Sine and cosine function graphs — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Key characteristics of parent sine and cosine graphs, amplitude, period, phase shift, vertical shift, graph transformations of sinusoidal functions, intercepts, extrema, and constructing sinusoidal functions from given graphs or parameters.

You should already know: Unit circle definitions of sine and cosine, basic function transformation rules, and linear equation solving.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Sine and cosine function graphs?

Sine and cosine function graphs are the graphical representations of periodic sinusoidal functions, the core periodic functions in AP Precalculus Unit 3: Trigonometric and Polar Functions. According to the AP Precalculus Course and Exam Description (CED), sinusoidal functions make up roughly 12-15% of Unit 3 exam content, and this topic appears regularly across both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone problem or embedded in real-world modeling questions. Unlike polynomial or exponential functions, sine and cosine are periodic, meaning their graphs repeat indefinitely at a fixed interval. This topic is foundational for all sinusoidal modeling, simple harmonic motion, and later polar graphing of circular and sinusoidal polar functions. At the exam level, you will be expected to identify key features from a graph, write the equation of a sinusoid from a given graph or description, and transform parent graphs to match given parameters.

2. Key Features of Parent Sine and Cosine Graphs

The untransformed (parent) sine and cosine functions are $y=\sin x$ and $y=\cos x$, where $x$ is measured in radians (the standard for AP Precalculus unless explicitly stated otherwise). Both functions share core properties: a domain of all real numbers $(-\infty ,\infty )$, a range of $[-1,1]$, and a fundamental period of $2\pi$, meaning one full cycle of oscillation completes over an interval of length $2\pi$.

For the parent sine function $y=\sin x$: the y-intercept is at $(0,0)$, x-intercepts occur at every $x=k\pi$ for any integer $k$, the maximum value of 1 occurs at $x=\frac{\pi }{2}+2k\pi$, and the minimum value of -1 occurs at $x=\frac{3\pi }{2}+2k\pi$. It is an odd function, symmetric about the origin. For the parent cosine function $y=\cos x$: the y-intercept is at $(0,1)$, x-intercepts occur at $x=\frac{\pi }{2}+k\pi$ for any integer $k$, the maximum value of 1 occurs at $x=2k\pi$, and the minimum value of -1 occurs at $x=\pi +2k\pi$. It is an even function, symmetric about the y-axis. These properties come directly from the unit circle definition: as the angle $x$ completes one full rotation of $2\pi$, sine (the y-coordinate on the unit circle) and cosine (the x-coordinate) complete one full oscillation.

Worked Example

Identify all maximum points of $y=\cos x$ on the interval $[-3\pi ,3\pi ]$.

1. Recall that for parent cosine, maxima occur at $x=2k\pi$ for all integers $k$, where the function equals its maximum value of 1.
2. Find all integers $k$ such that $-3\pi \le 2k\pi \le 3\pi$, which simplifies to $-1.5\le k\le 1.5$.
3. The valid integer values of $k$ are $k=-1,0,1$.
4. Substitute back to get the x-coordinates of the maxima: $x=-2\pi ,0,2\pi$, so the maximum points are $(-2\pi ,1),(0,1),(2\pi ,1)$.

Exam tip: Always confirm that your solutions lie within the interval specified in the question—AP exam questions regularly test your ability to restrict solutions to a given domain, and full credit is only given for solutions inside the interval.

3. Transformations of Sinusoidal Functions

Any translated, stretched, or reflected sine/cosine graph can be written in the standard general form: $$f(x)=A\sin \left(B(x-C)\right)+D\text{or}f(x)=A\cos \left(B(x-C)\right)+D$$ Each constant corresponds to a specific transformation that follows the same rules as function transformations for any function family:

• $|A|$ = Amplitude: the vertical distance from the midline (center line of the graph) to any maximum or minimum. If $A<0$, the graph is reflected over the midline.
• Period = $\frac{2\pi }{|B|}$: the length of one full cycle of the graph. Larger $|B|$ compresses the graph horizontally, resulting in a shorter period (faster oscillation). The sign of $B$ only reflects the graph horizontally, it does not change the period.
• $C$ = Phase Shift: the horizontal shift of the graph. If $C>0$, the graph shifts $C$ units right; if $C<0$, it shifts $C$ units left.
• $D$ = Vertical Shift: the midline of the graph is the horizontal line $y=D$.

A critical note: If the function is given in the form $\sin (Bx-C)$ (not factored), you must factor out $B$ to get the correct phase shift: $\sin (Bx-C)=\sin \left(B\left(x-\frac{C}{B}\right)\right)$, so the phase shift is $\frac{C}{B}$, not $C$.

Worked Example

Given $f(x)=3\sin \left(4x-\pi \right)-1$, find the amplitude, period, phase shift, and midline.

1. Factor $B=4$ out of the argument: $4x-\pi =4\left(x-\frac{\pi }{4}\right)$, so the function becomes $f(x)=3\sin \left(4\left(x-\frac{\pi }{4}\right)\right)-1$.
2. Amplitude is $|A|=|3|=3$.
3. Period is $\frac{2\pi }{|B|}=\frac{2\pi }{4}=\frac{\pi }{2}$.
4. Phase shift is $\frac{\pi }{4}$ units to the right, since $C=\frac{\pi }{4}>0$.
5. The midline is $y=D=-1$, which is also a vertical shift of 1 unit down from the parent midline.

Exam tip: If you are ever unsure of your phase shift calculation, plug the shifted starting point into the function to check that it matches the expected output for the parent function.

4. Constructing a Sinusoidal Function From a Graph

AP Precalculus regularly asks you to write the equation of a sinusoidal function given its graph or key features. The step-by-step method to solve for $A,B,C,D$ is consistent:

1. Find $D$ (midline/vertical shift): $D=\frac{\text{maximum value}+\text{minimum value}}{2}$
2. Find $|A|$ (amplitude): $|A|=\text{maximum value}-D=D-\text{minimum value}$
3. Find the period: measure the horizontal distance between two consecutive identical points (e.g., two consecutive maxima), then calculate $B=\frac{2\pi }{\text{period}}$
4. Find $C$ (phase shift): choose to use a sine or cosine base to simplify calculation. If a maximum/minimum is at $x=0$, use cosine with $C=0$ to avoid extra calculation. If a midline point with positive slope is at $x=0$, use sine with $C=0$.

Worked Example

A sinusoidal function has a minimum at $(0,-2)$ and the next maximum at $(2,4)$. Write a cosine function in standard form that matches this graph.

1. Calculate $D$: $D=\frac{-2+4}{2}=1$, so the midline is $y=1$.
2. Calculate amplitude: The minimum is at $y=-2$, so $|A|=D-(-2)=1-(-2)=3$. Since the minimum (not maximum) is at $x=0$, $A$ is negative: $A=-3$.
3. Calculate period: The horizontal distance from minimum to next maximum is half a period, so half-period = $2-0=2$, so full period = $4$. Then $B=\frac{2\pi }{4}=\frac{\pi }{2}$.
4. The minimum of $A\cos (B(x-C))+D$ is at $x=0=C$, so $C=0$.
5. The final function is $f(x)=-3\cos \left(\frac{\pi }{2}x\right)+1$. Check: $f(0)=-3(1)+1=-2$ (correct minimum), $f(2)=-3\cos (\pi )+1=-3(-1)+1=4$ (correct maximum).

Exam tip: Any sinusoidal function can be written as either a shifted sine or a shifted cosine—both are correct as long as they match the given features, but choosing the form that simplifies $C$ to zero reduces your chance of sign errors.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For $f(x)=\sin (2x-\pi )$, reading the phase shift as $\pi$ units right. Why: Students forget to factor out the horizontal scale factor $B$ from the argument, and confuse the form $(Bx-C)$ with $B(x-C)$. Correct move: Always factor $B$ out of the argument first: $2x-\pi =2\left(x-\frac{\pi }{2}\right)$, so the phase shift is $\frac{\pi }{2}$ units right.
• Wrong move: Calculating the period as $\frac{|B|}{2\pi }$ instead of $\frac{2\pi }{|B|}$. Why: Students mix up the inverse relationship between $B$ and period—larger $B$ means shorter period, but the reciprocal flips this relationship. Correct move: After calculating period, verify: if $|B|>1$, period should be less than $2\pi$; if $|B|<1$, period should be greater than $2\pi$ to confirm.
• Wrong move: Claiming amplitude is negative when $A<0$. Why: Students confuse the sign of $A$ (which indicates reflection) with the amplitude, which is a distance and always non-negative. Correct move: Amplitude is always reported as $|A|$; note the reflection separately if the question asks for transformations.
• Wrong move: Measuring the distance between a maximum and the next minimum as the full period when reading from a graph. Why: Maximum and minimum are half a cycle apart, not a full cycle. Correct move: Always measure between two consecutive identical points (maximum to maximum, minimum to minimum, or two increasing midline crossings) to get the full period.
• Wrong move: Setting $D$ equal to the maximum value when constructing an equation. Why: Students confuse vertical shift with the maximum value for vertically shifted graphs. Correct move: Always calculate $D$ as the average of the maximum and minimum values to get the midline.
• Wrong move: Using degrees to calculate period when no units are specified. Why: Introductory courses often mix degree and radian graphing, but AP Precalculus assumes radians for all unspecified cases. Correct move: Use radians for all period and phase shift calculations unless the question explicitly says to use degrees.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following gives the amplitude and period of the function $f(x)=-5\sin \left(\frac{\pi }{4}x\right)+2$? A) Amplitude 5, Period $\frac{\pi }{4}$ B) Amplitude 5, Period 8 C) Amplitude 5, Period 4 D) Amplitude $-5$, Period 8

Worked Solution: First, amplitude is the absolute value of the leading coefficient, so $|A|=|-5|=5$, eliminating option D. Next, period is calculated as $\frac{2\pi }{|B|}$ where $B=\frac{\pi }{4}$. Substituting gives $\frac{2\pi }{\pi /4}=8$. The negative sign of $A$ only indicates a reflection over the midline, which does not change amplitude. This eliminates options A and C. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=A\cos \left(B(x-C)\right)+D$ be a sinusoidal function with maximum value 10, minimum value 2, and period $6\pi$. (a) Find the amplitude $A$ (given $A>0$) and midline $y=D$. (b) If $f(x)$ has a minimum at $x=\frac{3\pi }{2}$, find the phase shift $C$. (c) Write the final equation for $f(x)$ that satisfies all conditions.

Worked Solution: (a) The midline $D$ is the average of the maximum and minimum: $D=\frac{10+2}{2}=6$. Amplitude is the distance from midline to maximum: $A=10-6=4$. So amplitude = 4, midline $y=6$. (b) Calculate $B=\frac{2\pi }{\text{period}}=\frac{2\pi }{6\pi }=\frac{1}{3}$. A minimum of cosine occurs when $B(x-C)=\pi$. Substitute $x=\frac{3\pi }{2}$ and $B=\frac{1}{3}$: $\frac{1}{3}\left(\frac{3\pi }{2}-C\right)=\pi$. Multiply both sides by 3: $\frac{3\pi }{2}-C=3\pi$, so $C=-\frac{3\pi }{2}$ (or a positive equivalent of $\frac{9\pi }{2}$ for the smallest positive phase shift). (c) Substituting all values gives $f(x)=4\cos \left(\frac{1}{3}\left(x+\frac{3\pi }{2}\right)\right)+6$, which satisfies all conditions.

Question 3 (Application / Real-World Style)

The average monthly temperature in a coastal city can be modeled by a sinusoidal function, where $t=0$ corresponds to January (the coldest month). The average minimum temperature in January is 40°F, and the average maximum temperature in July is 84°F. Write a sinusoidal function $T(t)$ for the average temperature in month $t$, and find the average temperature in April ($t=3$).

Worked Solution: Since the minimum is at $t=0$, we use a cosine function with negative amplitude and $C=0$. Calculate midline: $D=\frac{40+84}{2}=62$°F. Amplitude: $|A|=62-40=22$, so $A=-22$. The period is 12 months (one full year), so $B=\frac{2\pi }{12}=\frac{\pi }{6}$. The function is $T(t)=-22\cos \left(\frac{\pi }{6}t\right)+62$. Substitute $t=3$: $T(3)=-22\cos \left(\frac{\pi }{6}\cdot 3\right)+62=-22\cos \left(\frac{\pi }{2}\right)+62=62$°F. In context, the average temperature in April in this city is 62°F, which is the midpoint between the average winter low and summer high.

7. Quick Reference Cheatsheet

8. What's Next

Mastering sine and cosine function graphs is a non-negotiable prerequisite for the next core topics in Unit 3: tangent function graphs and sinusoidal real-world modeling. Without being able to quickly identify key features, transform graphs, and write sinusoidal equations from context, you will struggle to score full credit on periodic modeling FRQs, which make up a large portion of Unit 3 exam points. This topic also lays the foundation for polar graphing, where many common polar curves are defined using sinusoidal functions of the angle, and for inverse trigonometric functions, where understanding the domain and range restrictions of sine and cosine relies on familiarity with their full graphs. Next, continue your study with these topics: Tangent function graphs Sinusoidal modeling Polar coordinate graphs Inverse trigonometric functions