# Chapter 2: Review of Trigonometry

> Radians and the unit circle, ASTC signs, trig beyond [0, 2π), graphs of the six trig functions, special-angle values, the identities that power calculus, and simple trig equations.

**Canonical:** https://calculus.kitzo.work/chapter/trigonometry  
**Level:** Starter  
**Estimated time:** 40 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Radian measure
- Unit circle
- ASTC signs
- Extending trig beyond [0,2π)
- Graphs of trig functions
- Trig identities
- Special-angle values

**Why this matters**

Trig is the language of everything that oscillates — waves, springs, alternating current, orbits. Calculus asks how fast those oscillations change, so before derivatives can say anything about sin and cos, you need the circle, the angles and the identities cold.

## 1 · Radians — angles measured by arc length

Degrees are a human invention. Radians are a **mathematical** definition: one radian is the angle whose arc length equals the radius. A full turn is 2π radians, so `π rad = 180°`. Multiply degrees by π/180 to convert, or radians by 180/π.

Why bother? In radians, facts like d/dx sin x = cos x come out clean; in degrees they carry an ugly π/180 factor. Calculus assumes radians unless told otherwise.

## 2 · The unit circle — one picture, all six functions

Draw the circle of radius 1 centered at the origin. A point on it at angle θ has coordinates `(cos θ, sin θ)`. That single sentence generates every trig value you will ever need.

  - **sin θ** = the y-coordinate · **cos θ** = the x-coordinate · **tan θ** = sin θ / cos θ (the slope of the radius line!)

  - **csc θ** = 1/sin θ · **sec θ** = 1/cos θ · **cot θ** = cos θ / sin θ

  - Key coordinates: 0 → (1, 0), π/6 → (√3/2, 1/2), π/4 → (√2/2, √2/2), π/3 → (1/2, √3/2), π/2 → (0, 1), π → (−1, 0), 3π/2 → (0, −1).

Since every point sits on the circle x² + y² = 1, the identity **sin²θ + cos²θ = 1** is just the circle's equation in disguise. It is the most-used identity in this course — use it to swap sin² for 1 − cos² (and back) whenever an integral or limit gets stuck.

## 3 · ASTC — signs without memorizing 24 facts

Going around the four quadrants, the signs of sin, cos and tan follow the pattern **All Students Take Calculus**:

  - **Q1** (0 to π/2): All three positive.

  - **Q2** (π/2 to π): **S**in positive only (cos, tan negative).

  - **Q3** (π to 3π/2): **T**an positive only (sin, cos negative).

  - **Q4** (3π/2 to 2π): **C**os positive only (sin, tan negative).

### Extending beyond [0, 2π)

Trig functions are **periodic**: sin and cos repeat every 2π, tan and cot every π. Any angle reduces mod 2π to a coterminal angle in [0, 2π). Negative angles go clockwise. For example, sin(7π/3) = sin(7π/3 − 2π) = sin(π/3) = √3/2, and cos(−π/4) = cos(π/4) = √2/2 (cos is even).

**Worked example**

**Example 1 — Evaluate cos(5π/6).**

Step 1: find the reference angle — the acute angle to the x-axis. 5π/6 sits in Q2, and its reference angle is π − 5π/6 = π/6.

Step 2: cos(π/6) = √3/2.

Step 3: apply ASTC — cosine is negative in Q2.

So cos(5π/6) = −√3/2. This reference-angle routine works for every quadrant and every sign question.

## 4 · Graphs — amplitude, period, shift

The graph of y = A sin(Bx + C) + D is a sine wave with four dials:

  - **|A|** = amplitude (height above/below the midline).

  - **B** sets the period: T = 2π/|B| for sin and cos, T = π/|B| for tan.

  - **−C/B** = phase shift (horizontal slide).

  - **D** = vertical shift (the midline itself).

The other four are reciprocals in disguise: sec and csc are U-shaped curves sitting on the peaks and troughs of cos and sin; cot is tan flipped and shifted.

**Watch out**

The phase shift is −C/B, not C. For y = sin(2x − π/3), factor first: sin(2(x − π/6)) — the shift is π/6 to the right, not π/3.

## 5 · Special angles and solving simple equations

Memorize one skeleton: at 0, π/6, π/4, π/3, π/2 the values of sin go 0, ½, √2/2, √3/2, 1 (cos walks the list backwards). Solving sin x = 1/2 then means: find the reference angle (π/6), place it in the quadrants where sin is positive (Q1 and Q2), and write x = π/6 + 2πk or x = 5π/6 + 2πk for any integer k. On a restricted interval you simply keep the solutions that land inside it.

**Worked example**

**Example 2 — Solve 2 cos²θ − 1 = 0 for θ in [0, 2π).**

Step 1: isolate: cos²θ = 1/2, so cos θ = ±√2/2.

Step 2: reference angle π/4; cosine is negative in Q2 and Q3, positive in Q1 and Q4.

Step 3: collect all four: θ = π/4, 3π/4, 5π/4, 7π/4.

Check by plugging back: cos(3π/4) = −√2/2, squared is 1/2. ✓

## 6 · The identity toolkit

Three families do all the work in calculus:

  - **Pythagorean:** sin²θ + cos²θ = 1 (and its two siblings after dividing by sin²θ and cos²θ: 1 + cot²θ = csc²θ and tan²θ + 1 = sec²θ).

  - **Angle sum:** sin(A ± B) = sin A cos B ± cos A sin B; cos(A ± B) = cos A cos B ∓ sin A sin B.

  - **Double angle:** sin 2θ = 2 sin θ cos θ; cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1.

These let you expand sin(x + π/6), collapse cos²x into (1 + cos 2x)/2, and simplify limits that otherwise look impossible. When stuck on a trig limit or integral, try: (1) rewrite everything in sin and cos, (2) apply sin²θ + cos²θ = 1, (3) use angle-sum to expand.

**Chapter formula sheet**

π rad = 180° · T = 2π/|B| (sin, cos) · T = π/|B| (tan)

sin²θ + cos²θ = 1 · 1 + tan²θ = sec²θ · 1 + cot²θ = csc²θ

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

sin 2θ = 2 sin θ cos θ · cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ

ASTC: All · Sin · Tan · Cos (quadrants 1–4) · Q1 (1,0), Q2 (0,1), Q3 (−1,0), Q4 (0,−1)

## Common mistakes

  - Phase shift read as C instead of −C/B — always factor B out of Bx + C first.

  - Giving only the Q1 solution to sin x = c; sine is also positive in Q2.

  - Forgetting the +2πk (general solution) or listing solutions outside the requested interval.

  - Thinking cos is odd — cos is even; sin is odd; tan is odd.

  - Using degrees inside calculus formulas; switch to radians before differentiating or integrating.

  - cos 2θ written only as cos²θ − sin²θ when 1 − 2sin²θ would kill the problem faster (or vice versa).

## Practice exercises

1. Convert 150° to radians.

- A. 5π/6
- B. 2π/3
- C. 3π/4
- D. π/6

2. Compute sin(π/6) + cos(π/3). Give your answer as a decimal.

3. What is the period of y = 3 sin(2x)? Write your answer in terms of π, e.g. 2pi.

4. cos(5π/6) equals:

- A. √3/2
- B. −√3/2
- C. 1/2
- D. −1/2

5. A wave has amplitude 4 and period π. If its equation is y = 4 sin(Bx), what is B?

6. The phase shift of y = sin(2x − π/3) is:

- A. π/3 to the right
- B. π/6 to the right
- C. π/3 to the left
- D. π/6 to the left

7. Use the identity to rewrite cos²θ in terms of cos 2θ. Enter e.g. (1+cos 2theta)/2 using cos2t or similar.

8. How many solutions does sin x = 1/2 have in [0, 2π)?

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