Introduction to Sine and Cosine Function Values

Welcome to Topic 3.3! In the previous chapter (3.2), we established that the sine and cosine functions are defined by the coordinates of a point moving around a unit circle. In this chapter, we are going to learn how to find the exact values of these functions for specific angles. This is a vital skill for the AP Precalculus exam because many questions will require you to work without a calculator. Don't worry if the numbers look intimidating at first—once you see the patterns, it becomes much easier!

1. The Foundation: Coordinates on the Unit Circle

Recall that on a unit circle (a circle with a radius of \(1\)), the coordinates of any point \(P\) at an angle \(\theta\) are given by \((x, y) = (\cos(\theta), \sin(\theta))\). This means:

  • The cosine of an angle is the \(x\)-coordinate.
  • The sine of an angle is the \(y\)-coordinate.

Because the radius is \(1\), all values for sine and cosine must fall between \(-1\) and \(1\). If you ever get a value like \(1.5\), you know something went wrong!

2. Special Angles and Exact Values

The AP Exam expects you to know the exact values for "special angles" in radians. These are angles that are multiples of \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\).

The "Big Three" Values

There are three main numbers you will see over and over again. Learning these is the "secret code" to mastering trigonometry:

  1. Smallest: \(\frac{1}{2}\) (or \(0.5\))
  2. Middle: \(\frac{\sqrt{2}}{2}\) (approx \(0.707\))
  3. Largest: \(\frac{\sqrt{3}}{2}\) (approx \(0.866\))
Common Values in Quadrant I:

At \(0\) radians: \(\cos(0) = 1\), \(\sin(0) = 0\)
At \(\frac{\pi}{6}\) (\(30^\circ\)): \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)
At \(\frac{\pi}{4}\) (\(45^\circ\)): \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
At \(\frac{\pi}{3}\) (\(60^\circ\)): \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)
At \(\frac{\pi}{2}\) (\(90^\circ\)): \(\cos\left(\frac{\pi}{2}\right) = 0\), \(\sin\left(\frac{\pi}{2}\right) = 1\)

Quick Trick: Notice that as the angle increases from \(0\) to \(\frac{\pi}{2}\), the sine values increase (\(0 \to \frac{1}{2} \to \frac{\sqrt{2}}{2} \to \frac{\sqrt{3}}{2} \to 1\)) while the cosine values decrease.

3. Reference Angles and Quadrants

What happens when the angle is outside the first quadrant? We use reference angles. A reference angle is the positive acute angle between the terminal side of the angle and the x-axis.

Signs in Different Quadrants

The value of the sine or cosine will stay the same as its reference angle, but the sign (+ or -) depends on the quadrant:

  • Quadrant I (Top Right): All are positive. (\(x > 0, y > 0\))
  • Quadrant II (Top Left): Sine is positive, Cosine is negative. (\(x < 0, y > 0\))
  • Quadrant III (Bottom Left): Both are negative. (\(x < 0, y < 0\))
  • Quadrant IV (Bottom Right): Cosine is positive, Sine is negative. (\(x > 0, y < 0\))

Mnemonic: All Students Take Calculus
(All positive in Q1, Sine in Q2, Tangent in Q3, Cosine in Q4).

4. The Pythagorean Identity

One of the most important tools in Unit 3 is the Pythagorean Identity. Because the coordinates \((x, y)\) lie on a circle with radius \(1\), they must satisfy the equation \(x^2 + y^2 = 1\). Substituting our trig functions, we get:

\(\sin^2(\theta) + \cos^2(\theta) = 1\)

Example: If you know that \(\sin(\theta) = \frac{3}{5}\) and \(\theta\) is in Quadrant II, what is \(\cos(\theta)\)?

  1. Plug into the identity: \((\frac{3}{5})^2 + \cos^2(\theta) = 1\)
  2. Square the fraction: \(\frac{9}{25} + \cos^2(\theta) = 1\)
  3. Subtract: \(\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}\)
  4. Take the square root: \(\cos(\theta) = \pm\frac{4}{5}\)
  5. Since we are in Quadrant II, cosine must be negative: \(\cos(\theta) = -\frac{4}{5}\)

5. Working with Periodic Values

As discussed in Topic 3.1, sine and cosine are periodic. This means they repeat their values every \(2\pi\) radians. If you are asked to find \(\sin(\frac{13\pi}{6})\), you can subtract \(2\pi\) (which is \(\frac{12\pi}{6}\)) to find that it has the same value as \(\sin(\frac{\pi}{6})\).

Key Takeaway: \(\sin(\theta + 2\pi k) = \sin(\theta)\) and \(\cos(\theta + 2\pi k) = \cos(\theta)\) for any integer \(k\).

6. Calculator vs. No-Calculator

On the AP Precalculus exam, you must be careful about which section you are in:

  • Section I Part A & Section II Part B: No calculator allowed. You must provide exact values (like \(\frac{\sqrt{3}}{2}\)). Do not use decimals here!
  • Section I Part B & Section II Part A: Graphing calculator required. Make sure your calculator is in Radian Mode. Unless specified otherwise, round your decimal answers to three decimal places.

Common Mistakes to Avoid

1. Degree vs. Radian: The AP Exam defaults to radians. If you see \(\sin(30)\), it means 30 radians, not 30 degrees! Always check your calculator mode.
2. Mixing up Sine and Cosine: Remember, Cosine is X (think "alphabetical order": C comes before S, X comes before Y).
3. Forgetting the Sign: Always identify which quadrant your angle is in before finalizing your answer. A correct number with the wrong sign is still incorrect!

Summary Checklist

  • Do I know the sine and cosine values for \(\frac{\pi}{6}, \frac{\pi}{4}, \text{ and } \frac{\pi}{3}\)?
  • Can I find a reference angle for an angle in any quadrant?
  • Do I know the sign (+/-) of sine and cosine in all four quadrants?
  • Can I use \(\sin^2(\theta) + \cos^2(\theta) = 1\) to find a missing value?