Sine and Cosine Function Values
Why this matters
Imagine you're on a giant Ferris wheel, like the Centennial Wheel at Chicago's Navy Pier. You've stopped right at the top, and your friend on the ground wants to take the perfect photo. They yell up, "Where are you, exactly?"
It's not enough to say "at the top." For the perfect shot, they need your precise location. If you knew the wheel's radius and the angle your cart has rotated from the starting point, could you give them your exact (x, y) coordinates relative to the center of the wheel?
Absolutely. This is the core idea we're exploring today. We're moving beyond just angles and triangles to pinpoint exact locations in a circular path. You'll learn how sine and cosine are the keys to translating rotational motion into the familiar (x, y) grid we use for everything from maps to video games.
Concept overview
flowchart TD
A[Start: Given r and θ] --> B{1. Determine Quadrant of θ};
B --> C[2. Find Reference Angle θ_ref];
C --> D{3. Find cos(θ_ref) and sin(θ_ref) from Special Triangles};
D --> E[4. Apply +/- Signs based on Quadrant];
E --> F[5. Calculate x = r * cos(θ)];
E --> G[5. Calculate y = r * sin(θ)];
F & G --> H[End: Coordinates (x, y)];
Core explanation
From Rotation to Location: The Big Idea
Let's start with a circle centered at the origin (0, 0) in the xy-plane. This circle has a radius, which we'll call r.
Now, imagine drawing an angle, θ (theta), in standard position. Its initial side is on the positive x-axis, and it rotates counterclockwise. The terminal side of this angle will intersect our circle at some point, which we'll call P. This point P has coordinates (x, y).
Our goal is to find the values of x and y using only the radius r and the angle θ.
Building the Bridge with SOHCAHTOA
To connect r and θ to x and y, we can create a right triangle. Drop a perpendicular line from point P straight down to the x-axis.
Look at the triangle we just formed:
- The hypotenuse is the radius of the circle,
r. - The adjacent side (next to angle
θ) is the horizontal distance from the origin, which is ourx-coordinate. - The opposite side is the vertical distance from the x-axis, which is our
y-coordinate.
Remember our old friend SOHCAHTOA? Let's apply it.
- CAH
cos(θ) = Adjacent / Hypotenuse = x / r - SOH
sin(θ) = Opposite / Hypotenuse = y / r
Now, let's solve for x and y.
- If
cos(θ) = x / r, we can multiply both sides byrto get:x = r cos(θ) - If
sin(θ) = y / r, we can multiply both sides byrto get:y = r sin(θ)
And there it is. That's the formula. The coordinates of any point P on a circle of radius r at an angle θ are:
P(x, y) = (r cos(θ), r sin(θ))
This single formula is the heart of this topic. It directly connects an angle and a radius to a specific (x, y) point.
Finding Exact Values with Special Triangles
Your calculator can find cos(θ) and sin(θ), but it will give you long decimal approximations. The AP exam will often ask for exact values. This means no decimals—we need fractions and square roots.
To get these exact values, we rely on two special right triangles. You should have these memorized.
1. The 45°-45°-90° Triangle (Multiples of π/4)
This is an isosceles right triangle. If the two legs have a length of 1, the hypotenuse is √2.
sin(π/4)= Opposite/Hypotenuse =1/√2=√2 / 2cos(π/4)= Adjacent/Hypotenuse =1/√2=√2 / 2
2. The 30°-60°-90° Triangle (Multiples of π/6 and π/3)
This triangle comes from slicing an equilateral triangle in half. The sides are in a 1 : √3 : 2 ratio.
- For the
π/6(30°) angle:sin(π/6)= Opposite/Hypotenuse =1 / 2cos(π/6)= Adjacent/Hypotenuse =√3 / 2
- For the
π/3(60°) angle:sin(π/3)= Opposite/Hypotenuse =√3 / 2cos(π/3)= Adjacent/Hypotenuse =1 / 2
Don't Forget Your Signs: The Four Quadrants
Think about the xy-plane:
- Quadrant I (0 to π/2)
xis positive,yis positive.(+ , +) - Quadrant II (π/2 to π)
xis negative,yis positive.(- , +) - Quadrant III (π to 3π/2)
xis negative,yis negative.(- , -) - Quadrant IV (3π/2 to 2π)
xis positive,yis negative.(+ , -)
Since x = r cos(θ) and y = r sin(θ), the signs of cos(θ) and sin(θ) must match the signs of x and y.
A helpful mnemonic is All Students Take Calculus:
- All (Quadrant I): All functions (sin, cos, tan) are positive.
- Students (Quadrant II): Sine is positive (so cosine is negative).
- Take (Quadrant III): Tangent is positive (so sine and cosine are both negative).
- Calculus (Quadrant IV): Cosine is positive (so sine is negative).
When you're given an angle like θ = 3π/4, first figure out its quadrant (Quadrant II). Then, find its reference angle—the acute angle it makes with the x-axis. For 3π/4, the reference angle is π/4.
You'll use the special triangle values for π/4, but then apply the correct signs for Quadrant II: cosine is negative, sine is positive.
cos(3π/4) = -cos(π/4) = -√2 / 2sin(3π/4) = +sin(π/4) = +√2 / 2
By combining the formula (r cos θ, r sin θ) with your knowledge of special triangles and quadrant signs, you can find the exact coordinates for any of these key angles on any circle.
Worked examples
A Point in Quadrant I
Problem: Find the coordinates of the point P on a circle with radius r = 10 that corresponds to an angle of θ = π/3.
Step 1: Identify the formula and given values.
The formula to find the coordinates (x, y) is (r cos θ, r sin θ).
We are given r = 10 and θ = π/3.
Step 2: Determine the values for cos(θ) and sin(θ).
The angle π/3 is one of our special angles. From the 30-60-90 triangle, we know:
cos(π/3) = 1/2sin(π/3) = √3 / 2Sinceπ/3is in Quadrant I, both sine and cosine are positive. We don't need to change any signs.
Step 3: Calculate the x and y coordinates.
Now, plug the values into our formula.
x = r cos(θ) = 10 * (1/2) = 5y = r sin(θ) = 10 * (√3 / 2) = 5√3
Solution: The coordinates of point P are (5, 5√3).
A Point in Quadrant III
Problem: A point lies on a circle of radius 4 centered at the origin. Find the exact coordinates of the point corresponding to an angle of θ = 5π/4.
Step 1: Identify the formula, values, and quadrant.
Formula: (r cos θ, r sin θ)
Given: r = 4, θ = 5π/4.
The angle 5π/4 is slightly more than π (which is 4π/4), so it's in Quadrant III. In this quadrant, both x and y coordinates are negative. This is a critical check to make right away.
Step 2: Find the reference angle and the trig values.
The reference angle is the acute angle the terminal side makes with the x-axis.
Reference angle θ_ref = 5π/4 - π = π/4.
Now we find the values for the reference angle:
cos(π/4) = √2 / 2sin(π/4) = √2 / 2
Step 3: Apply the correct signs for Quadrant III.
Because our original angle 5π/4 is in Quadrant III, both cosine and sine are negative.
cos(5π/4) = -cos(π/4) = -√2 / 2sin(5π/4) = -sin(π/4) = -√2 / 2
This is the step where students most often make a mistake. They use the positive values from the reference angle and forget to adjust for the quadrant.
Step 4: Calculate the final coordinates.
x = r cos(θ) = 4 * (-√2 / 2) = -2√2y = r sin(θ) = 4 * (-√2 / 2) = -2√2
Solution: The coordinates of the point are (-2√2, -2√2).
Try it yourself
Ready to try a couple on your own? Don't worry about getting it perfect on the first try. Focus on the process.
Problem 1:
Find the exact coordinates of the point on a circle of radius 6 that is intersected by the terminal ray of the angle θ = 11π/6.
Hint: What quadrant is 11π/6 in? What does that tell you about the signs of your x and y coordinates?
Problem 2:
A point P is on a circle of radius 20 centered at the origin. If P is the intersection of the terminal ray of θ = 2π/3, what are its coordinates?
Hint: What is the reference angle for 2π/3? Remember which trig function is positive in Quadrant II.
Practice — 8 questions
In simple terms, this topic is about finding the exact (x, y) coordinates of a point on any circle, just by knowing the circle's radius and an angle.
- 3.3.A: Determine coordinates of points on a circle centered at the origin.
- 3.3.A.1
- Given an angle of measure θ in standard position and a circle with radius r centered at the origin, there is a point, P, where the terminal ray intersects the circle. The coordinates of point P are (r cos θ, r sin θ).
- 3.3.A.2
- The geometry of isosceles right and equilateral triangles, while attending to the signs of the values based on the quadrant of the angle, can be used to find exact values for the cosine and sine of angles that are multiples of π/4 and π/3 radians and whose terminal rays do not lie on an axis.
flowchart TD
A[Start: Given r and θ] --> B{1. Determine Quadrant of θ};
B --> C[2. Find Reference Angle θ_ref];
C --> D{3. Find cos(θ_ref) and sin(θ_ref) from Special Triangles};
D --> E[4. Apply +/- Signs based on Quadrant];
E --> F[5. Calculate x = r * cos(θ)];
E --> G[5. Calculate y = r * sin(θ)];
F & G --> H[End: Coordinates (x, y)];
Read what Saavi narrates
Hey everyone, it's Saavi.
Imagine you're on a giant Ferris wheel, like the one at Chicago's Navy Pier. You've stopped somewhere on the ride, and your friend on the ground wants to take the perfect photo. They need your precise location. If you knew the wheel's radius and how far you've rotated, could you give them your exact x and y coordinates?
Today, we're going to learn how to do just that. We're building a bridge between angles and the familiar x, y coordinate grid. You'll see how any point on a circle can be described by just two things: the circle's radius, and an angle.
Let's walk through an example. Say we need to find the coordinates of a point on a circle with a radius of ten, at an angle of pi over three radians.
First, we need our formula, which is that the x coordinate is r times cosine of theta, and the y coordinate is r times sine of theta.
We know r is ten and theta is pi over three. This is one of our special angles. From our special triangles, we know that cosine of pi over three is one-half, and sine of pi over three is the square root of three over two.
Now, we just plug everything in. The x-coordinate is ten times one-half, which is five. The y-coordinate is ten times the square root of three over two, which simplifies to five times the square root of three. So, our final coordinates are (5, 5√3).
Now, here's a really common mistake I see every year. Students will correctly find the reference angle, but then forget to adjust the sign for the quadrant. For example, an angle in Quadrant three, where x and y are both negative, requires you to make both the sine and cosine values negative. Always check your quadrant first! It's a small step that saves a lot of points.
You're building a really powerful tool here. Keep practicing, and you'll see how this idea connects to so much of what's coming next. You've got this.
This flips the x and y coordinates. Cosine is always associated with the horizontal `x` value (adjacent side), and sine is always associated with the vertical `y` value (opposite side).
Always remember the alphabetical and geometrical order: `x` comes before `y`, `cos` comes before `sin` in the coordinate pair: `(r cos θ, r sin θ)`.
Writing the coordinates as just `(cos θ, sin θ)` is only correct for the *unit circle*, where the radius is 1. For any other circle, you are finding a point on the wrong circle.
Always start by writing the full formula `(r cos θ, r sin θ)` and then substitute your `r` value. Don't leave it out.
An angle of `3π/4` (in Q2) and `π/4` (in Q1) have the same reference angle, but their `x` coordinates have opposite signs. Ignoring the quadrant gives you the wrong location.
Before you calculate, determine the quadrant of your angle `θ`. Use the "All Students Take Calculus" rule to determine if `cos(θ)` and `sin(θ)` should be positive or negative.
It's easy to swap the values for 30° and 60° angles. This leads to incorrect `x` and `y` values.
Remember that `sin(π/6)` (the smaller angle) corresponds to the smaller value, `1/2`. `sin(π/3)` (the larger angle) corresponds to the larger value, `√3 / 2`.