Year 11 Paper 4: Trigonometric Methods
Covers surds and indices, simultaneous equations and inequalities, trigonometric identities and equations, non-right-angled triangle trigonometry, and sequences.
Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [2 marks]
Sequences
An arithmetic sequence has first term 12 and common difference -3.
Find the 15th term of the sequence.
Question 2 [4 marks]
Simultaneous Equations and Inequalities
Solve the simultaneous equations 2x + 3y = 13 and 3x - y = 3.
Question 3 [4 marks]
Trigonometric Identities and Equations
Show that (sin(x) + cos(x))^2 can be simplified to 1 + 2sin(x)cos(x).
Question 4 [4 marks]
Surds and Indices
Simplify sqrt(45) + sqrt(80) - sqrt(20), giving your answer in the form k*sqrt(5).
Question 5 [4 marks]
Non-Right-Angled Triangle Trigonometry
In triangle DEF, DE = 11 cm, EF = 8 cm, and DF = 6 cm.
Find the size of angle DEF, giving your answer to 1 decimal place.
Question 6 [4 marks]
Trigonometric Identities and Equations
Prove the identity (sin(x) + cos(x))(sin(x) - cos(x)) = 2sin^2(x) - 1.
Question 7 [5 marks]
Non-Right-Angled Triangle Trigonometry
Triangle XYZ has XY = 10 cm, XZ = 13 cm, and area 55 cm^2.
Find the possible size(s) of angle YXZ, giving your answer(s) to 1 decimal place.
Question 8 [5 marks]
Surds and Indices
Rationalise the denominator of 12/sqrt(18), giving your answer in its simplest form.
Question 9 [5 marks]
Sequences
A geometric sequence has third term 20 and sixth term 160.
Find the common ratio and the first term of the sequence.
Question 10 [5 marks]
Simultaneous Equations and Inequalities
A rectangle has length (x + 3) cm and width (x - 1) cm, where x > 1.
The area of the rectangle is greater than 45 cm^2.
Form and solve an inequality to find the range of possible values of x.
Question 11 [6 marks]
Sequences
The first three terms of a geometric sequence are (k + 1), (k + 4), and (3k + 2), where k is a positive constant.
Find the value of k, and find the common ratio.
Question 12 [6 marks]
Non-Right-Angled Triangle Trigonometry
A ship sails from port A on a bearing of 070 degrees for 25 km to reach point B.
It then sails from B on a bearing of 160 degrees for 18 km to reach point C.
Find angle ABC, and hence find the distance AC using the cosine rule, giving your answer to 3 significant figures.
Question 13 [6 marks]
Surds and Indices
Show that (4 + sqrt(3))/(4 - sqrt(3)) can be written in the form a + b*sqrt(3), where a and b are rational numbers to be found.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| using the formula a + (n - 1)d with n = 15 | M1 |
| T15 = -30 | A1 |
| Final answer: -30 | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging 3x - y = 3 to make y the subject, y = 3x - 3 | M1 |
| substituting into 2x + 3y = 13 | M1 |
| simplifying to 11x = 22 | M1 |
| x = 2 and y = 3 | A1 |
| Final answer: x = 2, y = 3 | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| expanding the brackets to give sin^2(x) + 2sin(x)cos(x) + cos^2(x) | M1 |
| stating the identity sin^2(x) + cos^2(x) = 1 | B1 |
| substituting sin^2(x) + cos^2(x) = 1 | M1 |
| 1 + 2sin(x)cos(x) | A1 |
| Final answer: 1 + 2sin(x)cos(x), shown using sin^2(x) + cos^2(x) = 1 | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(45) = 3sqrt(5) | M1 |
| writing sqrt(80) = 4sqrt(5) | M1 |
| writing sqrt(20) = 2sqrt(5) | M1 |
| 5sqrt(5) | A1 |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the cosine rule DF^2 = DE^2 + EF^2 - 2(DE)(EF)cos(E) | M1 |
| substituting to get 36 = 185 - 176cos(E) | M1 |
| rearranging to cos(E) = 149/176 | M1 |
| angle DEF = 32.2 degrees | A1 |
| Final answer: angle DEF = 32.2 degrees (1 d.p.) | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| expanding the brackets to give sin^2(x) - cos^2(x) | M1 |
| stating the identity sin^2(x) + cos^2(x) = 1, i.e. cos^2(x) = 1 - sin^2(x) | B1 |
| substituting to get sin^2(x) - (1 - sin^2(x)) | M1 |
| a complete and correctly reasoned proof to 2sin^2(x) - 1 | A1 |
| Final answer: Shown: (sin(x) + cos(x))(sin(x) - cos(x)) = sin^2(x) - cos^2(x) = 2sin^2(x) - 1 | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| using Area = (1/2)(XY)(XZ)sin(X) | M1 |
| substituting to get 55 = (1/2)(10)(13)sin(X) | M1 |
| rearranging to sin(X) = 55/65 | M1 |
| X = 57.8 degrees | A1 |
| X = 122.2 degrees | A1 |
| Final answer: angle YXZ = 57.8 degrees or 122.2 degrees (1 d.p.) | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(18) = 3sqrt(2) | M1 |
| simplifying to 4/sqrt(2) | M1 |
| multiplying numerator and denominator by sqrt(2) | M1 |
| simplifying the numerator to 4sqrt(2) | M1 |
| 2sqrt(2) | A1 |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing T3 = ar^2 = 20 and T6 = ar^5 = 160 | M1 |
| dividing to eliminate a, giving r^3 = 8 | M1 |
| r = 2 | A1 |
| substituting back into ar^2 = 20 to find a | M1 |
| a = 5 | A1 |
| Final answer: common ratio r = 2; first term a = 5 | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| forming the area expression (x + 3)(x - 1) | M1 |
| expanding to x^2 + 2x - 3 | M1 |
| forming the inequality x^2 + 2x - 48 > 0 | M1 |
| factorising as (x + 8)(x - 6) > 0 and identifying the critical values -8 and 6 | M1 |
| x > 6 | A1 |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| using the geometric sequence property (k + 4)^2 = (k + 1)(3k + 2) | M1 |
| expanding the left side to k^2 + 8k + 16 | M1 |
| expanding the right side to 3k^2 + 5k + 2 | M1 |
| rearranging to 2k^2 - 3k - 14 = 0 and solving, rejecting k = -2 since k is positive | M1 |
| k = 3.5 | A1 |
| common ratio = 5/3 | A1 |
| Final answer: k = 3.5; common ratio = 5/3 | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| finding the bearing of A from B as 070 + 180 = 250 degrees | M1 |
| finding angle ABC as 250 - 160 = 90 degrees | M1 |
| angle ABC = 90 degrees | A1 |
| using the cosine rule AC^2 = AB^2 + BC^2 - 2(AB)(BC)cos(ABC) | M1 |
| substituting to get AC^2 = 625 + 324 - 0, since cos(90) = 0 | M1 |
| AC = 30.8 km (3 s.f.) | A1 |
| Final answer: angle ABC = 90 degrees; AC = 30.8 km (3 s.f.) | |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying numerator and denominator by (4 + sqrt(3)) | M1 |
| the denominator simplifying to 16 - 3 = 13 | M1 |
| expanding the numerator (4 + sqrt(3))^2 | M1 |
| the numerator simplifying to 19 + 8sqrt(3) | A1 |
| a = 19/13 | A1 |
| b = 8/13 | A1 |
| Final answer: a = 19/13, b = 8/13, so 19/13 + (8/13)sqrt(3) | |