AS Level · Mathematics 9709 · Pure Mathematics 2

Trigonometry: AS Level Mathematics questions with answers

Free Trigonometry questions for AS Level Mathematics (9709), part of Pure Mathematics 2. Each question has the answer and the marking points an examiner would look for.

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Trigonometry practice questions

Question 1Structured · 9 marksA* challenge
(a)

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.

[3]
(b)

Hence solve the equation 12 cos θ + 5 sin θ = 6 for 0° < θ < 360°.

[4]
(c)

State the greatest value of 114 − 12 cos θ − 5 sin θ and the value of θ, for 0° < θ < 360°, at which it occurs.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(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)1 at θ = 22.62°
1 B1
22.62° B1
2
Question 2Structured · 6 marksA* challenge
(a)

Prove the identity sin 2θ1 + cos 2θ ≡ tan θ.

[3]
(b)

Hence find the exact value of ∫0π/3 sin 2θ1 + cos 2θ dθ.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(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
Question 3Structured · 9 marksA* challenge
(a)

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.

[3]
(b)

Hence solve the equation 12 cos θ + 5 sin θ = 3 for 0° < θ < 360°.

[4]
(c)

State the greatest value of 115 − 12 cos θ − 5 sin θ and the value of θ, for 0° < θ < 360°, at which it occurs.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)13 cos(θ − 22.62°)
R = 13 B1
tan α = 512 M1
α = 22.62° A1
3
(b)99.3°, 306.0°
cos(θ − 22.62) = 313 M1
θ − α = 76.66° A1
second solution method (360° − or negative angle) M1
99.3°, 306.0° A1
4
(c)12 at θ = 22.62°
12 B1
22.62° B1
2
Question 4Structured · 5 marksA* challenge
(a)

Show that cos 2θ + 3 sin θ = 2 can be written as 2 sin2 θ − 3 sin θ + 1 = 0.

[2]
(b)

Hence solve the equation for 0° ≤ θ ≤ 360°.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(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
Question 5Structured · 4 marks
(a)

Prove the identity 1 − cos 2θsin 2θ ≡ tan θ.

[3]
(b)

Hence find the exact value of 1 − cos(π/3)sin(π/3).

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)shown
1 − cos 2θ = 2 sin2θ M1
sin 2θ = 2 sin θ cos θ M1
simplify to tan θ A1
3
(b)1√3
tan π6 = 1√3 B1
1
Question 6Structured · 6 marks
(a)

Express 3 sin θ + 4 cos θ in the form R sin(θ + α), where R > 0 and 0° < α < 90°. Give α correct to 2 decimal places.

[3]
(b)

Hence solve 3 sin θ + 4 cos θ = 3 for 0° < θ < 360°.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)R = 5, α = 53.13°
R correct B1
tan α = 4/3 M1
α = 53.13 A1
3
(b)θ = 343.7°, 90.0°
sin(θ + α) = c/R M1
One correct angle A1
Second correct angle A1
3

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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