Trigonometric Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 1.2: “Basic Classes of Functions” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Reading the Unit Circle
On the unit circle (radius 1, centered at origin), a point at angle $\theta$ measured counterclockwise from the positive $x$-axis has coordinates $(\cos\theta, \sin\theta)$.
Predict without computing:
- What are $\cos(0)$ and $\sin(0)$?
- What are $\cos(\pi/2)$ and $\sin(\pi/2)$?
- At what second angle $\theta > 0$ does $\sin\theta = 0$?
Reason from the picture: $\cos$ is the $x$-coordinate, $\sin$ is the $y$-coordinate.
Prerequisite Check
Quick Reference
The six trigonometric functions.
| Function | Definition | Domain |
|---|---|---|
| $\sin\theta$ | $y$-coordinate on unit circle | All reals |
| $\cos\theta$ | $x$-coordinate on unit circle | All reals |
| $\tan\theta = \sin\theta/\cos\theta$ | $\cos\theta \neq 0$ | |
| $\csc\theta = 1/\sin\theta$ | $\sin\theta \neq 0$ | |
| $\sec\theta = 1/\cos\theta$ | $\cos\theta \neq 0$ | |
| $\cot\theta = \cos\theta/\sin\theta$ | $\sin\theta \neq 0$ |
Key values.
| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
| $0$ | $0$ | $1$ | $0$ |
| $\pi/6$ | $1/2$ | $\sqrt{3}/2$ | $1/\sqrt{3}$ |
| $\pi/4$ | $\sqrt{2}/2$ | $\sqrt{2}/2$ | $1$ |
| $\pi/3$ | $\sqrt{3}/2$ | $1/2$ | $\sqrt{3}$ |
| $\pi/2$ | $1$ | $0$ | undefined |
Key Concepts
1. The Unit Circle Definition
$\cos\theta$ and $\sin\theta$ are the $x$- and $y$-coordinates of the point at angle $\theta$ on the unit circle. This definition:
- Applies to all real $\theta$, not just acute angles.
- Makes $\sin^2\theta + \cos^2\theta = 1$ geometric (the unit circle equation).
- Makes $\sin(\theta + 2\pi) = \sin\theta$ geometric (one full rotation).
Prediction check. $\cos(0) = 1$, $\sin(0) = 0$ (rightmost point). $\cos(\pi/2) = 0$, $\sin(\pi/2) = 1$ (top). $\sin\theta = 0$ again at $\theta = \pi$ (leftmost point).
2. Two Representations: Unit Circle and Graph
Unit circle. As $\theta$ increases counterclockwise, the point moves around the circle. The height ($\sin$) and horizontal distance ($\cos$) oscillate between $-1$ and $1$.
Graph. $y = \sin\theta$ is a wave: starts at 0, rises to 1 at $\pi/2$, returns to 0 at $\pi$, falls to $-1$ at $3\pi/2$, completes at $2\pi$. $y = \cos\theta$ is the same wave shifted left by $\pi/2$.
Translation prompt. For $y = \sin(B\theta)$: to complete one full cycle, $B\theta$ must increase by $2\pi$, so $\theta$ must increase by $2\pi/B$. The period is $2\pi/B$. Verify from the unit circle: one full rotation of the point requires the argument to change by $2\pi$.
3. Graph Properties of $y = A\sin(B\theta + C) + D$
- Amplitude: $|A|$.
- Period: $2\pi/|B|$.
- Phase shift: $-C/B$ (horizontal displacement).
- Vertical shift: $D$.
Example 1. $y = 3\sin(2\theta - \pi)$: amplitude 3, period $\pi$, phase shift $\pi/2$ (right).
4. Key Identities
- Pythagorean: $\sin^2\theta + \cos^2\theta = 1$; dividing by $\cos^2\theta$: $\tan^2\theta + 1 = \sec^2\theta$
- Even/odd: $\cos(-\theta) = \cos\theta$ (even), $\sin(-\theta) = -\sin\theta$ (odd)
- Double angle: $\sin(2\theta) = 2\sin\theta\cos\theta$; $\cos(2\theta) = \cos^2\theta - \sin^2\theta$
5. Ask Why: Why Are Radians Natural?
Radians define arc length $= r\theta$ on a circle of radius $r$. This makes $\dfrac{d}{dx}[\sin x] = \cos x$ (no conversion factor). If degrees were used, the derivative would be $\dfrac{\pi}{180}\cos x$, cluttering every calculus formula. Calculus always uses radians.
Named Misconception: trig-as-algebra-symbols
Students sometimes write $\sin(A + B) = \sin A + \sin B$, treating $\sin$ as a multiplier that distributes over addition.
One example shows this breaks: $\sin(\pi/6 + \pi/6) = \sin(\pi/3) = \sqrt{3}/2 \approx 0.866$, but $\sin(\pi/6) + \sin(\pi/6) = 1$. The two values differ.
The correct formula is $\sin(A+B) = \sin A\cos B + \cos A\sin B$ -- a product, not a sum.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Distributing trig over addition | $\sin(x + \pi/6) = \sin x + \sin(\pi/6)$ | Use the addition formula: $\sin x\cos(\pi/6) + \cos x\sin(\pi/6)$ |
| Using degrees in calculus formulas | $(\sin x)' = \cos x$ with $x$ in degrees | Calculus formulas require $x$ in radians |
| Confusing $\sin^2 x$ and $\sin(x^2)$ | “Sin squared equals sin of x squared” | $\sin^2 x = (\sin x)^2$; $\sin(x^2)$ is a composition |
Common Misconceptions
trigonometric expressions can be simplified using ordinary algebra rules, such as $\sin(x + y) = \sin x + \sin y$.
This is the trig-as-algebra-symbols error. Sine is not a multiplier that distributes over addition. $\sin(x + y)$ is a composition of the sine function with the sum $x + y$; it evaluates to $\sin x \cos y + \cos x \sin y$ by the addition formula. The equation $\sin(30° + 60°) = \sin 90° = 1$, whereas $\sin 30° + \sin 60° = 0.5 + 0.866 \approx 1.366 \neq 1$. This error is the most common source of wrong answers in trigonometric calculations.
a radian is just another unit for measuring angles, like a degree, so the conversion is simply a scale factor.
This is the radian-as-arc-length-in-radius-units error -- in reverse. A radian is not an arbitrary unit: it is defined as the angle that subtends an arc length equal to the radius. This makes radian measure dimensionless (arc length divided by radius, both in meters, cancels the units), which is exactly why calculus formulas such as $(\sin x)' = \cos x$ hold only in radians. In degrees, the formula becomes $(\sin x°)' = \tfrac{\pi}{180} \cos x°$, with an extra factor from the conversion. The value of radian measure is that it fuses angle to arc length in the simplest possible ratio.
Leveled Practice
Level 1 -- Key Values and Symmetry
Problem 1. Without a calculator: (a) $\sin(5\pi/6)$, (b) $\cos(3\pi/4)$, (c) $\tan(\pi/3)$.
Show answer
(a) $\sin(\pi - \pi/6) = \sin(\pi/6) = 1/2$.
(b) $\cos(\pi - \pi/4) = -\cos(\pi/4) = -\sqrt{2}/2$.
(c) $\sqrt{3}/2 \div 1/2 = \sqrt{3}$.
Problem 2. Verify $1 + \tan^2\theta = \sec^2\theta$ by dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$.
Show answer
$\dfrac{\sin^2\theta + \cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}$ gives $\tan^2\theta + 1 = \sec^2\theta$.
Level 2 -- Graph Properties
Problem 3. State the amplitude, period, and phase shift of $y = -2\cos(3x + \pi/2)$.
Show answer
Amplitude $= 2$, period $= 2\pi/3$, phase shift $= -(\pi/2)/3 = -\pi/6$ (left by $\pi/6$).
Problem 4. Solve $\tan\theta = -1$ on $[0, 2\pi)$.
Show answer
$\tan\theta = -1$ at $\theta = 3\pi/4$ (second quadrant, where $\sin > 0$, $\cos < 0$) and $\theta = 7\pi/4$ (fourth quadrant, where $\sin < 0$, $\cos > 0$).
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 5 (Extension).
(a) (Floor) Show that $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$ from the double-angle formula.
(b) (Mid) Prove $\sin(-\theta) = -\sin\theta$ using the unit-circle definition.
(c) (Ceiling) Show that $A\sin x + B\cos x = R\sin(x+\phi)$ where $R = \sqrt{A^2+B^2}$ by expanding the right side and matching coefficients.
Show answer
(a) $\cos(2\theta) = 1 - 2\sin^2\theta \Rightarrow \sin^2\theta = \frac{1-\cos(2\theta)}{2}$.
(b) Angle $-\theta$ reflects the unit-circle point across the $x$-axis: $y$-coordinate flips sign, $x$-coordinate stays. So $\sin(-\theta) = -\sin\theta$.
(c) $R\sin(x+\phi) = R(\sin x\cos\phi + \cos x\sin\phi)$. Match $A = R\cos\phi$, $B = R\sin\phi$. Then $A^2+B^2 = R^2$, so $R = \sqrt{A^2+B^2}$, and $\tan\phi = B/A$.
Mastery Checklist
Mental Model
The unit circle is the complete picture. Every trig value, every identity, every symmetry follows from the geometry of a circle of radius 1. Sine is height, cosine is horizontal distance, and they oscillate as the angle grows.
The graphs of $\sin$ and $\cos$ are the unit-circle motion “unrolled” into the $\theta$-axis: trace the height (or width) of the moving point as $\theta$ increases steadily, and the oscillating wave appears.
Connections
Within MATH161
- Continuity: $\sin$ and $\cos$ are continuous everywhere; the derived functions have excluded points at zeros of $\sin$ or $\cos$.
- Derivatives: $(\sin x)' = \cos x$, $(\cos x)' = -\sin x$ -- proved using special trig limits and the radian definition.
- Integration: $\int \sin x\,dx = -\cos x + C$, directly from the derivative rule in reverse.