IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read

Relationship Between Trigonometric Ratios

Sin, cos, and tan aren’t independent — they’re locked together by the Pythagorean identity and the complementary angle relationship sin θ = cos(90° − θ). Give me one ratio and the quadrant, and I can find every other ratio (and every double-angle ratio too).

📘 What you need to know

Complementary angles: sin θ = cos(90° − θ) and cos θ = sin(90° − θ).
This comes from a right triangle — the two non-right angles are 90° − θ and θ.
From one ratio to all others: Method 1 — draw a right triangle with the given ratio’s sides; Method 2 — use sin²θ + cos²θ = 1 and tan θ = sin θ / cos θ.
The square root in sin² = 1 − cos² gives ± — the quadrant tells you which sign.
Quadrant signs (CAST): Q1 all positive; Q2 sin only; Q3 tan only; Q4 cos only.
Once you have sin θ and cos θ, you can find every double-angle ratio (sin 2θ, cos 2θ, tan 2θ).
Equations like sin(x + a) = cos x can be solved by rewriting cos as sin of a complementary angle.

The complementary angle relationship

Complementary angle identities sin θ = cos(90° − θ) and cos θ = sin(90° − θ)

In a right triangle, the two non-right angles add to 90°. Whichever angle you call θ, the side that’s opposite to it is adjacent to (90° − θ), and vice versa. So opposite/hypotenuse for one angle equals adjacent/hypotenuse for the other — that’s why sin θ = cos(90° − θ).

In radians: sin θ = cos(π/2 − θ) and cos θ = sin(π/2 − θ). Same idea, π/2 instead of 90°.

Quadrant signs (CAST)

Quadrant	Range	sin θ	cos θ	tan θ
Q1	0° < θ < 90°	+	+	+
Q2	90° < θ < 180°	+	−	−
Q3	180° < θ < 270°	−	−	+
Q4	270° < θ < 360°	−	+	−

“CAST” — start at Q4 and go anticlockwise: Cos, All, Sin, Tan. That tells you which ratio is positive in each quadrant. Everything else is negative.

🧭 Recipe — find every ratio from one given value

Identify the quadrant from the given range — write down the signs of sin, cos, tan there.
Draw a right triangle using the magnitudes of the given ratio (e.g., sin θ = 3/5 → opp = 3, hyp = 5).
Find the third side using Pythagoras.
Read off the other ratios using SOHCAHTOA (or use sin²θ + cos²θ = 1 and tan = sin/cos).
Apply the quadrant signs to each ratio. Verify with the calculator if allowed.

Worked examples

WE 1

Simplify using complementary angles

Without using a calculator, simplify the expression cos 70° + sin 20°.

Step 1: Use cos θ = sin(90° − θ) on cos 70° cos 70° = sin(90° − 70°) = sin 20° Step 2: Substitute cos 70° + sin 20° = sin 20° + sin 20° 2 sin 20° decimal check: cos 70° + sin 20° ≈ 0.342 + 0.342 = 0.684 = 2 × 0.342 ✓

WE 2

Find cos θ and tan θ from sin θ (acute)

Given that sin θ = 7/25 and θ is acute, find the exact values of cos θ and tan θ.

Step 1: θ acute (Q1) → cos θ and tan θ are both positive Step 2: Right triangle with opp = 7, hyp = 25 → adj = √(25² − 7²) = √576 = 24 cos θ = adj/hyp = 24/25 tan θ = opp/adj = 7/24 cos θ = 2425, tan θ = 724 classic 7-24-25 right triangle

WE 3

Find sin α and tan α with α in Q3

Given that cos α = −8/17 and α is in the third quadrant (180° < α < 270°), find the exact values of sin α and tan α.

Step 1: Quadrant signs (Q3): sin α negative, tan α positive Step 2: Use sin²α = 1 − cos²α sin²α = 1 − 64/289 = 225/289 sin α = ±15/17 → take negative (Q3) sin α = −15/17 Step 3: Use tan α = sin α / cos α tan α = (−15/17) / (−8/17) = 15/8 sin α = −1517, tan α = 158 tan α positive in Q3 ✓ (negative ÷ negative = positive)

WE 4

Find sin 2β and cos 2β from tan β

Given that tan β = −5/12 and β is in the fourth quadrant (270° < β < 360°), find the exact values of sin 2β and cos 2β.

Step 1: Q4 → sin β negative, cos β positive Step 2: Right triangle with |opp| = 5, |adj| = 12 → hyp = √(25 + 144) = 13 sin β = −5/13, cos β = 12/13 Step 3: sin 2β = 2 sin β cos β = 2 · (−5/13) · (12/13) = −120/169 Step 4: cos 2β = 1 − 2 sin²β = 1 − 2(25/169) = 1 − 50/169 = 119/169 sin 2β = −120169, cos 2β = 119169 check: 120² + 119² = 14400 + 14161 = 28561 = 169² ✓

WE 5

Prove a complementary-angle identity

Show that sin²75° + sin²15° = 1.

Step 1: Use sin θ = cos(90° − θ) on sin 15° sin 15° = cos(90° − 15°) = cos 75° Step 2: Substitute into the LHS sin²75° + sin²15° = sin²75° + cos²75° Step 3: Apply the Pythagorean identity = 1 ✓ proved works for any θ + (90°−θ): they’re always complementary, so the squares sum to 1

WE 6

Solve sin(x + 20°) = cos x

Solve the equation sin(x + 20°) = cos x for 0° ≤ x ≤ 360°.

Step 1: Rewrite cos x using the complementary identity cos x = sin(90° − x) sin(x + 20°) = sin(90° − x) Step 2: For sin A = sin B: either A = B + 360°k, or A = 180° − B + 360°k Step 3: First case x + 20° = 90° − x → 2x = 70° → x = 35° Add 180° → x = 215° (also valid) Step 4: Second case x + 20° = 180° − (90° − x) = 90° + x 20° = 90° → no solution x = 35°, 215° check x = 35°: sin 55° = cos 35° ≈ 0.819 ✓

💡 Top tips

Always state the quadrant signs first before doing any algebra. That tells you which sign to take after a square root.
Right triangle method is fastest for rational ratios (like 7/25). Pythagorean identity is fastest when the answer involves surds.
Use Pythagorean triples when you can: 3-4-5, 5-12-13, 7-24-25, 8-15-17, 20-21-29 — they appear constantly.
Complementary angle trick: anytime you see sin and cos of angles that add to 90° (or π/2), one can be rewritten as the other.
For sin A = sin B equations: A = B or A = 180° − B (then add 360°k for periodicity).

⚠ Common mistakes

Forgetting the ± after a square root. sin α = ±√(1 − cos²α) — the quadrant tells you which sign.
Wrong quadrant signs. In Q3, both sin and cos are negative — but tan = sin/cos is positive (negative ÷ negative).
Confusing sin θ = cos(90° − θ) with sin(90° − θ) = sin θ. The second is wrong — sin(90° − θ) equals cos θ, not sin θ.
Mixing radians and degrees. The complementary identity in radians uses π/2 − θ, not 90° − θ.
Stopping after one solution in equations like sin(x + 20°) = cos x. Always check both cases A = B and A = 180° − B.

Next note: Linear Trigonometric Equations. The full method for solving sin(x) = k, cos(x) = k, tan(x) = k, and the transformed versions like sin(ax + b) = k — using the substitution trick and a careful interval shift.

Need help with Trig Ratio Relationships?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →