Of Trigonometric FunctionsQuestion 1 of 20

To solve the equation 2sin⁡2(x)+3cos⁡(x)−3=02\sin^2(x) + 3\cos(x) - 3 = 0, it is useful to first rewrite it as an equation involving a single trigonometric function. Which of the following equations is an equivalent form that achieves this?

2cos⁡2(x)−3cos⁡(x)+1=02\cos^2(x) - 3\cos(x) + 1 = 0
2cos⁡2(x)+3cos⁡(x)−5=02\cos^2(x) + 3\cos(x) - 5 = 0
2sin⁡2(x)−3sin⁡(x)−1=02\sin^2(x) - 3\sin(x) - 1 = 0
−2sin⁡2(x)+3sin⁡(x)+1=0-2\sin^2(x) + 3\sin(x) + 1 = 0
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AP Precalculus Quiz

AP Precalculus Quiz: Equivalent Representations Of Trigonometric Functions

Practice Equivalent Representations Of Trigonometric Functions in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Equivalent Representations Of Trigonometric Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

To solve the equation 2sin⁡2(x)+3cos⁡(x)−3=02\sin^2(x) + 3\cos(x) - 3 = 0, it is useful to first rewrite it as an equation involving a single trigonometric function. Which of the following equations is an equivalent form that achieves this?

  1. 2cos⁡2(x)−3cos⁡(x)+1=02\cos^2(x) - 3\cos(x) + 1 = 0 (correct answer)
  2. 2cos⁡2(x)+3cos⁡(x)−5=02\cos^2(x) + 3\cos(x) - 5 = 0
  3. 2sin⁡2(x)−3sin⁡(x)−1=02\sin^2(x) - 3\sin(x) - 1 = 0
  4. −2sin⁡2(x)+3sin⁡(x)+1=0-2\sin^2(x) + 3\sin(x) + 1 = 0
Explanation: Using the Pythagorean identity sin⁡2(x)=1−cos⁡2(x)\sin^2(x) = 1 - \cos^2(x), substitute this into the original equation: 2(1−cos⁡2(x))+3cos⁡(x)−3=02(1 - \cos^2(x)) + 3\cos(x) - 3 = 0. This simplifies to 2−2cos⁡2(x)+3cos⁡(x)−3=02 - 2\cos^2(x) + 3\cos(x) - 3 = 0, or −2cos⁡2(x)+3cos⁡(x)−1=0-2\cos^2(x) + 3\cos(x) - 1 = 0. Multiplying the entire equation by -1 yields 2cos⁡2(x)−3cos⁡(x)+1=02\cos^2(x) - 3\cos(x) + 1 = 0.

Question 2

A spring's position is modeled by f(t)=10sin⁡ ⁣(2t+π3)f(t)=10\sin\!\left(2t+\frac{\pi}{3}\right) (centimeters), where tt is seconds. Letting t=θt=\theta gives the polar form r=10sin⁡ ⁣(2θ+π3)r=10\sin\!\left(2\theta+\frac{\pi}{3}\right). On the coordinate plane, the sinusoid has amplitude 1010, period π\pi, and its first maximum occurs at t=π12t=\frac{\pi}{12}. This periodic motion matches simple harmonic oscillation. Based on the description, how does the phase shift change when converting the function to polar form?

  1. It becomes a vertical shift of π3\frac{\pi}{3}
  2. It changes from π3\frac{\pi}{3} to π6\frac{\pi}{6}
  3. It remains the same phase shift, since only tt is renamed θ\theta (correct answer)
  4. It becomes a period change from π\pi to 2π2\pi
  5. It disappears because polar form cannot include phase shifts
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically how phase shifts behave when converting to polar form. When converting f(t) = 10sin(2t + π/3) to polar form by substituting t = θ, the result is r = 10sin(2θ + π/3). The phase shift, represented by the constant π/3 inside the sine function, remains unchanged because we're simply renaming the variable from t to θ. Choice C correctly identifies that the phase shift remains the same since only t is renamed θ. Choices A, B, D, and E incorrectly suggest the phase shift transforms into other parameters or disappears, misunderstanding that polar conversion is merely a coordinate system change. To help students: Emphasize that converting to polar form is a relabeling process, not a mathematical transformation. Practice identifying which parameters change (variable names) versus which remain constant (amplitude, frequency, phase) during conversion.

Question 3

A vibration sensor reads f(x)=3sin⁡ ⁣(x2−π4)f(x)=3\sin\!\left(\frac{x}{2}-\frac{\pi}{4}\right) (volts), where xx is time in seconds. Converting by setting x=θx=\theta gives r=3sin⁡ ⁣(θ2−π4)r=3\sin\!\left(\frac{\theta}{2}-\frac{\pi}{4}\right). On the coordinate plane, the sinusoid has amplitude 33, period 4π4\pi, and crosses the midline at x=π2x=\frac{\pi}{2}. This periodic signal can represent a steady machine hum. Based on the description, what is the amplitude of the function given in polar form?

  1. 32\frac{3}{2}
  2. 33 (correct answer)
  3. 4π4\pi
  4. π4\frac{\pi}{4}
  5. 12\frac{1}{2}
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically identifying amplitude in polar form. The function f(x) = 3sin(x/2 - π/4) has an amplitude of 3, which is the coefficient multiplying the entire sine function. When converted to polar form r = 3sin(θ/2 - π/4), the amplitude remains 3 because it represents the maximum distance from the midline. Choice B correctly identifies the amplitude as 3. Choices A and E confuse amplitude with other parameters, while C and D incorrectly incorporate π or the frequency coefficient. To help students: Emphasize that amplitude is always the positive coefficient in front of the sine or cosine function. Practice identifying amplitude in various forms and stress that it remains constant through coordinate transformations.

Question 4

A tide height model is f(t)=5sin⁡ ⁣(π6t)+2f(t)=5\sin\!\left(\frac{\pi}{6}t\right)+2 (meters), with tt in hours. Using t=θt=\theta gives the polar form r=5sin⁡ ⁣(π6θ)+2r=5\sin\!\left(\frac{\pi}{6}\theta\right)+2. On a coordinate plane, the midline is y=2y=2 and peaks reach 77 meters. This sinusoid represents periodic ocean tides. Refer to the passage above. Identify the period of the trigonometric function in the graphical representation.

  1. 1212 (correct answer)
  2. 66
  3. π6\frac{\pi}{6}
  4. 12π\frac{12}{\pi}
  5. 2π2\pi
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically calculating the period from a given function. The function f(t) = 5sin(π/6·t) + 2 has a coefficient of π/6 multiplying t inside the sine function. Using the period formula P = 2π/B where B = π/6, we get P = 2π/(π/6) = 12 hours. Choice A correctly identifies the period as 12. The vertical shift of +2 affects the midline but not the period, and choices B through E represent common calculation errors or confusion between period and other function parameters. To help students: Reinforce that vertical shifts don't affect period, only horizontal behavior does. Practice extracting the coefficient of the variable from various function forms and applying the period formula systematically.

Question 5

A rotating fan's vibration is f(t)=9sin⁡ ⁣(3π2t)f(t)=9\sin\!\left(\frac{3\pi}{2}t\right) (millimeters), where tt is seconds. Using t=θt=\theta gives the polar form r=9sin⁡ ⁣(3π2θ)r=9\sin\!\left(\frac{3\pi}{2}\theta\right). On the coordinate plane, the graph has amplitude 99 and repeats every 43\frac{4}{3} seconds. This periodic model matches steady mechanical vibration. Based on the description, what is the amplitude of the function given in polar form?

  1. 3π2\frac{3\pi}{2}
  2. 43\frac{4}{3}
  3. 99 (correct answer)
  4. 23π\frac{2}{3\pi}
  5. 1818
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically identifying amplitude from a polar representation. The function f(t) = 9sin(3π/2·t) has an amplitude of 9, which is the coefficient in front of the sine function. When converted to polar form r = 9sin(3π/2·θ), the amplitude remains 9 as it represents the maximum radial distance. Choice C correctly identifies the amplitude as 9. The other choices incorrectly involve the frequency coefficient 3π/2 or attempt calculations with it, demonstrating confusion between amplitude and other function parameters. To help students: Stress that amplitude is always the positive coefficient multiplying the entire trigonometric function. Practice identifying amplitude across different representations and emphasize it's independent of frequency or period.

Question 6

A metronome's side-to-side position is f(t)=2sin⁡ ⁣(πt+π2)f(t)=2\sin\!\left(\pi t+\frac{\pi}{2}\right) (centimeters), with tt in seconds. Letting t=θt=\theta gives r=2sin⁡ ⁣(πθ+π2)r=2\sin\!\left(\pi\theta+\frac{\pi}{2}\right). On the coordinate plane, the sinusoid has amplitude 22, period 22, and starts at a maximum when t=0t=0. This matches a steady beat. Based on the description, identify the period of the trigonometric function in the graphical representation.

  1. π\pi
  2. 22 (correct answer)
  3. 12\frac{1}{2}
  4. 2π2\pi
  5. 2π\frac{2}{\pi}
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically determining period from a trigonometric equation. The function f(t) = 2sin(πt + π/2) has π as the coefficient of t, giving a period of 2π/π = 2 seconds using the formula P = 2π/B. The passage explicitly confirms the period is 2, making Choice B correct. Choice A confuses the coefficient π with the period, while choices C through E represent various calculation errors involving π. To help students: Reinforce the period formula P = 2π/B and practice extracting B from functions where it's multiplied by π. Use graphical representations to verify calculated periods match the visual cycle length.

Question 7

If cos⁡(2θ)=725\cos(2\theta) = \frac{7}{25} and angle θ\theta is in Quadrant I, what is the value of cos⁡(θ)\cos(\theta)?

  1. 35\frac{3}{5}
  2. 45\frac{4}{5} (correct answer)
  3. 1625\frac{16}{25}
  4. 925\frac{9}{25}
Explanation: Use the double-angle identity cos⁡(2θ)=2cos⁡2(θ)−1\cos(2\theta) = 2\cos^2(\theta) - 1. Substitute the given value: 725=2cos⁡2(θ)−1\frac{7}{25} = 2\cos^2(\theta) - 1. Add 1 to both sides: 1+725=3225=2cos⁡2(θ)1 + \frac{7}{25} = \frac{32}{25} = 2\cos^2(\theta). Divide by 2: cos⁡2(θ)=1625\cos^2(\theta) = \frac{16}{25}. Take the square root: cos⁡(θ)=±45\cos(\theta) = \pm\frac{4}{5}. Because θ\theta is in Quadrant I, cos⁡(θ)\cos(\theta) must be positive, so cos⁡(θ)=45\cos(\theta) = \frac{4}{5}.

Question 8

Which of the following expressions is equivalent to sin⁡(2x)2sin⁡(x)\frac{\sin(2x)}{2\sin(x)} for all values of xx for which the expression is defined?

  1. cos⁡(x)\cos(x) (correct answer)
  2. sin⁡(x)\sin(x)
  3. tan⁡(x)\tan(x)
  4. 11
Explanation: Using the double-angle identity for sine, sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x). Substituting this into the numerator of the given expression yields 2sin⁡(x)cos⁡(x)2sin⁡(x)\frac{2\sin(x)\cos(x)}{2\sin(x)}. For values of xx where sin⁡(x)≠0\sin(x) \neq 0, the term 2sin⁡(x)2\sin(x) can be canceled from the numerator and denominator, leaving cos⁡(x)\cos(x).

Question 9

The expression 1−cos⁡(2α)sin⁡(2α)\frac{1-\cos(2\alpha)}{\sin(2\alpha)} is equivalent to which of the following for all values of α\alpha for which the expression is defined?

  1. sin⁡(α)\sin(\alpha)
  2. cos⁡(α)\cos(\alpha)
  3. tan⁡(α)\tan(\alpha) (correct answer)
  4. cot⁡(α)\cot(\alpha)
Explanation: Use the double-angle identities cos⁡(2α)=1−2sin⁡2(α)\cos(2\alpha) = 1 - 2\sin^2(\alpha) and sin⁡(2α)=2sin⁡(α)cos⁡(α)\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha). Substitute these into the expression. The numerator becomes 1−(1−2sin⁡2(α))=2sin⁡2(α)1 - (1 - 2\sin^2(\alpha)) = 2\sin^2(\alpha). The expression becomes 2sin⁡2(α)2sin⁡(α)cos⁡(α)\frac{2\sin^2(\alpha)}{2\sin(\alpha)\cos(\alpha)}. After canceling the common factor of 2sin⁡(α)2\sin(\alpha), the expression simplifies to sin⁡(α)cos⁡(α)\frac{\sin(\alpha)}{\cos(\alpha)}, which is equal to tan⁡(α)\tan(\alpha).

Question 10

A sound wave is modeled by f(x)=7sin⁡ ⁣(4x−2π)f(x)=7\sin\!\left(4x-2\pi\right) (arbitrary units), where xx is time in seconds. Converting by setting x=θx=\theta gives r=7sin⁡ ⁣(4θ−2π)r=7\sin\!\left(4\theta-2\pi\right). On the coordinate plane, the sinusoid has amplitude 77, period π2\frac{\pi}{2}, and the phase shift is π2\frac{\pi}{2} to the right. This periodic model matches a stable tone. Refer to the passage above. How does the phase shift change when converting the function to polar form?

  1. It becomes a vertical shift of 2π2\pi
  2. It changes to a left shift of π2\frac{\pi}{2}
  3. It stays the same, since xx is simply relabeled as θ\theta (correct answer)
  4. It doubles because polar angles are measured differently
  5. It becomes the amplitude because −2π-2\pi is outside the sine
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically phase shift behavior in polar conversion. The function f(x) = 7sin(4x - 2π) has a phase shift that can be found by solving 4x - 2π = 0, giving x = π/2 (shift right). When converting to polar form r = 7sin(4θ - 2π), the phase shift calculation remains identical: 4θ - 2π = 0 gives θ = π/2. Choice C correctly states the phase shift stays the same since x is simply relabeled as θ. Choices A, B, D, and E incorrectly suggest the phase shift transforms or relates to other parameters, misunderstanding the nature of coordinate conversion. To help students: Practice calculating phase shifts before and after conversion to verify they remain constant. Emphasize that relabeling variables doesn't change the function's behavior or characteristics.

Question 11

A rotating beacon's brightness is f(t)=6sin⁡ ⁣(π4t−π6)f(t)=6\sin\!\left(\frac{\pi}{4}t-\frac{\pi}{6}\right) (lumens), with tt in seconds. Using t=θt=\theta, an equivalent polar form is r=6sin⁡ ⁣(π4θ−π6)r=6\sin\!\left(\frac{\pi}{4}\theta-\frac{\pi}{6}\right). On a coordinate plane, the sinusoid has amplitude 66, period 88, and is shifted right by 23\frac{2}{3} seconds from sin⁡ ⁣(π4t)\sin\!\left(\frac{\pi}{4}t\right). This periodicity matches a steady rotating light. Refer to the passage above. Identify the period of the trigonometric function in the graphical representation.​

  1. 88 (correct answer)
  2. π4\frac{\pi}{4}
  3. 2π8\frac{2\pi}{8}
  4. 4π\frac{4}{\pi}
  5. 8π\frac{8}{\pi}
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically identifying the period from a given trigonometric function. The function f(t) = 6sin(π/4·t - π/6) has a coefficient of π/4 multiplying t inside the sine function. The period of a sine function in the form sin(Bt) is calculated as 2π/B, so here the period is 2π/(π/4) = 8 seconds. The passage explicitly confirms the period is 8, making Choice A correct. Choices B through E represent common errors in period calculation, such as confusing the coefficient with the period or incorrectly manipulating π. To help students: Memorize the period formula P = 2π/B for sin(Bt) and cos(Bt). Practice extracting B from various function forms and emphasize that the period represents one complete cycle of the trigonometric function.

Question 12

A speaker cone displacement is f(x)=4sin⁡(3x+π)f(x)=4\sin(3x+\pi) (millimeters), where xx is time in seconds. Converting by letting x=θx=\theta gives the polar form r=4sin⁡(3θ+π)r=4\sin(3\theta+\pi). On the coordinate plane, the graph has amplitude 44, period 2π3\frac{2\pi}{3}, and crosses the midline at x=−π3x=-\frac{\pi}{3}. This periodic model matches a steady musical tone. Based on the description, which of the following represents the function f(x)=4sin⁡(3x+π)f(x)=4\sin(3x+\pi) in polar coordinates?​

  1. r=4cos⁡(3θ+π)r=4\cos(3\theta+\pi)
  2. r=4sin⁡ ⁣(θ+π3)r=4\sin\!\left(\theta+\frac{\pi}{3}\right)
  3. r=4sin⁡(3θ+π)r=4\sin(3\theta+\pi) (correct answer)
  4. r=3sin⁡(4θ+π)r=3\sin(4\theta+\pi)
  5. r=4sin⁡(3θ)+πr=4\sin(3\theta)+\pi
Explanation: This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically converting from algebraic to polar form. When converting f(x) = 4sin(3x + π) to polar coordinates, we simply replace the independent variable x with θ, maintaining all other parameters. The passage explicitly states that converting by letting x = θ gives the polar form r = 4sin(3θ + π). Choice C correctly shows this direct substitution where amplitude (4), frequency coefficient (3), and phase shift (π) remain unchanged. Choice E incorrectly places π outside the sine function as addition rather than inside as a phase shift, a common algebraic error. To help students: Practice direct variable substitution in trigonometric conversions. Emphasize that polar conversion primarily changes the variable name and interpretation, not the function's mathematical structure.

Question 13

Let α\alpha be an angle in Quadrant II such that sin⁡(α)=35\sin(\alpha) = \frac{3}{5}, and let β\beta be an angle in Quadrant I such that cos⁡(β)=513\cos(\beta) = \frac{5}{13}. What is the value of cos⁡(α+β)\cos(\alpha + \beta)?

  1. −1665-\frac{16}{65}
  2. −5665-\frac{56}{65} (correct answer)
  3. 1665\frac{16}{65}
  4. −3365-\frac{33}{65}
Explanation: First, find cos⁡(α)\cos(\alpha) and sin⁡(β)\sin(\beta). Since α\alpha is in QII, cos⁡(α)\cos(\alpha) is negative. Using sin⁡2(α)+cos⁡2(α)=1\sin^2(\alpha) + \cos^2(\alpha) = 1, cos⁡(α)=−1−(3/5)2=−4/5\cos(\alpha) = -\sqrt{1-(3/5)^2} = -4/5. Since β\beta is in QI, sin⁡(β)\sin(\beta) is positive. Using sin⁡2(β)+cos⁡2(β)=1\sin^2(\beta) + \cos^2(\beta) = 1, sin⁡(β)=1−(5/13)2=12/13\sin(\beta) = \sqrt{1-(5/13)^2} = 12/13. Now use the sum identity: cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)=(−45)(513)−(35)(1213)=−2065−3665=−5665\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) = (-\frac{4}{5})(\frac{5}{13}) - (\frac{3}{5})(\frac{12}{13}) = -\frac{20}{65} - \frac{36}{65} = -\frac{56}{65}.

Question 14

Which of the following is equivalent to the expression (sin⁡(x)+cos⁡(x))2(\sin(x)+\cos(x))^2?

  1. 11
  2. 1+sin⁡(2x)1+\sin(2x) (correct answer)
  3. 1+cos⁡(2x)1+\cos(2x)
  4. sin⁡2(x)+cos⁡2(x)\sin^2(x)+\cos^2(x)
Explanation: Expanding the square gives (sin⁡(x)+cos⁡(x))(sin⁡(x)+cos⁡(x))=sin⁡2(x)+2sin⁡(x)cos⁡(x)+cos⁡2(x)(\sin(x)+\cos(x))(\sin(x)+\cos(x)) = \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x). Rearranging the terms yields (sin⁡2(x)+cos⁡2(x))+2sin⁡(x)cos⁡(x)(\sin^2(x) + \cos^2(x)) + 2\sin(x)\cos(x). By the Pythagorean identity, sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1. By the double-angle identity for sine, 2sin⁡(x)cos⁡(x)=sin⁡(2x)2\sin(x)\cos(x) = \sin(2x). Therefore, the expression is equivalent to 1+sin⁡(2x)1+\sin(2x).

Question 15

To solve the equation cos⁡(2θ)=3sin⁡(θ)−1\cos(2\theta) = 3\sin(\theta) - 1, which of the following identities would be the most useful first step to express the equation in terms of a single trigonometric function?

  1. cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta) (correct answer)
  2. cos⁡(2θ)=2cos⁡2(θ)−1\cos(2\theta) = 2\cos^2(\theta) - 1
  3. cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)
  4. sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1
Explanation: The original equation contains both cos⁡(2θ)\cos(2\theta) and sin⁡(θ)\sin(\theta). The goal is to have an equation with only one type of trigonometric function. Using the identity cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta) replaces cos⁡(2θ)\cos(2\theta) with an expression solely in terms of sin⁡(θ)\sin(\theta), resulting in a quadratic equation 1−2sin⁡2(θ)=3sin⁡(θ)−11 - 2\sin^2(\theta) = 3\sin(\theta) - 1, which can then be solved for sin⁡(θ)\sin(\theta). The other identities would introduce cos⁡(θ)\cos(\theta) or would not be as direct.

Question 16

Which of the following is equivalent to sin⁡(2arcsin⁡(x))\sin(2\arcsin(x)) for −1≤x≤1-1 \le x \le 1?

  1. 2x2x
  2. 2x22x^2
  3. 2x1−x22x\sqrt{1-x^2} (correct answer)
  4. 1−x2\sqrt{1-x^2}
Explanation: Let θ=arcsin⁡(x)\theta = \arcsin(x), which means sin⁡(θ)=x\sin(\theta) = x. The expression becomes sin⁡(2θ)\sin(2\theta). Using the double-angle identity, sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta). We know sin⁡(θ)=x\sin(\theta) = x. To find cos⁡(θ)\cos(\theta), we use the Pythagorean identity: cos⁡(θ)=1−sin⁡2(θ)=1−x2\cos(\theta) = \sqrt{1-\sin^2(\theta)} = \sqrt{1-x^2} (cosine is positive because the range of arcsin⁡\arcsin is [−π/2,π/2][-\pi/2, \pi/2], where cosine is non-negative). Substituting back, we get 2sin⁡(θ)cos⁡(θ)=2x1−x22\sin(\theta)\cos(\theta) = 2x\sqrt{1-x^2}.

Question 17

The expression cos⁡(x−π2)\cos(x - \frac{\pi}{2}) is equivalent to which of the following?

  1. sin⁡(x)\sin(x) (correct answer)
  2. −sin⁡(x)-\sin(x)
  3. cos⁡(x)\cos(x)
  4. −cos⁡(x)-\cos(x)
Explanation: Using the cosine difference identity, cos⁡(A−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B), we have cos⁡(x−π2)=cos⁡(x)cos⁡(π2)+sin⁡(x)sin⁡(π2)\cos(x - \frac{\pi}{2}) = \cos(x)\cos(\frac{\pi}{2}) + \sin(x)\sin(\frac{\pi}{2}). Since cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0 and sin⁡(π2)=1\sin(\frac{\pi}{2}) = 1, the expression simplifies to cos⁡(x)(0)+sin⁡(x)(1)=sin⁡(x)\cos(x)(0) + \sin(x)(1) = \sin(x).

Question 18

The expression csc⁡2(α)−1cos⁡2(α)\frac{\csc^2(\alpha)-1}{\cos^2(\alpha)} is equivalent to which of the following for all values of α\alpha for which the expression is defined?

  1. csc⁡2(α)\csc^2(\alpha) (correct answer)
  2. sec⁡2(α)\sec^2(\alpha)
  3. cot⁡2(α)\cot^2(\alpha)
  4. 11
Explanation: Using the Pythagorean identity 1+cot⁡2(α)=csc⁡2(α)1 + \cot^2(\alpha) = \csc^2(\alpha), the numerator can be rewritten as csc⁡2(α)−1=cot⁡2(α)\csc^2(\alpha) - 1 = \cot^2(\alpha). The expression then becomes cot⁡2(α)cos⁡2(α)\frac{\cot^2(\alpha)}{\cos^2(\alpha)}. Rewriting cot⁡2(α)\cot^2(\alpha) as cos⁡2(α)sin⁡2(α)\frac{\cos^2(\alpha)}{\sin^2(\alpha)}, the expression simplifies to cos⁡2(α)/sin⁡2(α)cos⁡2(α)=1sin⁡2(α)\frac{\cos^2(\alpha)/\sin^2(\alpha)}{\cos^2(\alpha)} = \frac{1}{\sin^2(\alpha)}, which is equal to csc⁡2(α)\csc^2(\alpha).

Question 19

Which of the following expressions is equivalent to (sec⁡(x)−tan⁡(x))(sec⁡(x)+tan⁡(x))(\sec(x) - \tan(x))(\sec(x) + \tan(x)) for all values of xx for which the expression is defined?

  1. 11 (correct answer)
  2. −1-1
  3. tan⁡2(x)\tan^2(x)
  4. 2sec⁡2(x)−12\sec^2(x) - 1
Explanation: The expression is a difference of squares, which expands to sec⁡2(x)−tan⁡2(x)\sec^2(x) - \tan^2(x). Using the Pythagorean identity 1+tan⁡2(x)=sec⁡2(x)1 + \tan^2(x) = \sec^2(x), we can rearrange it to sec⁡2(x)−tan⁡2(x)=1\sec^2(x) - \tan^2(x) = 1. Therefore, the expression is equivalent to 1.

Question 20

Which of the following expressions is equivalent to cos⁡4(θ)−sin⁡4(θ)\cos^4(\theta) - \sin^4(\theta)?

  1. 11
  2. cos⁡(2θ)\cos(2\theta) (correct answer)
  3. sin⁡(2θ)\sin(2\theta)
  4. (cos⁡(θ)−sin⁡(θ))4(\cos(\theta)-\sin(\theta))^4
Explanation: The expression can be factored as a difference of squares: (cos⁡2(θ)−sin⁡2(θ))(cos⁡2(θ)+sin⁡2(θ))(\cos^2(\theta) - \sin^2(\theta))(\cos^2(\theta) + \sin^2(\theta)). By the Pythagorean identity, cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1. The other factor, cos⁡2(θ)−sin⁡2(θ)\cos^2(\theta) - \sin^2(\theta), is the double-angle identity for cosine. Therefore, the expression simplifies to cos⁡(2θ)\cos(2\theta).