Lesson 6: The Unit Circle and Identities

The unit circle — radius 1, centre the origin — is the machine that generates all of trigonometry (AA Topic 3). A point P on it at angle θ from the positive x-axis has coordinates (cos θ, sin θ): cosine and sine are defined as the coordinates themselves, which is why they make sense for any angle, not just the acute angles of a triangle. Then tan θ = sin θ / cos θ is the slope of the radius OP. Every identity and every graph in the course is a shadow of this picture.

Exact values and quadrants

Angles land in four quadrants; the signs of sin, cos, tan follow "All, Sin, Tan, Cos" (positive in Q1, Q2, Q3, Q4 respectively). Reference-angle thinking turns any exact-value question into a first-quadrant one:

Worked Example 1 — one value determines the rest (past-paper style). Given sin θ = 3/5 with π/2 ≤ θ ≤ π, find cos θ and tan θ exactly.

Solution. On the unit circle, x² + y² = 1 with y = 3/5 gives cos²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5. The quadrant decides: θ is in Q2, where cosine is negative, so cos θ = −4/5. Then tan θ = (3/5)/(−4/5) = −3/4. The ± and its resolution by quadrant is the entire method — examiner reports show candidates routinely drop the negative root.

The identities

The Pythagorean identity sin²θ + cos²θ = 1 is the unit circle equation itself (the notation sin²θ means (sin θ)²). Dividing it by cos²θ or sin²θ yields the HL companions tan²θ + 1 = sec²θ and 1 + cot²θ = csc²θ, where sec, csc, cot are the reciprocal functions (HL). Double-angle formulas (core at both levels): sin 2θ = 2 sin θ cos θ and cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ — three forms of the last, chosen to match what the problem hands you. At HL the compound-angle formulas sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B generate the double angles and exact values like sin(π/12).

Worked Example 2 — proving an identity. Prove that (sin x + cos x)² = 1 + sin 2x.

Solution. Work from the more complicated side and aim at the other. LHS = sin²x + 2 sin x cos x + cos²x (expanding the square). Group: (sin²x + cos²x) + 2 sin x cos x = 1 + sin 2x, using the Pythagorean identity and then the double-angle formula read *backwards* — recognising 2 sin x cos x as sin 2x is the skill being tested. = RHS. ∎ A proof must transform one side into the other; substituting a test value verifies nothing in general, and IB awards zero for it.

Worked Example 3 — an equation that needs an identity. Solve 2cos²x + 3 sin x = 0 for 0 ≤ x ≤ 2π.

Solution. The equation mixes cos² and sin, so convert: cos²x = 1 − sin²x gives −2sin²x + 3 sin x + 2 = 0, i.e. 2sin²x − 3 sin x − 2 = 0. Factor: (2 sin x + 1)(sin x − 2) = 0, so sin x = −1/2 or sin x = 2. The second is impossible (|sin x| ≤ 1) — reject it in writing. sin x = −1/2 in Q3/Q4: x = π + π/6 = 7π/6 and x = 2π − π/6 = 11π/6. Solutions: 7π/6, 11π/6. Note the two-step discipline: identity to unify, algebra to factor, range to reject, circle to finish.

Graphs and inverse functions (HL)

y = sin x and y = cos x are the unit circle unwrapped: period 2π, range [−1, 1], cosine an even function (cos(−x) = cos x), sine odd. y = tan x has period π and vertical asymptotes where cos x = 0. Transformations follow Lesson 1's rules. Because trig functions are not one-to-one, their inverses (HL) need restricted domains: arcsin: [−1, 1] → [−π/2, π/2]; arccos: [−1, 1] → [0, π]; arctan: ℝ → (−π/2, π/2). When a GDC returns arcsin(0.5) = 0.524 it returns the *principal value* only — the equation sin x = 0.5 has infinitely many solutions, and recovering them all is your job, not the calculator's.

Real-world modelling: combining waves

Two sound waves of the same frequency add via the compound-angle formulas: sin x + cos x can be rewritten (compare with sin(x + π/4) expanded) as √2 sin(x + π/4) — one wave, bigger amplitude, shifted phase. This is how noise-cancelling headphones are tuned and how AC voltages combine: identities are engineering tools, not just exam decoration. In the exam experience below you will prove and use exactly this rewrite.

Common mistakes

• Writing sin²x as sin(x²) or "cancelling" sin from sin 2x / sin x.
• Proving identities by assuming the conclusion — working both sides independently to different expressions, or cross-multiplying as if it were an equation.
• Forgetting the impossible root in identity-based equations (sin x = 2 accepted silently).
• Dividing an equation by cos x (e.g. sin x = cos x → tan x = 1) without checking the lost case cos x = 0 separately — here x = π/2, 3π/2 are not solutions, but in other problems they are, and dividing erases them.
• Principal-value blindness: reporting arctan or arcsin outputs as the complete solution set.
• Degrees sneaking back in for identity work — identities are unit-free but your GDC evaluations are not.

IB Exam Experience

Question 1 [2 marks]. Given cos θ = −2/3 with π ≤ θ ≤ 3π/2, find tan θ exactly. *Marking notes:* M1 for sin²θ = 1 − 4/9 = 5/9 with the Q3 sign decision (sine negative in Q3); A1 for tan θ = (−√5/3)/(−2/3) = √5/2. Both sin and cos negative, tan positive — the sign journey is the mark scheme's focus.

Question 2 [3 marks]. Show that (1 − cos 2x)/(sin 2x) = tan x for x where both sides are defined. *Marking notes:* M1 for using cos 2x = 1 − 2sin²x (choosing the form that creates a factor); M1 for sin 2x = 2 sin x cos x and cancelling sin x; A1 for reaching sin x / cos x = tan x. Choosing the *wrong* double-angle form (2cos²x − 1) leads to a dead end — form selection is the tested skill.

Question 3 [4 marks]. Solve sin 2x = cos x for 0 ≤ x ≤ 2π. *Marking notes:* M1 for sin 2x = 2 sin x cos x; M1 for factorising cos x (2 sin x − 1) = 0 rather than dividing by cos x; A1 for cos x = 0: x = π/2, 3π/2; A1 for sin x = 1/2: x = π/6, 5π/6. Four solutions. Candidates who divide find only two and lose half the marks — a fixture of examiner reports.

Extended problem (Paper 2 style) [9 marks]. Let f(x) = sin x + cos x, 0 ≤ x ≤ 2π. (a) Show that f(x) = √2 sin(x + π/4). [3] (b) Hence state the maximum value of f and the x at which it occurs. [3] (c) Solve f(x) = 1. [3] *Marking notes:* (a) M1 for expanding √2 sin(x + π/4) with the compound formula: √2[sin x cos(π/4) + cos x sin(π/4)] = √2 × (√2/2)(sin x + cos x); A1 for the exact values cos(π/4) = sin(π/4) = √2/2; A1 for a clean concluding line "= sin x + cos x = f(x)". (b) A1: maximum √2; M1 for sin(x + π/4) = 1 → x + π/4 = π/2; A1: x = π/4. (c) M1: sin(x + π/4) = 1/√2; A1: x + π/4 = π/4 or 3π/4; A1: within the interval, x + π/4 = π/4, 3π/4 or 9π/4, giving x = 0, π/2 and 2π — the endpoint 2π also satisfies f(x) = 1 and must be included. Award the mark only for the complete set; endpoint blindness is the deliberate trap of this question.

Exploration (IA) idea: analyse a musical chord or a tide gauge record as a sum of sinusoids; use identities to combine components and compare the reconstruction with the original signal — a first taste of Fourier thinking with honest evaluation.

Going deeper: R sin(x + α) thinking and phase

Combining a sin x + b cos x into a single sinusoid (via the compound-angle formula read backwards) is HL-flavoured but illuminates every trig graph: amplitude √(a² + b²), phase shift from tan α = b/a with quadrant care. It converts "max/min of a trig combination" from a calculus problem into a reading-off exercise — though calculus (Lesson 11) also solves it, and comparing the two methods in revision is instructive.

Extra worked example. Express 3 sin x + 4 cos x as a single sine function, and hence find its maximum value and the smallest positive x at which it occurs.

Solution. Seek R sin(x + α) = R sin x cos α + R cos x sin α. Match: R cos α = 3, R sin α = 4. Then R = √(9 + 16) = 5 and tan α = 4/3 with α in Q1, α ≈ 0.927 rad. So 3 sin x + 4 cos x = 5 sin(x + 0.927). Maximum 5 when x + α = π/2, i.e. x ≈ 0.644 rad. Notice: no derivative taken. (Calculus check: derivative 3 cos x − 4 sin x = 0 → tan x = 3/4 → x ≈ 0.644 ✓ — two routes, one answer.)

Extra exam practice [3 marks]. Solve cos 2x = sin x for 0 ≤ x ≤ 2π. *Marking notes:* M1 for cos 2x = 1 − 2sin²x (choosing the sine form); M1: 2sin²x + sin x − 1 = 0 factorises as (2 sin x − 1)(sin x + 1) = 0; A1 for the solution set: sin x = 1/2 → π/6, 5π/6; sin x = −1 → 3π/2. Answer: π/6, 5π/6, 3π/2.