Pre-Calculus
Trigonometry

Degrees → Radians: multiply by π/180
Radians → Degrees: multiply by 180/π

0° = 0 rad  |  30° = π/6  |  45° = π/4
60° = π/3  |  90° = π/2  |  180° = π  |  360° = 2π

s = rθ    A = ½r²θ   (θ in radians)

On the unit circle (r=1):   P(θ) = (cos θ, sin θ)
tan θ = sin θ / cos θ  (when cos θ ≠ 0)

|A| = Amplitude  |  Period = 2π/|B|
Phase shift = C/B  |  Vertical shift = D

sin and cos: period 2π, domain ℝ, range [−1, 1]
tan: period π, asymptotes at x = π/2 + nπ
csc, sec: range (−∞,−1] ∪ [1,+∞)

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

csc θ = 1/sin θ  |  sec θ = 1/cos θ  |  cot θ = 1/tan θ
tan θ = sin θ/cos θ  |  cot θ = cos θ/sin θ

cos(−θ) = cos θ  (even)  |  sin(−θ) = −sin θ  (odd)
tan(−θ) = −tan θ  (odd)

sin(A±B) = sinA cosB ± cosA sinB
cos(A±B) = cosA cosB ∓ sinA sinB
tan(A±B) = (tanA ± tanB)/(1 ∓ tanA tanB)

sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ − sin²θ = 1−2sin²θ = 2cos²θ−1
tan 2θ = 2tanθ/(1−tan²θ)

arcsin: [−π/2, π/2]  |  arccos: [0, π]
arctan: (−π/2, π/2)

a/sin A = b/sin B = c/sin C

c² = a² + b² − 2ab cos C
(Generalization of Pythagorean theorem)

If θ₁ = arcsin(k), then θ₂ = π − θ₁
Both θ₁ and θ₂ are solutions (if in range)