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): Sin positive only (cos, tan negative).
- Q3 (π to 3π/2): Tan positive only (sin, cos negative).
- Q4 (3π/2 to 2π): Cos 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).
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.
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.
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.
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).