Free Trigonometry questions for A Level Mathematics (9709), part of Pure Mathematics 3. Each question has the answer and the marking points an examiner would look for.
Generate a full Trigonometry worksheet →All Pure Mathematics 3 questionsProve the identity sin 2θ1 + cos 2θ ≡ tan θ.
Hence find the exact value of ∫0π/3 sin 2θ1 + cos 2θ dθ.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 2 sin θ cos θ ÷ 2 cos2 θ sin 2θ = 2 sin θ cos θ B1 1 + cos 2θ = 2cos2 θ B1 cancelled to tan θ A1 | 3 |
| (b) | ln 2 ∫ tan θ dθ = −ln|cos θ| M1 correct limits M1 ln 2 A1 | 3 |
Express 12 cos θ + 5 sin θ in the form R cos(θ − α), where R > 0 and 0° < α < 90°. Give α correct to 2 decimal places.
Hence solve the equation 12 cos θ + 5 sin θ = 6 for 0° < θ < 360°.
State the greatest value of 116 − 12 cos θ − 5 sin θ and the value of θ, for 0° < θ < 360°, at which it occurs.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 13 cos(θ − 22.62°) R = 13 B1 tan α = 512 M1 α = 22.62° A1 | 3 |
| (b) | 85.1°, 320.1° cos(θ − 22.62) = 613 M1 θ − α = 62.51° A1 second solution method (360° − or negative angle) M1 85.1°, 320.1° A1 | 4 |
| (c) | 13 at θ = 22.62° 13 B1 22.62° B1 | 2 |
Solve cos 2θ + 3 sin θ = 2 for 0° < θ < 360°.
| Part | Answer and marking points | Marks |
|---|---|---|
| 30°, 90°, 150° double angle formula M1 2 sin2θ − 3 sin θ + 1 = 0 A1 sin θ = ½ or 1 A1 30°, 90°, 150° A1 | 4 |
[For a cubic equation ax3 + bx2 + cx + d = 0 with roots p, q and r: p + q + r = −ba, pq + qr + rp = ca and pqr = −da.]
Show that tan 3θ ≡ 3t − t31 − 3t2, where t = tan θ.
Show that tan π9 − tan 2π9 + tan 4π9 = 3√3.
Find the exact value of tan2 π9 + tan2 2π9 + tan2 4π9.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | tan(2θ + θ) with tan 2θ = 2t1 − t2 tan 3θ = tan 2θ + tan θ1 − tan 2θ tan θ M1 tan 2θ = 2t1 − t2 substituted M1 simplified to the given result A1 | 3 |
| (b) | roots of t3 − 3√3t2 − 3t + √3 = 0 are tan π9, tan 4π9, tan 7π9 considers tan 3θ = √3, with 3θ = π3, 4π3, 7π3 M1 t3 − 3√3t2 − 3t + √3 = 0 A1 roots tan π9, tan 4π9, tan 7π9, distinct A1 tan 7π9 = −tan 2π9 B1 sum of roots = 3√3, given result A1 | 5 |
| (c) | 33 sum of products in pairs = −3 B1 sum of squares = (3√3)2 − 2(−3) M1 33 A1 | 3 |
Show that cos 2θ + 3 sin θ = 2 can be written as 2 sin2 θ − 3 sin θ + 1 = 0.
Hence solve the equation for 0° ≤ θ ≤ 360°.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | cos 2θ = 1 − 2 sin2 θ correct double-angle identity M1 rearranged to the given form A1 | 2 |
| (b) | 30°, 90°, 150° factorise: (2 sin θ − 1)(sin θ − 1) = 0 M1 sin θ = ½ or 1 A1 30°, 90°, 150° only A1 | 3 |
Express 7 cos θ + 24 sin θ in the form R cos(θ − α), where R > 0 and 0° < α < 90°. Give α correct to 2 decimal places.
Hence solve the equation 7 cos θ + 24 sin θ = 4 for 0° < θ < 360°.
State the greatest value of 126 − 7 cos θ − 24 sin θ and the value of θ, for 0° < θ < 360°, at which it occurs.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 25 cos(θ − 73.74°) R = 25 B1 tan α = 247 M1 α = 73.74° A1 | 3 |
| (b) | 154.5°, 352.9° cos(θ − 73.74) = 425 M1 θ − α = 80.79° A1 second solution method (360° − or negative angle) M1 154.5°, 352.9° A1 | 4 |
| (c) | 1 at θ = 73.74° 1 B1 73.74° B1 | 2 |
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in