Quick: what's sin(π/12)? It's not on the special-angle table. But π/12 = π/3 − π/4 — a difference of two angles you know cold. The sum/difference formulas turn unknown angles into arithmetic on known ones:
sin(π/3 − π/4) = sin(π/3)cos(π/4) − cos(π/3)sin(π/4) = (√6 − √2)/4
Identities are the algebra of trigonometry: rules for rewriting expressions into equivalent, more useful forms. Today's toolkit has exactly three drawers — Pythagorean, sum/difference, double-angle — and the exam opens all three.
From x² + y² = 1 on the unit circle:
sin²θ + cos²θ = 1
Divide through by cos²θ, or by sin²θ, for the two siblings:
tan²θ + 1 = sec²θ 1 + cot²θ = csc²θ
Primary use: one trig value + a quadrant → all trig values. Given sin θ = 3/5 with θ in QII: cos²θ = 1 − 9/25 = 16/25 → cos θ = −4/5 (negative in QII) → tan θ = −3/4, and reciprocals follow.
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B ← note the flipped sign!
Memory anchors: sine's expansion mixes the functions (sin·cos + cos·sin) and keeps the sign; cosine's matches them (cos·cos, sin·sin) and flips the sign. Check with a known value: cos(π/2) = cos(π/4 + π/4) = cos²(π/4) − sin²(π/4) = 1/2 − 1/2 = 0 ✓.
Uses: exact values at π/12, 5π/12, 7π/12 (combinations of π/6, π/4, π/3); proving co-function and shift identities (sin(θ + π/2) = cos θ falls right out); rewriting expressions to solve equations.
Set A = B = θ in the sum formulas:
sin(2θ) = 2 sin θ cos θ
cos(2θ) = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
Cosine's three versions are the point: pick the one whose variable matches the rest of your equation. Solving cos(2θ) + sin θ = 0? Use 1 − 2sin²θ, and the equation becomes a quadratic in sin θ alone:
1 − 2sin²θ + sin θ = 0 → 2sin²θ − sin θ − 1 = 0 → (2sin θ + 1)(sin θ − 1) = 0
Example: (1 − cos²θ)/sin θ = sin²θ/sin θ = sin θ (for sin θ ≠ 0 — note the exclusion; the original and simplified forms differ where sin θ = 0).
Problem: Given cos θ = −5/13 with θ in QIII, find sin θ and tan θ.
Solution: sin²θ = 1 − 25/169 = 144/169 → sin θ = ±12/13 → QIII: sin θ = −12/13. tan θ = sin/cos = (−12/13)/(−5/13) = 12/5 (positive — correct for QIII ✓).
Interpretation: The identity gives the magnitude; the quadrant gives the sign. Two-step ritual, no triangle required (though the 5-12-13 triangle is hiding in there).
Problem: Find the exact value of sin(7π/12).
Solution: 7π/12 = π/3 + π/4:
sin(π/3 + π/4) = sin(π/3)cos(π/4) + cos(π/3)sin(π/4)
= (√3/2)(√2/2) + (1/2)(√2/2) = (√6 + √2)/4
Sanity: 7π/12 ≈ 105°, sine near its peak → value should be just under 1: (√6+√2)/4 ≈ (2.449 + 1.414)/4 ≈ 0.966 ✓.
Interpretation: Twelfths of π decompose into sixths, quarters, thirds. Find the decomposition, run the formula, sanity-check the size.
Problem: θ is acute with sin θ = 3/5. Find sin(2θ) and cos(2θ).
Solution: Acute → cos θ = 4/5 (3-4-5).
sin(2θ) = 2(3/5)(4/5) = 24/25
cos(2θ) = 1 − 2(9/25) = 7/25
Check: (24/25)² + (7/25)² = (576 + 49)/625 = 625/625 = 1 ✓ — the double angle is still on the unit circle.
Interpretation: The Pythagorean check on your two answers takes seconds and catches sign/arithmetic slips — 2θ's values must themselves satisfy the circle equation.
Problem: Solve cos(2θ) = sin θ on [0, 2π).
Solution: Convert cos(2θ) to the sine version: 1 − 2sin²θ = sin θ →
2sin²θ + sin θ − 1 = 0 → (2sin θ − 1)(sin θ + 1) = 0
sin θ = 1/2 → θ = π/6, 5π/6 sin θ = −1 → θ = 3π/2
Solutions: {π/6, 5π/6, 3π/2}.
Interpretation: The identity choice did the work: matching the double angle to the equation's existing sin θ produced a one-variable quadratic. Choosing cos²θ − sin²θ instead would have left a two-variable mess.
1. (B). cos²θ = 1 − 9/25 = 16/25; QII → negative: −4/5. (A) ignores the quadrant.
2. (A). Divide sin² + cos² = 1 by cos²: sec²θ. (B) comes from dividing by sin².
3. (C). Mixed functions, sign kept: sin A cos B + cos A sin B. (A) is sin(A − B); (B) is cos(A + B); (D) is the distribution fantasy.
4. (A). Worked in Example 2: (√6 + √2)/4 ≈ 0.966 (near the 105° peak ✓). (B) is sin(π/12) — the difference version.
5. (B). 1 − 2sin²θ. (A) is its negative (that's 2cos²θ − 1 rewritten wrongly); (D) is backwards (cos² − sin², not sin² − cos²).
6. (A). 2(3/5)(4/5) = 24/25. (B) forgets the 2; (C) adds instead of multiplying.
7. (D). 16/25 − 9/25 = 7/25. (B) subtracts in the wrong order.
8. (B). 1 − cos²θ = sin²θ; divided by sin θ leaves sin θ. (C) would need a cos θ denominator.
9. (C). Difference → sign flips to plus: cos A cos B + sin A sin B. (A) is cos(A + B).
10. (D). cos(π/3 − π/4) = cos π/3 cos π/4 + sin π/3 sin π/4 = (√2 + √6)/4 = (√6 + √2)/4 ≈ 0.966 (15° — cosine near 1 ✓). (A) is cos(5π/12).
11. (C). 1 − 2sin²θ matches the equation's sin θ, giving a quadratic in one variable. (A)/(B) leave mixed variables; (D) is the sin(2θ) formula, not cos(2θ).
12. (FRQ-style, 6 points) (i) [2 pts] sin²θ = 1 − 64/289 = 225/289 → QIV: sin θ = −15/17; tan θ = −15/8. (ii) [2 pts] sin(2θ) = 2(−15/17)(8/17) = −240/289; cos(2θ) = 2(64/289) − 1 = 128/289 − 289/289 = −161/289. Check: 240² + 161² = 57,600 + 25,921 = 83,521 = 289² ✓. (iii) [2 pts] sin(2θ) < 0 and cos(2θ) < 0 → 2θ is in Quadrant III (both coordinates negative).
12. (FRQ-style) 🚫 θ is in Quadrant IV with cos θ = 8/17. (i) Find sin θ and tan θ. (ii) Find sin(2θ) and cos(2θ), and verify they satisfy sin² + cos² = 1. (iii) Determine the quadrant of the angle 2θ. Justify using your signs from (ii).
(a) Using the sum formula, show that sin(θ + π/2) = cos θ for all θ.
(b) Find the exact value of cos(5π/12). (Hint: 5π/12 = π/6 + π/4.)
(c) Solve on [0, 2π): sin(2θ) = cos θ. (Express sin(2θ) with the double-angle formula, then factor — do not divide by cos θ.)
(a) [2 pts] sin(θ + π/2) = sin θ cos(π/2) + cos θ sin(π/2) [1 pt formula] = sin θ·0 + cos θ·1 = cos θ [1 pt evaluation].
(b) [2 pts] cos(π/6 + π/4) = cos(π/6)cos(π/4) − sin(π/6)sin(π/4) [1 pt — with the minus sign] = (√3/2)(√2/2) − (1/2)(√2/2) = (√6 − √2)/4 [1 pt]. (Size check: 5π/12 ≈ 75°, cosine small and positive: ≈ 0.259 ✓)
(c) [2 pts] 2 sin θ cos θ = cos θ → 2 sin θ cos θ − cos θ = 0 → cos θ(2 sin θ − 1) = 0 [1 pt factoring, not dividing] → cos θ = 0: θ = π/2, 3π/2; sin θ = 1/2: θ = π/6, 5π/6. Solutions: {π/6, π/2, 5π/6, 3π/2} [1 pt all four]. (Dividing by cos θ would have silently discarded π/2 and 3π/2.)
1. (B). cos²θ = 1 − 9/25 = 16/25; QII → negative: −4/5. (A) ignores the quadrant.
2. (A). Divide sin² + cos² = 1 by cos²: sec²θ. (B) comes from dividing by sin².
3. (C). Mixed functions, sign kept: sin A cos B + cos A sin B. (A) is sin(A − B); (B) is cos(A + B); (D) is the distribution fantasy.
4. (A). Worked in Example 2: (√6 + √2)/4 ≈ 0.966 (near the 105° peak ✓). (B) is sin(π/12) — the difference version.
5. (B). 1 − 2sin²θ. (A) is its negative (that's 2cos²θ − 1 rewritten wrongly); (D) is backwards (cos² − sin², not sin² − cos²).
6. (A). 2(3/5)(4/5) = 24/25. (B) forgets the 2; (C) adds instead of multiplying.
7. (D). 16/25 − 9/25 = 7/25. (B) subtracts in the wrong order.
8. (B). 1 − cos²θ = sin²θ; divided by sin θ leaves sin θ. (C) would need a cos θ denominator.
9. (C). Difference → sign flips to plus: cos A cos B + sin A sin B. (A) is cos(A + B).
10. (D). cos(π/3 − π/4) = cos π/3 cos π/4 + sin π/3 sin π/4 = (√2 + √6)/4 = (√6 + √2)/4 ≈ 0.966 (15° — cosine near 1 ✓). (A) is cos(5π/12).
11. (C). 1 − 2sin²θ matches the equation's sin θ, giving a quadratic in one variable. (A)/(B) leave mixed variables; (D) is the sin(2θ) formula, not cos(2θ).
12. (FRQ-style, 6 points) (i) [2 pts] sin²θ = 1 − 64/289 = 225/289 → QIV: sin θ = −15/17; tan θ = −15/8. (ii) [2 pts] sin(2θ) = 2(−15/17)(8/17) = −240/289; cos(2θ) = 2(64/289) − 1 = 128/289 − 289/289 = −161/289. Check: 240² + 161² = 57,600 + 25,921 = 83,521 = 289² ✓. (iii) [2 pts] sin(2θ) < 0 and cos(2θ) < 0 → 2θ is in Quadrant III (both coordinates negative).
🎯 Exam tip: You get no formula sheet on AP Precalculus. Rebuild rather than recall: the two Pythagorean siblings come from dividing the original; the double angles come from setting A = B in the sum formulas; any half-remembered sign gets tested against θ = 0 or π/2 in five seconds. Nothing in this lesson needs to be trusted to memory alone.