IB Maths AI HL Trig Identities & Equations Paper 1 & 2 Pythagorean & tan ID ~8 min read

Simple Identities

Two identities run through every trig question: the Pythagorean sin²θ+cos²θ=1 and the tan identity tan θ = sin θ / cos θ. Together they let you find missing trig values, rewrite mixed equations as quadratics, and prove other identities. Both are in the formula booklet.

📘 What you need to know

Pythagorean identity: sin²θ + cos²θ = 1 (formula booklet). The notation sin²θ means (sin θ)².
tan identity: tan θ = sin θcos θ (formula booklet). Undefined where cos θ = 0.
Useful rearrangements: sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ.
Finding the missing trig value: given one of sin/cos, use the Pythagorean ID to find the other (the quadrant decides the sign).
Rewriting mixed equations: an equation with both sin²x and cos x (or both cos²x and sin x) can be turned into a quadratic in a single trig function using sin² = 1 − cos² (or the other way round).
Proving identities: start from one side (usually the more complicated), substitute one of the two identities, simplify until it matches the other side.

Where the two identities come from

Draw a right triangle with hypotenuse 1 and an acute angle θ. By right-angled trig (SOHCAHTOA), the side opposite θ has length sin θ and the side adjacent has length cos θ. Pythagoras’ theorem on this triangle gives sin²θ+cos²θ = 1² = 1; the tan identity follows from tan θ = opposite/adjacent = sin θ/cos θ. Although the derivation uses an acute angle, both identities hold for every angle (positive, negative, beyond 360°) because of how the unit circle extends sin and cos.

Right-triangle derivation: with hypotenuse 1, the two legs are sin θ and cos θ. Pythagoras yields sin²θ + cos²θ = 1; tan θ = opposite/adjacent gives tan θ = sin θ / cos θ.

The two identities sin²θ + cos²θ = 1 · tan θ = sin θcos θ rearrangements: sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ

Three things the identities let you do

Each identity has a typical use case. Find the missing trig value: given sin θ (or cos θ), use sin²+cos²=1 to find the other; the quadrant fixes the sign. Rewrite an equation as a quadratic: if the equation has both squared and unsquared trig terms (e.g. sin²x alongside cos x), use sin²x = 1 − cos²x to put everything in cos x, then solve as a quadratic. Prove or simplify identities: take the messier side, swap in one of the two identities, simplify until it matches the other side.

“Show that” technique: when you must convert one expression into another, look at what’s missing. If tan has disappeared from the target, it was probably replaced by sin/cos. If sin² disappeared, it was probably replaced by 1 − cos².

🧭 Recipe — using the simple identities

Spot the mix: sin² + cos? Or both sin and cos with tan involved? Identify which identity will unify the terms.
Substitute: replace sin²x with 1 − cos²x (or vice versa); replace tan x with sin x/cos x.
Simplify: expand, collect like terms, often into a quadratic in a single trig function.
Solve or rearrange: factor the quadratic; for unknown trig values, decide signs by quadrant.
Check: substitute one value back into the original to confirm.

Worked examples

WE 1

Find cos θ from sin θ (acute)

Given that sin θ = 513 and that θ is acute, find the exact value of cos θ.

apply sin²θ + cos²θ = 1 cos²θ = 1 − (5/13)² = 1 − 25/169 = 144/169 take the square root cos θ = ±√(144/169) = ±12/13 acute → Q1 → cos θ positive cos θ = 12/13 5-12-13 Pythagorean triple — clean rational answer.

WE 2

Find tan θ from cos θ (obtuse)

Given that cos θ = −35 and that 90° < θ < 180°, find the exact value of tan θ.

find sin θ via Pythagorean ID sin²θ = 1 − (−3/5)² = 1 − 9/25 = 16/25 sin θ = ±4/5 Q2 → sin θ positive sin θ = 4/5 tan = sin/cos tan θ = (4/5) / (−3/5) tan θ = −4/3 3-4-5 triple; sign of tan is negative in Q2 — consistent with ASTC.

WE 3

Rewrite as a quadratic in cos x

Show that the equation 3 sin²x − 7 cos x = 2 can be written in the form a cos²x + b cos x + c = 0, where a, b, and c are integers to be found.

substitute sin²x = 1 − cos²x 3(1 − cos²x) − 7cos x = 2 expand 3 − 3cos²x − 7cos x = 2 move all to one side −3cos²x − 7cos x + 1 = 0 multiply by −1 to clear 3cos²x + 7cos x − 1 = 0 a = 3, b = 7, c = −1

WE 4

Solve an equation using the Pythagorean identity

Solve the equation 2 cos²x = 1 + sin x for x in the interval 0° ≤ x ≤ 360°.

substitute cos²x = 1 − sin²x 2(1 − sin²x) = 1 + sin x 2 − 2sin²x = 1 + sin x rearrange 2sin²x + sin x − 1 = 0 factor as a quadratic in sin x (2sin x − 1)(sin x + 1) = 0 two cases sin x = 1/2 → x = 30°, 150° sin x = −1 → x = 270° x = 30°, 150°, 270° three solutions because sin x = −1 contributes only one in [0°, 360°].

WE 5

Use both identities together — find sin & cos from tan

Given that tan θ = 724 and that θ is acute, use the two identities to find the exact values of sin θ and cos θ.

parametrise using tan = sin/cos let sin θ = 7k, cos θ = 24k (so sin/cos = 7/24 ✓) apply Pythagorean identity (7k)² + (24k)² = 1 49k² + 576k² = 1 625k² = 1 → k² = 1/625 acute → both sin, cos positive → k positive k = 1/25 sin θ = 7/25, cos θ = 24/25 7-24-25 Pythagorean triple; check: (7/25)² + (24/25)² = 49/625 + 576/625 = 1 ✓

WE 6

Prove an identity

Show that sin θ · tan θ + cos θ = 1cos θ.

start from LHS — use tan = sin/cos LHS = sin θ · (sin θ/cos θ) + cos θ = sin²θ/cos θ + cos θ common denominator = sin²θ/cos θ + cos²θ/cos θ = (sin²θ + cos²θ)/cos θ apply Pythagorean identity = 1/cos θ = RHS ✓ identity proved classic two-step proof: substitute the tan identity, then the Pythagorean one.

💡 Top tips

Look at what’s missing in the target to spot which identity to apply: tan gone? It was rewritten as sin/cos. sin² gone? It was swapped for 1 − cos².
Quadrant first, then sign: after taking a square root, the quadrant (or stated range of θ) tells you whether to pick + or −.
Watch for hidden Pythagorean triples: 3-4-5, 5-12-13, 7-24-25, 8-15-17 — common in identity problems with nice rational answers.
Don’t forget the ± when you square-root sin² or cos² — both signs are mathematically valid; the quadrant chooses.
Prove by starting from one side (usually the messier one), substitute, then simplify; don’t manipulate both sides at once.

⚠ Common mistakes

Dropping the ± when square-rooting sin² θ or cos² θ — remember to consider both signs and use the quadrant to choose.
Mis-reading sin²θ as sin(θ²) — the notation means (sin θ)².
Sign errors in Q2/Q3/Q4 — check with ASTC after computing.
Forgetting tan = sin/cos when “tan” disappears in a “show that” target equation — that’s usually exactly the substitution you need.
Mixing degrees and radians when verifying numerically — match the calculator mode to the question.

Next up — Graphs of Trigonometric Functions. With the identities sorted, the focus shifts to visualising sin, cos, and tan as periodic graphs: amplitude, period, asymptotes, and how the key features sit on the axes.

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