AP Precalculus Quiz: Sine And Cosine Function Graphs

What this quiz covers

This quiz focuses on Sine And Cosine Function Graphs, 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.

Question 1

1. Vertical shift is $2$
2. Vertical shift is $-6$
3. Vertical shift is $-2$ (correct answer)
4. Vertical shift is $5$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically identifying vertical shift from a constant term. In the form A·cos(Bt) + D, the vertical shift D moves the entire graph up or down, changing the midline from y = 0 to y = D. For I(t) = 6cos(5t) - 2, the constant term is -2, indicating a downward shift of 2 units, so the function oscillates around y = -2 instead of y = 0. Choice C correctly identifies the vertical shift as -2, showing the midline has moved down by 2 units. Choice B incorrectly uses -6, confusing the amplitude with vertical shift, while choice A gives the wrong sign. To help students: identify vertical shift as the constant term added or subtracted at the end of the expression. Practice finding the new range by adding the vertical shift to the standard range of [-amplitude, amplitude].

Question 2

1. Shift left $\pi/4$
2. Shift right $\pi/4$ (correct answer)
3. Shift right $4\pi$
4. No phase shift

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically phase shift from horizontal translations. For cos(t - k), the phase shift is k units to the right, as the function reaches any given value k units later than the standard cosine. In V(t) = cos(t - π/4), the expression (t - π/4) indicates a horizontal shift of π/4 units to the right. Choice B correctly identifies this as a right shift of π/4, showing the cosine wave starts π/4 units later than usual. Choice A would be correct if the sign were positive inside the parentheses, while choice D fails to recognize the transformation. To help students: remember that f(t - h) shifts right by h units, consistent with general function transformations. Verify phase shifts by checking specific points, such as where the function equals its maximum value.

Question 3

1. Period is $3\pi$
2. Period is $6\pi$ (correct answer)
3. Period is $\pi/3$
4. Period is $2\pi$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically calculating period with fractional coefficients. The period of cos(Bt) is 2π/B, and when B is a fraction, the period becomes larger than the standard 2π. For V(t) = 2cos(t/3), we can rewrite as 2cos((1/3)t), so B = 1/3, giving period = 2π/(1/3) = 6π. Choice B correctly identifies the period as 6π, showing that the function takes three times longer to complete one cycle compared to standard cosine. Choice C incorrectly inverts the relationship, while choice A misapplies the coefficient 3. To help students: practice rewriting expressions like cos(t/n) as cos((1/n)t) to clearly identify B. Emphasize that dividing t by a number stretches the graph horizontally, increasing the period.

Question 4

1. Period doubles
2. Period halves (correct answer)
3. Period becomes $2\pi$
4. No period change

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically how the coefficient of t affects the period. The standard cosine function cos(t) has period 2π, but cos(Bt) has period 2π/B, so larger B values create shorter periods and more rapid oscillations. For V(t) = cos(2t), B = 2, giving period = 2π/2 = π, which is half the standard period of 2π. Choice B correctly states that the period halves, as the function now completes two full cycles in the time it previously took for one cycle. Choice A incorrectly suggests the period doubles, reversing the relationship between B and period, while choice D misses the transformation entirely. To help students: demonstrate graphically how cos(2t) compresses the standard cosine horizontally. Practice with various B values to build intuition that B > 1 compresses (shorter period) while 0 < B < 1 stretches (longer period).

Question 5

1. The period is $6\pi$.
2. The period is $3\pi$.
3. The period is $2\pi/3$. (correct answer)
4. The period is $\pi/3$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically period changes. The period of y = cos(Bx) is calculated using period = 2π/|B|, where B is the coefficient of x inside the cosine function. For y = cos(3x), B = 3, so period = 2π/3, meaning the circuit completes three full cycles in the space of 2π radians. Choice C correctly identifies the period as 2π/3, showing the function oscillates three times faster than standard cosine. Choice A incorrectly suggests 6π; Choice B suggests 3π; Choice D suggests π/3, all misapplying the period formula. To help students: consistently use period = 2π/|B| and remember that the coefficient B tells you how many complete cycles occur in 2π radians. Verify by checking that cos(3(x + 2π/3)) = cos(3x + 2π) = cos(3x).

Question 6

1. The period is $8\pi$.
2. The period is $\pi/2$. (correct answer)
3. The period is $4\pi$.
4. The period is $2\pi$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically determining period from the equation. The period of y = sin(Bx) is calculated using the formula period = 2π/|B|, where B is the coefficient of x inside the sine function. For y = sin(4x), B = 4, so the period = 2π/4 = π/2, meaning the function completes one full cycle in π/2 radians. Choice B correctly identifies the period as π/2, showing the function oscillates four times faster than standard sine. Choice A incorrectly suggests 8π; Choice C suggests 4π; Choice D suggests the standard 2π period, all failing to apply the period formula correctly. To help students: consistently apply the formula period = 2π/|B| and remember that larger B values create smaller periods (faster oscillations). Verify by checking that sin(4(x + π/2)) = sin(4x + 2π) = sin(4x).

Question 7

1. The period is $\pi$.
2. The period is $4\pi$. (correct answer)
3. The period is $2\pi$.
4. The period is $1/2$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically period changes with fractional coefficients. The period formula is period = 2π/|B| for y = sin(Bx), where smaller B values create larger periods (slower oscillations). For y = sin(1/2·x), B = 1/2, so period = 2π/(1/2) = 2π × 2 = 4π, meaning the pendulum takes twice as long to complete one cycle. Choice B correctly identifies the period as 4π, showing the function oscillates half as fast as standard sine. Choice A incorrectly suggests π; Choice C suggests the standard 2π; Choice D confuses period with the coefficient. To help students: when B is a fraction, remember that dividing by a fraction means multiplying by its reciprocal. Visualize that smaller B values stretch the graph horizontally.

Question 8

1. The period doubles to $4\pi$.
2. The period becomes $\pi$. (correct answer)
3. The period stays $2\pi$.
4. The period becomes $2$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically period changes. The period of a trigonometric function is determined by the coefficient of x inside the function, where period = 2π/|B| for y = cos(Bx). For y = cos(2x), B = 2, so the period = 2π/2 = π, which is half the standard period of 2π. Choice B correctly identifies the period as π, showing that when the coefficient inside is 2, the function completes its cycle twice as fast. Choice A incorrectly suggests the period doubles; Choice C incorrectly states it stays the same; Choice D confuses period with the coefficient. To help students: use the formula period = 2π/|B| consistently and visualize how larger B values compress the graph horizontally. Practice with various coefficients and verify by graphing to see complete cycles.

Question 9

1. Amplitude is $\tfrac{1}{3}$.
2. Amplitude is $3$. (correct answer)
3. Amplitude is $2\pi$.
4. Amplitude is $-3$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically amplitude. The amplitude of a trigonometric function is the absolute value of the coefficient in front of the sine or cosine function, determining the maximum distance from the midline to the peak or trough. For the equation y = 3sin(x), the coefficient is 3, so the amplitude is |3| = 3, meaning the wave oscillates between -3 and 3. Choice B correctly identifies the amplitude as 3, which matches the coefficient of the sine function. Choice A incorrectly suggests 1/3, possibly confusing reciprocal relationships; Choice C incorrectly uses 2π, which is the period of standard sine; Choice D incorrectly uses -3, failing to take the absolute value. To help students: emphasize that amplitude is always positive and equals the absolute value of the coefficient. Practice identifying transformations by comparing equations to the standard form y = A·sin(Bx + C) + D, where |A| is the amplitude.

Question 10

1. Amplitude is $60\pi$
2. Amplitude is $10$ (correct answer)
3. Amplitude is $\pi/10$
4. Amplitude is $-10$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically identifying amplitude. In the general form A·cos(Bt + C) + D, the amplitude is |A|, which represents the maximum distance from the midline to the peak or trough of the function. For the equation I(t) = 10cos(60πt), the coefficient of cosine is 10, making the amplitude |10| = 10. Choice B correctly identifies the amplitude as 10, which determines how far the current oscillates above and below its center value. Choice A incorrectly uses 60π, confusing the frequency coefficient with amplitude, while choices C and D misinterpret other components of the equation. To help students: emphasize that amplitude is always the absolute value of the coefficient directly multiplying the trigonometric function. Encourage students to identify each transformation parameter systematically by comparing to the standard form.

Question 11

1. Period is $4\pi$
2. Period is $\pi/2$ (correct answer)
3. Period is $2\pi$
4. Period is $8\pi$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically calculating the period. The period of cos(Bt) is 2π/B, where B is the coefficient of t inside the cosine function, determining how quickly the function completes one full cycle. For I(t) = 5cos(4πt), B = 4π, so the period is 2π/(4π) = 1/2, which equals π/2. Choice B correctly identifies the period as π/2, representing the time for one complete oscillation of the alternating current. Choice A incorrectly calculates 4π, possibly confusing period with the coefficient, while choice C gives the standard period without accounting for the frequency change. To help students: practice the period formula systematically and verify by checking that cos(B(t + period)) = cos(Bt). Watch for errors in algebraic simplification when dividing by coefficients containing π.

Question 12

1. $y=\sin\left(x+\tfrac{\pi}{2}\right)$
2. $y=\sin\left(x-\tfrac{\pi}{2}\right)$ (correct answer)
3. $y=\cos\left(x-\tfrac{\pi}{2}\right)$
4. $y=\sin\left(\tfrac{\pi}{2}x\right)$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically phase shifts. A phase shift moves the entire graph horizontally, where y = sin(x - h) shifts the graph h units to the right, and y = sin(x + h) shifts it h units to the left. To shift y = sin(x) right by π/2, we need y = sin(x - π/2), subtracting π/2 from x inside the function. Choice B correctly shows y = sin(x - π/2), which shifts the graph π/2 units to the right. Choice A incorrectly adds π/2, which would shift left; Choice C uses cosine instead of sine; Choice D incorrectly multiplies x by π/2, which changes the period instead. To help students: remember that horizontal shifts work opposite to intuition - subtract to go right, add to go left. Use the mnemonic 'minus means right' and verify by checking key points like where the function equals zero.

Question 13

1. Vertical shift is $-2$.
2. Vertical shift is $2$. (correct answer)
3. Vertical shift is $1$.
4. Vertical shift is $2\pi$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically vertical shifts. A vertical shift moves the entire graph up or down, where y = cos(x) + D shifts the graph D units vertically - positive D shifts up, negative D shifts down. For y = cos(x) + 2, the +2 outside the cosine function shifts the entire graph up by 2 units, so the midline moves from y = 0 to y = 2. Choice B correctly identifies the vertical shift as 2, meaning all y-values increase by 2. Choice A incorrectly suggests -2, confusing the sign; Choice C suggests 1; Choice D confuses vertical shift with period. To help students: identify vertical shifts as the constant added outside the trigonometric function in the form y = f(x) + D. Practice by finding new maximum, minimum, and midline values after the shift.

Question 14

1. Amplitude is $-2$.
2. Amplitude is $2$. (correct answer)
3. Amplitude is $\pi$.
4. Amplitude is $\tfrac{1}{2}$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically amplitude with negative coefficients. The amplitude is the absolute value of the coefficient in front of the trigonometric function, regardless of sign - the negative sign reflects the graph vertically but doesn't change the amplitude. For y = -2sin(x), the amplitude is |-2| = 2, meaning the function oscillates between -2 and 2 (but inverted due to the negative). Choice B correctly identifies the amplitude as 2, recognizing that amplitude is always positive. Choice A incorrectly uses -2, failing to take absolute value; Choice C incorrectly suggests π; Choice D suggests 1/2. To help students: emphasize that amplitude = |A| for y = A·sin(x), and the negative sign only flips the graph vertically. Practice identifying both amplitude and reflection transformations separately.

Question 15

1. Vertical shift is $4$.
2. Vertical shift is $-4$. (correct answer)
3. Vertical shift is $-1$.
4. Vertical shift is $4\pi$.

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically vertical shifts with negative values. A vertical shift is determined by the constant added to or subtracted from the trigonometric function: y = sin(x) + D shifts up by D if positive, down by |D| if negative. For y = sin(x) - 4, the -4 shifts the entire graph down by 4 units, moving the midline from y = 0 to y = -4. Choice B correctly identifies the vertical shift as -4, indicating a downward shift of 4 units. Choice A incorrectly suggests +4, confusing the sign; Choice C suggests -1; Choice D incorrectly includes π. To help students: identify the vertical shift as the constant term outside the trigonometric function, keeping its sign. The new range becomes [-1, 1] shifted down by 4, giving [-5, -3].

Question 16

1. $y=\cos\left(x+\tfrac{\pi}{3}\right)$ (correct answer)
2. $y=\cos\left(x-\tfrac{\pi}{3}\right)$
3. $y=\sin\left(x+\tfrac{\pi}{3}\right)$
4. $y=\cos\left(\tfrac{\pi}{3}x\right)$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically phase shifts to the left. A horizontal shift left by h units is achieved by adding h to x inside the function: y = cos(x + h) shifts the graph h units to the left. To shift y = cos(x) left by π/3, we need y = cos(x + π/3), adding π/3 to x inside the cosine function. Choice A correctly shows y = cos(x + π/3), which shifts the graph π/3 units to the left. Choice B incorrectly subtracts π/3, which would shift right; Choice C uses sine instead of cosine; Choice D incorrectly multiplies x by π/3, changing the period. To help students: remember that horizontal shifts work opposite to vertical shifts - add to go left, subtract to go right. Verify by checking key points like where cos(0 + π/3) = cos(π/3) = 1/2.

Question 17

1. Vertical shift is $7$
2. Vertical shift is $-3$
3. Vertical shift is $3$ (correct answer)
4. Vertical shift is $\pi$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically identifying vertical shift. In the form A·cos(Bt) + D, the vertical shift is D, which moves the entire graph up (if positive) or down (if negative) from the standard position. For I(t) = 7cos(πt) + 3, the constant term +3 indicates a vertical shift of 3 units upward, meaning the midline of oscillation is at y = 3 instead of y = 0. Choice C correctly identifies the vertical shift as 3, representing how the average current value is displaced from zero. Choice A incorrectly uses the amplitude value 7, while choice B suggests a negative shift that doesn't match the positive constant. To help students: emphasize that vertical shift is always the constant term added outside the trigonometric function. Encourage visualizing how this shifts the midline and affects the range of the function.

Question 18

1. Amplitude is $-4$
2. Amplitude is $4$ (correct answer)
3. Amplitude is $\omega$
4. Amplitude is $1/4$

Explanation: This question tests AP Precalculus skills: understanding sine and cosine function transformations, specifically that amplitude is always positive. The amplitude of A·cos(ωt) is |A|, the absolute value of the coefficient, representing the maximum distance from the midline regardless of whether the graph is reflected. For I(t) = -4cos(ωt), the amplitude is |-4| = 4, even though the negative sign flips the graph vertically. Choice B correctly identifies the amplitude as 4, understanding that amplitude measures magnitude of oscillation, not direction. Choice A incorrectly gives -4, failing to take the absolute value, which is a common error when students confuse amplitude with the coefficient itself. To help students: emphasize that amplitude is always positive and represents a distance. Demonstrate how negative coefficients reflect the graph but don't change the height of oscillations from the midline.