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More questions on this topic: Level 2 FM Shape & Space: Exact Trig Values & Graphs (17) AS Pure: Trigonometry (107)
14 \(ABC\) is a triangle.
Work out the length of \(AB\). (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | for either \(\sin 30 = 0.5\) oe or \(\sin 45 = \dfrac{1}{\sqrt{2}}\) or \(\dfrac{\sqrt{2}}{2}\) |
| M1 | for \(\dfrac{AB}{\sin 45} = \dfrac{3\sqrt{2}}{\sin 30}\) oe | |
| A1 | cao |
22
| Answer | Mark | Mark scheme |
|---|---|---|
| Sketch | M1 | for correct shape for \(0 \le x \le 360\) |
| A1 | for fully correct sketch with labels at 1 and \(-1\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| 210, 330 | M1 | for one correct solution or for \(\sin 30 = \dfrac{1}{2}\) |
| A1 | cao |
Can be implied by \(x = -30\)
19 Here is the graph of \(y = \sin x^\circ\) for \(-180 \leqslant x \leqslant 180\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 18 and 162 | M1 | for a correct method, eg line drawn at 0.3 to sine curve or point(s) marked on sine curve or for one correct value |
| A1 | for answers in the range 15 to 18 and 162 to 165 and no incorrect values |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-20\) or 160 | B1 | cao |
30 Write down the exact value of \(\cos 60^\circ\) (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{1}{2}\) | B1 | for \(\dfrac{1}{2}\) oe |
22 Here is a triangle \(ABC\).
Work out the value of \(\sin ABC\)
Give your answer in the form \(\dfrac{m}{n}\) where \(m\) and \(n\) are integers. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{65}{214}\) | B1 | for \(\sin 30 = 0.5\) |
| P1 | for use of the sine rule with values substituted, eg \(\dfrac{6.5}{\sin ABC} = \dfrac{10.7}{\sin 30}\) oe | |
| P1 | for \((\sin ABC =)\ \dfrac{6.5 \times \sin 30}{10.7}\) oe or for a complete process to find \(\sin ABC\), eg \((\sin ABC =)\ \dfrac{6.5 \times [0.5]}{10.7}\) oe | |
| A1 | for \(\dfrac{65}{214}\) oe eg \(\dfrac{325}{1070}\) |
Answer of \(\dfrac{3.25}{10.7}\) or \(\dfrac{6.5}{21.4}\) gets 3 marks
Where [0.5] is their value of \(\sin 30\)
Answer must be in the form \(\dfrac{m}{n}\) where \(m\) and \(n\) are integers
21
The diagram shows a sketch of part of the curve with equation \(y = \cos x^\circ\)
\(P\) is a minimum point on the curve.
Write down the coordinates of \(P\). (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| (180, −1) | B1 | for 180(o) |
| B1 | for −1 SC B1 if B0 scored for answer of (−1, 180) |
30 Write down the value of \(\sin 30^\circ\) (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 0.5 | B1 | for 0.5 or \(\dfrac{1}{2}\) oe |
18 Here is trapezium \(ABCD\).
The area of the trapezium is 66 cm2
the length of \(AB\) : the length of \(CD = 2 : 3\)
Find the length of \(AB\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 17.6 | P1 | for correct trig statement, eg \(\sin 30 = \dfrac{h}{6}\) |
| P1 | for complete process to find \(h\), eg \(6 \times \dfrac{1}{2}\ (= 3)\) | |
| P1 | for correct substitution into the area of a trapezium formula, eg \(\dfrac{1}{2}(a + b) \times \text{``}3\text{''} = 66\) or \(a + b = 44\) or \(\dfrac{1}{2}(2x + 3x) \times h = 66\) | |
| P1 | for complete correct process to find the length of \(AB\), eg \(\left[\dfrac{66 \times 2}{3} \div (2 + \text{``}3\text{''})\right] \times 2\) | |
| A1 | cao |
An answer of \(\dfrac{88}{5}\) gets P4 A0
14 Find the exact value of \(\tan 30^\circ \times \sin 60^\circ\)
Give your answer in its simplest form. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{1}{2}\) | M1 | for \(\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{2}\) or \(\dfrac{\sqrt{3}}{3} \times \dfrac{\sqrt{3}}{2}\) or \(\left(\dfrac{1}{2} \div \dfrac{\sqrt{3}}{2}\right) \times \dfrac{\sqrt{3}}{2}\) OR \(\tan 30 = \dfrac{1}{\sqrt{3}}\) oe or \(\sin 60 = \dfrac{\sqrt{3}}{2}\) |
| A1 | for \(\dfrac{1}{2}\) or 0.5 |
11 Sketch the graph of \(y = \tan x^\circ\) for \(0 \leqslant x \leqslant 360\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| Graph drawn | C2 | for fully correct sketch between 0 and 360 |
| (C1 | for a graph with clear asymptotes at 90 and 270 only or the correct graph translated along the \(x\)-axis must have a period of 180) |
8
Here is a right-angled triangle.
\(\cos 60^\circ = 0.5\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 8 | M1 | starts process, eg \(\cos(60) = \dfrac{4}{x}\) or \(0.5 = \dfrac{4}{x}\) oe or \(\sin 30 = \dfrac{4}{x}\) or \(\dfrac{\sin 30}{4} = \dfrac{\sin 90}{x}\) oe |
| A1 | cao |
All three elements of cos, 4, \(x\) must be present in an equation. eg \(\cos = 4/x\) is acceptable but \(\cos(4/x)\) is insufficient
20 The table shows some values of \(x\) and \(y\) that satisfy the equation \(y = a\cos x^\circ + b\)
| \(x\) | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
|---|---|---|---|---|---|---|---|
| \(y\) | 3 | \(1 + \sqrt{3}\) | 2 | 1 | 0 | \(1 - \sqrt{3}\) | \(-1\) |
Find the value of \(y\) when \(x = 45\) (4)
| Answer | Mark | Notes |
|---|---|---|
| \(1+\sqrt{2}\) | B1 | for a value for a known trigonometric ratio stated |
| P1 | for process to form 2 equations in \(a\) and \(b\) or one correct value stated | |
| P1 | for complete process to solve to reach \(a = 2\) and \(b = 1\) | |
| A1 | for \(1+\sqrt{2}\) oe |
22 The diagram shows a hexagon \(ABCDEF\).
\(ABEF\) and \(CBED\) are congruent parallelograms where \(AB = BC = x\) cm.
\(P\) is the point on \(AF\) and \(Q\) is the point on \(CD\) such that \(BP = BQ = 10\) cm.
Given that angle \(ABC = 30^\circ\),
prove that \(\cos PBQ = 1 - \dfrac{(2 - \sqrt{3})}{200}x^2\) (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| Proof | B1 | (indep) for stating \(\cos 30 = \dfrac{\sqrt{3}}{2}\) | |
| M1 | for \(PQ^2 = 10^2 + 10^2 - 2 \times 10 \times 10 \times \cos PBQ\) or \(AC^2 = x^2 + x^2 - 2 \times x \times x \times \cos 30\ (= x^2(2 - \sqrt{3}))\) oe | ||
| M1 | for \(\cos PBQ = \dfrac{10^2 + 10^2 - PQ^2}{2 \times 10 \times 10}\) (implies previous M1) | ||
| \(\cos PBQ = {}\) \(\dfrac{10^2 + 10^2 - x^2(2 - \sqrt{3})}{200}\) \({= \dfrac{200 - x^2(2 - \sqrt{3})}{200}}\) | M1 | for \(\cos PBQ = \dfrac{10^2 + 10^2 - (x^2 + x^2 - 2 \times x \times x \times \cos 30)}{2 \times 10 \times 10}\) | |
| A1 | conclusion of proof with all working seen |
23 Here is the graph of \(y = \sin x^\circ\) for \(-180 \leqslant x \leqslant 180\)
Here is the graph of \(y = \cos x^\circ\) for \(-180 \leqslant x \leqslant 180\)
| Answer | Mark | Notes |
|---|---|---|
| Graph drawn | 2 | B2 correct graph drawn (B1 for a graph translated up/down) |
| Answer | Mark | Notes |
|---|---|---|
| Graph drawn | 2 | B2 for correct graph drawn (B1 for a graph reflected in the \(x\) axis or stretched by sf 2 parallel to the \(y\) axis) |
26 The diagram shows part of a sketch of the curve \(y = \sin x^\circ\).
Here is a sketch of the curve \(y = a\cos bx^\circ + c\), \(0 \leqslant x \leqslant 360\)
| Answer | Mark | Notes |
|---|---|---|
| 180, 0 | 1 | B1 for 180, 0 Accept \(\pi\), 0 |
| Answer | Mark | Notes |
|---|---|---|
| 270, \(-1\) | 1 | B1 for 270, \(-1\) accept \(\dfrac{3}{2}\pi\), \(-1\) |
| Answer | Mark | Notes |
|---|---|---|
| \(a = 2\) \(b = 3\) \(c = 1\) | 3 | B1 cao B1 cao B1 cao |
28 The diagram shows a sketch of the graph of \(y = \cos x^\circ\)
| Answer | Mark | Notes |
|---|---|---|
| (90, 0) | 1 | B1 for (90, 0) (condone \(\left(\frac{\pi}{2}, 0\right)\)) |
| Answer | Mark | Notes |
|---|---|---|
| Correct graph | 1 | B1 for graph through (0, 2) (90, 0) \((180, -2)\) (270, 0) (360, 2) professional judgement |
25 Circle the value of \(\quad \sin 90^\circ\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 1 | B1 |
18 Show that \(\quad \dfrac{\sin 30^\circ \times \cos 45^\circ}{\tan 60^\circ} \quad\) can be written in the form \(\dfrac{\sqrt{a}}{b}\)
where \(a\) and \(b\) are integers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 30 = \dfrac{1}{2}\) and \(\cos 45 = \dfrac{1}{\sqrt{2}}\) or \(\dfrac{\sqrt{2}}{2}\) and \(\tan 60 = \sqrt{3}\) | M2 | oe eg \(\dfrac{\frac{1}{2} \times \frac{1}{\sqrt{2}}}{\sqrt{3}}\) or \(\dfrac{\frac{1}{2} \times \frac{\sqrt{2}}{2}}{\sqrt{3}}\) or \(\dfrac{1}{2\sqrt{6}}\) may be seen in a table implied by position in a calculation eg \(\dfrac{1}{2} \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{3}}\) is correct for M2 M1 1 or 2 correct values |
| \(\dfrac{\sqrt{6}}{12}\) with all three correct values seen | A1 | oe in the form \(\dfrac{\sqrt{a}}{b}\) eg \(\dfrac{\sqrt{24}}{24}\) condone \(a = 6\) and \(b = 12\) with all three correct values seen |
| Allow \(\sqrt{1}\) for 1 throughout | |
| Allow, eg \(\dfrac{\sqrt{3}/2}{1/2}\) for tan 60 | |
| Correct answer from \(\dfrac{\frac{1}{\sqrt{2}} \times \frac{1}{2}}{\sqrt{3}}\) scores only M1 for tan 60 unless the correct values are attributed to sin 30 and cos 45 elsewhere in the working | |
| Do not allow further work eg \(\dfrac{\sqrt{6}}{12} = \dfrac{\sqrt{1}}{2}\) (with all three correct values seen) | M2A0 |
25 Write down the value of \(\sin 90^\circ\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 1 | B1 | condone 1° |
21 Here is a sketch of the graph of \(\quad y = \tan x \quad\) where \(\quad 0^\circ \leqslant x \leqslant 360^\circ\)
You are given that \(\quad\) tan 78.69° = 5
| Answer | Mark | Comments |
|---|---|---|
| 258.69 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(180 - 78.69\) or 101.31 or \(360 - 78.69\) or 281.31 | M1 | oe |
| 101.31 and 281.31 | A1 | either order |
| −78.69, 101.31 and 281.31 | M1A0 |
25 Show that the value of \(\quad 6\sin 30^\circ + 2\cos 30^\circ \times 4\tan 30^\circ \quad\) is an integer. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes values | ||
| \((\sin 30^\circ =)\ \dfrac{1}{2}\) or \(6\sin 30^\circ = 3\) or \((\cos 30^\circ =)\ \dfrac{\sqrt{3}}{2}\) or \(2\cos 30^\circ = \sqrt{3}\) or \((\tan 30^\circ =)\ \dfrac{1}{\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{3}\) or \(4\tan 30^\circ = \dfrac{4}{\sqrt{3}}\) or \(\dfrac{4\sqrt{3}}{3}\) | M1 | may be seen beside the given expression or in a table |
| \(6\left(\dfrac{1}{2}\right)\) and \(2\left(\dfrac{\sqrt{3}}{2}\right)\) and \(4\left(\dfrac{1}{\sqrt{3}}\right)\) or \(6\left(\dfrac{1}{2}\right)\) and \(2\left(\dfrac{\sqrt{3}}{2}\right)\) and \(4\left(\dfrac{\sqrt{3}}{3}\right)\) or \(\dfrac{6}{2}\) and \(\dfrac{2\sqrt{3}}{2}\) and \(\dfrac{4\sqrt{3}}{3}\) | M1dep | oe |
| Processing at least as far as \(\dfrac{6}{2} + \dfrac{8\sqrt{3}}{2\sqrt{3}}\) or \(\dfrac{6}{2} + \dfrac{8\sqrt{3}\sqrt{3}}{6}\) or \(\dfrac{6}{2} + \dfrac{24}{6}\) | M1dep | oe |
| 7 from correct working | A1 | SC2 \(4 + 4\sqrt{3}\) oe |
| Alternative method 2: uses a trig identity | ||
| \(6\sin 30^\circ + 2\cos 30^\circ \times 4\dfrac{\sin 30^\circ}{\cos 30^\circ}\) | M1 | oe |
| \(6\sin 30^\circ + 8\sin 30^\circ\) or \(14\sin 30^\circ\) | M1dep | oe |
| \(14 \times \dfrac{1}{2}\) | M1dep | oe |
| 7 from correct working | A1 | SC2 \(4 + 4\sqrt{3}\) oe |
| Alt 2 is not on this specification, but may be seen if other qualifications have been studied, eg AQA Certificate – Level 2 Further Maths | |
| Incorrect order of operations gives \(4 + 4\sqrt{3}\) oe | SC2 |
| Allow \(\sqrt{1}\) for 1 throughout |
29 Here are the equations of three graphs.
\(y = \sin x \qquad\qquad y = \cos x \qquad\qquad y = \tan x\)
Tick one box. [1 mark]
Tick one box. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(y = \cos x\) passes through \((180^\circ, -1)\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| None of the graphs pass through \((270^\circ, 1)\) | B1 |
21 Write down the value of \(\cos 90^\circ\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 0 or zero | B1 |
20 \(x\) and \(y\) are acute angles.
\(\sin x = \dfrac{\sqrt{3}}{2} \qquad\qquad \tan y = 1\)
\(w = 3x - 2y\)
Work out the value of \(\quad \cos w\)
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x =)\ 60\) or \((3x =)\ 180\) or \((y =)\ 45\) or \((2y =)\ 90\) | M1 | implied by sin 60 or tan 45 must be selected and not just in a table |
| \(3 \times 60 - 2 \times 45\) or \(180 - 90\) or \((w =)\ 90\) or cos 90 | M1dep | |
| 0 with M2 awarded | A1 |
| 0 with no working | M0M0A0 |
| Condone degrees sign on answer with correct working | M1M1A1 |
25 Show that \(\quad \dfrac{4\sin 30^\circ - \tan 45^\circ}{2\cos 30^\circ} \quad\) can be written as \(\tan x\), where \(x\) is an acute angle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 30 = \dfrac{1}{2}\) or \(\tan 45 = 1\) or \(\cos 30 = \dfrac{\sqrt{3}}{2}\) | M1 | oe eg \(\tan 45 = \dfrac{\sqrt{2}}{\sqrt{2}}\) or \(4\sin 30 = 2\) or \(2\cos 30 = \sqrt{3}\) implied by position in the expression may be seen in a table |
| substitution of all three correct values | M1dep | eg \(\dfrac{4 \times \frac{1}{2} - 1}{2 \times \frac{\sqrt{3}}{2}}\) or \(\dfrac{2 - 1}{2 \times \frac{\sqrt{3}}{2}}\) or \(\dfrac{2 - 1}{\sqrt{3}}\) |
| \(\dfrac{1}{\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{3}\) | M1dep | |
| \(\left(\dfrac{1}{\sqrt{3}}\right.\) or \(\left.\dfrac{\sqrt{3}}{3} =\right)\ \tan 30\) or \(x = 30\) with full working seen for M3 | A1 |
Allow \(\sqrt{1}\) for 1 throughout
Reference to 30° being an acute angle is not required
23 \(4 \times \sin 30^\circ \times \tan 30^\circ \times \cos 30^\circ = \sin y\)
Work out one possible value of \(y\).
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((\sin 30 =)\ \dfrac{1}{2}\) or \((\tan 30 =)\ \dfrac{1}{\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{3}\) or \((\cos 30 =)\ \dfrac{\sqrt{3}}{2}\) | M1 | oe may be implied by \((4 \times \sin 30 =)\ 2\) may be implied by correct position in a multiplication string |
| \(4 \times \dfrac{1}{2} \times \dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{2}\) | M1dep | oe with all trig values correct condone any order unless error seen |
| 1 with all three values seen | A1 | implied by 90 with all three values seen |
| 90 with M1M1A1 scored | A1 | accept any angle of the form \(90 + 360n\), where \(n\) is an integer |
| Alternative method 2 | ||
| \(4 \times \sin 30^\circ \times \dfrac{\sin 30^\circ}{\cos 30^\circ} \times \cos 30^\circ\) | M1 | |
| \(4 \times \dfrac{1}{2} \times \dfrac{1/2}{\sqrt{3}/2} \times \dfrac{\sqrt{3}}{2}\) | M1dep | oe eg \(4 \times \left(\dfrac{1}{2}\right)^2\) |
| 1 with \(\dfrac{\sin 30^\circ}{\cos 30^\circ}\) and \(\dfrac{1}{2}\) and \(\dfrac{\sqrt{3}}{2}\) seen | A1 | if \(\cos 30^\circ\) is cancelled out only \(\dfrac{1}{2}\) need be seen |
| 90 with M1M1A1 scored | A1 | accept any angle of the form \(90 + 360n\), where \(n\) is an integer |
Condone a square root sign on 1 up to M1M1
28 Work out the value of \(\quad (\cos 30^\circ \times \sin 45^\circ \times \tan 60^\circ)^2\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((\cos 30 =)\ \dfrac{\sqrt{3}}{2}\) or \((\sin 45 =)\ \dfrac{\sqrt{2}}{2}\) or \(\dfrac{1}{\sqrt{2}}\) or \((\tan 60 =)\ \sqrt{3}\) | M1 | oe correct trig function may be implied by position in multiplication string may be seen in a table |
| \(\left(\dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{2}}{2} \times \sqrt{3}\right)^2\) or \(\left(\dfrac{\sqrt{3}}{2}\right)^2 \times \left(\dfrac{\sqrt{2}}{2}\right)^2 \times \left(\sqrt{3}\right)^2\) | M1dep | oe with all values correct |
| or \(\dfrac{3\sqrt{2}}{4}\) or \(\dfrac{3}{2\sqrt{2}}\) or \(\dfrac{\sqrt{18}}{4}\) | oe single term not squared | |
| \(\left(\dfrac{3\sqrt{2}}{4}\right)^2\) or \(\left(\dfrac{3}{2\sqrt{2}}\right)^2\) or \(\left(\dfrac{\sqrt{18}}{4}\right)^2\) | M1dep | oe with all values correct oe single term squared |
| or \(\dfrac{3}{4} \times \dfrac{1}{2} \times 3\) | oe multiplication string without surds | |
| or \(\dfrac{\sqrt{324}}{16}\) | oe single fraction with one surd | |
| \(\dfrac{9}{8}\) or \(1\dfrac{1}{8}\) or 1.125 | A1 | oe fraction, mixed number or decimal |
| Ignore an incorrect attempt to simplify or convert a correct answer eg \(\dfrac{9}{8} = 1.8\) | M1M1M1A1 |
12 Here is a right-angled triangle.
Not drawn accurately
Use trigonometry to work out the value of \(x\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin 30 = \dfrac{x}{10}\) or (\(x\) =) \(10 \sin 30\) | M1 | oe eg \(\dfrac{x}{\sin 30} = \dfrac{10}{\sin 90}\) |
| \(\sin 30 = 0.5\) | M1 | oe may be seen in a table \(0.5 = \dfrac{x}{10}\) oe scores M1M1 |
| 5 | A1 | |
| Alternative method 2 | ||
| Correct trigonometric method to show that the length of the missing side is \(5\sqrt{3}\) | M1 | oe |
| \(\sqrt{(5\sqrt{3})^2 + x^2} = 10\) | M1dep | oe |
| 5 | A1 | |
Accept use of cos 60 instead of sin 30
24 \(y\) is an obtuse angle.
Which statement is true?
Tick one box. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin y > 0\) and \(\cos y < 0\) | B1 |
22 One of these is the graph of \(\quad y = 1 + \sin x \quad\) for \(\quad 0^\circ \leqslant x \leqslant 360^\circ\)
Circle the letter above the correct graph. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| A | B1 |
29 Show that the value of \(\quad 5\sin 30^\circ \times \cos 30^\circ \times 8\tan 30^\circ \quad\) is an integer. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((\sin 30^\circ =)\ \dfrac{1}{2}\) or \((\cos 30^\circ =)\ \dfrac{\sqrt{3}}{2}\) or \((\tan 30^\circ =)\ \dfrac{1}{\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{3}\) or \(\left(\dfrac{1/2}{\sqrt{3}/2}\right)\) | M1 | may be seen beside question |
| \(5\left(\dfrac{1}{2}\right) \times \dfrac{\sqrt{3}}{2} \times 8\left(\dfrac{1}{\sqrt{3}}\right)\) or \(5\left(\dfrac{1}{2}\right) \times \dfrac{\sqrt{3}}{2} \times 8\left(\dfrac{1/2}{\sqrt{3}/2}\right)\) or \(\dfrac{5}{2} \times \dfrac{\sqrt{3}}{2} \times \dfrac{8\sqrt{3}}{3}\) | M1dep | oe multiplication string with all correct values |
| \(\dfrac{40\sqrt{3}}{4\sqrt{3}}\) or \(\dfrac{40\sqrt{3}\sqrt{3}}{12}\) | M1dep | oe single fraction with roots rationalised or able to be cancelled |
| 10 from correct working | A1 | |
| Alternative method 2: substituting \(\dfrac{\sin}{\cos}\) for tan and cancelling | ||
| \(5\sin 30^\circ \times \cos 30^\circ \times 8\dfrac{\sin 30^\circ}{\cos 30^\circ}\) | M1 | |
| \(40\sin^2 30^\circ\) | M1dep | oe cancels \(\cos 30^\circ\) |
| \(40\left(\dfrac{1}{2}\right)^2\) | M1dep | oe |
| 10 from correct working | A1 | |
19 Circle the word that describes the graph \(\quad y = \sin x\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| periodic | B1 |
27 Angle \(x\) is acute.
\(\cos x = \sin 60^\circ \times \tan 30^\circ\)
Work out the size of angle \(x\).
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\) or \(\tan 30^\circ = \dfrac{\sqrt{3}}{3}\) or \(\dfrac{1}{\sqrt{3}}\) or \(\tan 30^\circ \left(= \dfrac{\sin 30}{\cos 30}\right) = \dfrac{1/2}{\sqrt{3}/2}\) | M1 | oe may be in a table may be implied by position in multiplication |
| \(\dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{2}\) or \(\cos x = \dfrac{1}{2}\) or \((x =)\ \cos^{-1}\dfrac{1}{2}\) | M1dep | oe works out the value of \(\cos x\) as a fraction or decimal with no surd values |
| 60 with M2 awarded | A1 |
\(\cos x = 60\) does not score the final mark
24 \(\mathrm{f}(x) = \sin(x - 90^\circ)\)
Circle the value of \(\quad \mathrm{f}(0^\circ)\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(-1\) | B1 |
27 Here is a sketch of \(\quad y = \cos x \quad\) for values of \(x\) from 0° to 360°
Not drawn accurately
\(\alpha°\) is an acute angle.
\(\cos \alpha° = k\)
| Answer | Mark | Comments |
|---|---|---|
| \(-k\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(k\) | B1 |
26 The area of this triangle is \(25\sqrt{3}\) cm2
Not drawn accurately
Work out the value of \(w\).
Give your answer in the form \(\ a\sqrt{b}\ \) where \(a\) and \(b\) are integers greater than 1 [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(0.5 \times 20 \times x \times \sin 60\) or \(10x \sin 60\) or \(5\sqrt{3}\,x\) | M1 | oe |
| \(0.5 \times 20 \times x \times \sin 60 = 25\sqrt{3}\) or \(x = 5\) | M1dep | oe equation |
| (their 5)\(^2 + 20^2\) \(-\ 2 \times\) their \(5 \times 20 \times \cos 60\) or \(25 + 400 - 200\cos 60\) or 325 | M1 | oe their 5 must be their value of \(x\) |
| \(\sqrt{\text{their } 325}\) | M1dep | dep on 3rd M1 their 325 can be unsimplified |
| \(5\sqrt{13}\) | A1 | |
| Alternative method 2 | ||
| \(0.5 \times 20 \times h = 25\sqrt{3}\) or \(h = \dfrac{5\sqrt{3}}{2}\) | M1 | oe any letter \(h\) is perpendicular height for 20 cm base |
| \(\sin 60 = \dfrac{\text{their } \frac{5\sqrt{3}}{2}}{x}\) or \(x = 5\) | M1dep | oe |
| (their 5)\(^2 + 20^2\) \(-\ 2 \times\) their \(5 \times 20 \times \cos 60\) or \(25 + 400 - 200\cos 60\) or 325 | M1 | oe their 5 must be their value of \(x\) |
| \(\sqrt{\text{their } 325}\) | M1dep | dep on 3rd M1 their 325 can be unsimplified |
| \(5\sqrt{13}\) | A1 | |
| Alternative method 3 | ||
| \(0.5 \times 20 \times h = 25\sqrt{3}\) or \(h = \dfrac{5\sqrt{3}}{2}\) | M1 | oe any letter \(h\) is perpendicular height for 20 cm base |
| \(\tan 60 = \dfrac{\text{their } h}{c}\) or \(c = \dfrac{5}{2}\) | M1dep | oe any letter \(c\) is part of 20 cm base |
| \(\left(\text{their } \dfrac{5\sqrt{3}}{2}\right)^2 + \left(20 - \text{their } \dfrac{5}{2}\right)^2\) or \(\left(\text{their } \dfrac{5\sqrt{3}}{2}\right)^2 + \left(\dfrac{35}{2}\right)^2\) or 325 | M1dep | |
| \(\sqrt{\left(\text{their } \frac{5\sqrt{3}}{2}\right)^2 + \left(20 - \text{their } \frac{5}{2}\right)^2}\) or \(\sqrt{\text{their } 325}\) | M1dep | |
| \(5\sqrt{13}\) | A1 | |
| Omitting 0.5 in area formula can score a maximum of M0M0M1M1A0 | |
| \(\sqrt{(\text{their } 5)^2 + 20^2 - 2 \times \text{their } 5 \times 20 \times \cos 60}\) | M0M0M1M1A0 |
24 Circle the value of \(\cos 30^\circ\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{\sqrt{3}}{2}\) | B1 |
4 Circle the value of cos 30° [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{\sqrt{3}}{2}\) | B1 |
30 Show that the value of \(\quad \cos 30^\circ \times \tan 60^\circ + \sin 30^\circ \quad\) is an integer. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{\sqrt{3}}{2} \times \sqrt{3} + \dfrac{1}{2}\) \(= \dfrac{3}{2} + \dfrac{1}{2}\) \(= 2\) | B3 | B2 \(\dfrac{\sqrt{3}}{2} \times \sqrt{3} + \dfrac{1}{2}\) B1 \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\) or \(\tan 60^\circ = \sqrt{3}\) or \(\sin 30^\circ = \dfrac{1}{2}\) |
For B3 all steps must be shown
Allow \(\dfrac{\sqrt{3}}{2} \times \sqrt{3} + \dfrac{1}{2}\) given as \(\dfrac{\sqrt{3}}{2} \times \sqrt{3}\), followed by their \(\dfrac{3}{2} + \dfrac{1}{2}\)
Allow equivalent expressions for all trig values
eg
\(\cos 30^\circ = \sqrt{\dfrac{3}{4}} \quad \sin 30^\circ = \dfrac{\sqrt{1}}{2} \quad \tan 60^\circ = \dfrac{\sqrt{3}}{\sqrt{1}}\)
For B1 allow the trig value(s) given in a table unless contradicted in working
25 Here is a sketch of the graph of \(\quad y = \cos x \quad\) for values of \(x\) from \(0^\circ\) to \(360^\circ\)
Work out the value of \(x\) when \(\;90^\circ \leqslant x \leqslant 360^\circ\) [1 mark]
Work out the value of \(x\) when \(\;180^\circ \leqslant x \leqslant 360^\circ\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 300 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 240 | B1 |
29 Here is a sketch of \(\quad y = \sin x° \quad\) for \(\;-360 \leqslant x \leqslant 360\)
| Answer | Mark | Comments |
|---|---|---|
| (180, 0) | B1 |
| Condone degrees symbol on 180 | |
| Condone \((\pi, 0)\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| (−270, 1) | B1 |
| Condone degrees symbol on 270 | |
| Condone \(\left(\dfrac{-3\pi}{2}, 1\right)\) | B1 |
29 Simplify \(\qquad \dfrac{2\sin 45^\circ - \tan 45^\circ}{4\tan 60^\circ}\)
Give your answer in the form \(\qquad \dfrac{\sqrt{a} - \sqrt{b}}{c} \qquad\) where \(a\), \(b\) and \(c\) are integers. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 45 = \dfrac{\sqrt{2}}{2}\) or \(\dfrac{1}{\sqrt{2}}\) or \(\tan 45 = 1\) or \(\dfrac{1}{1}\) or \(\tan 60 = \sqrt{3}\) or \(\dfrac{\sqrt{3}}{1}\) | B1 | oe stated or in correct place in expression or implied by multiplier of 2 or 4 |
| \(\sin 45 = \dfrac{\sqrt{2}}{2}\) or \(\dfrac{1}{\sqrt{2}}\) and \(\tan 45 = 1\) or \(\dfrac{1}{1}\) and \(\tan 60 = \sqrt{3}\) or \(\dfrac{\sqrt{3}}{1}\) | B1 | oe stated or in correct place in expression or implied by multiplier of 2 or 4 eg \(\;\dfrac{2 \times \dfrac{1}{\sqrt{2}} - 1}{4 \times \dfrac{\sqrt{3}}{1}}\) |
| \(\dfrac{\sqrt{2} - 1}{4\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}\) | M1 | oe rationalisation of their denominator eg \(\;\dfrac{\dfrac{2}{\sqrt{2}} - 1}{4\sqrt{3}} \times \dfrac{4\sqrt{3}}{4\sqrt{3}}\) |
| \(\dfrac{\sqrt{6} - \sqrt{3}}{12}\) | A1 | oe in the form \(\dfrac{\sqrt{6a^2} - \sqrt{3a^2}}{12a}\) where \(a\) is a positive integer eg \(\;\dfrac{\sqrt{24} - \sqrt{12}}{24}\) (when \(a = 2\)) |
| \(\dfrac{2 \times \dfrac{1}{\sqrt{2}} - 1}{4\sqrt{3}}\) or \(\dfrac{\sqrt{2} - 1}{4\sqrt{3}}\) or \(\dfrac{\sqrt{2} - 1}{\sqrt{48}}\) | B1B1 |
| \(\dfrac{\sqrt{48}(\sqrt{2} - 1)}{\sqrt{48}\sqrt{48}}\) or \(\dfrac{\sqrt{48}(\sqrt{2} - 1)}{48}\) | B1B1M1 |
| \(\dfrac{\sqrt{96} - \sqrt{48}}{48}\) | B1B1M1A1 |
| B1B1 awarded, incorrect simplification, then correct method to rationalise | B1B1M1 |
27 Circle the value of \(\;\cos 90^\circ\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 0 | B1 |
16 Work out the exact value of \(x\) in this triangle. [4]
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\sqrt{31}\) final answer | 4 | B3 for answer 31 nfww OR B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\) and M2 for \(x^2 = (6\sqrt{3})^2 + 7^2 - 2 \times 6\sqrt{3} \times 7\cos 30\) or better or M1 for other correct arrangements of cosine rule where \(x\) or \(x^2\) are not the subject | Do not accept work in decimals rounded to answers of \(\sqrt{31}\) or 31 Award if seen as clear statement of fact or in use; once seen condone replacement by 0.866… or 0.87 in subsequent work Condone missing brackets Condone use of 10.39… or 10.4 and 0.866… or 0.87 in M2 and M1 |
20 The shape below is part of a circle, centre O and radius 12 cm.
Angle OAB = 30°.
Not to scale
Work out the perimeter of the shape.
Give your answer in its simplest terms in the form \(a\sqrt{b} + k\pi\).
You must show your working. [7]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(12\sqrt{3} + 16\pi\) final answer with correct working | 7 | Correct working requires evidence of at least M1M1M2 | |
| M1 for angle of major sector = 240° or \(\frac{240}{360}\) oe soi | For M1 accept e.g. \(\frac{2}{3}\) soi 240 may be on diagram | ||
| M1 for \(\frac{\textit{their } 240}{360} \times 2 \times \pi \times 12\) A1 for \(16\pi\) AND | For M1 \(120 \leqslant \textit{their } 240 \leqslant 300\) but not 180 \(24\pi \div 3 \times 2\) oe eg done in stages implies M1M1 | ||
| M2 for [ 2 ×] 12 cos30° oe | M2 oe accept e.g. [\(\text{AB}^2 =\) ] \(12^2 + 12^2 - 2.12.12.\cos 120\) or \(\dfrac{12\sin 120}{\sin 30}\), [2×] \(\sqrt{12^2 - (12\sin 30)^2}\) oe | ||
| or M1 for \(\frac{x}{12} = \cos 30\) oe | where \(x\) is ½ AB M1 for any correct implicit method for finding AB or ½ AB | ||
| B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\) soi A1 for \(12\sqrt{3}\) | In this method and other methods used, B1 is awarded for the correct trig value[s] associated with their method for find [½]AB, even if their method is incorrect, not just seen in a table of trig values unless selected Award maximum of 6 marks if answer incorrect | ||
| If 0, 1 or 2, scored instead award SC3 for answer \(12\sqrt{3} + 16\pi\) If 0 or 1 scored, instead award SC2 for \(12\sqrt{3}\) or \(16\pi\) in answer | SEE AG | ||
The length AB or \(\frac{1}{2}\) AB (\(12\sqrt{3}\) oe or \(6\sqrt{3}\) oe may be seen unsimplified e.g. \(2\sqrt{108}\) or \(\sqrt{108}\)) can be obtained by a number of different methods.
M2 is for the explicit method leading to AB or \(\frac{1}{2}\)AB and examples of this are in the guidance column of the scheme
M1 is for a correct implicit method
The B mark is awarded for a correct trig value[s] for their chosen method
Implicit methods may include
| \(\text{AB}^2 + (12\sin 30)^2 = 12^2\) oe or better | \(\cos 120 = \dfrac{12^2 + 12^2 - \text{AB}^2}{2 \times 12 \times 12}\) oe or better | \(\dfrac{\sin 120}{AB} = \dfrac{\sin 30}{12}\) oe or better |
| Award B1 for \(\sin 30 = \frac{1}{2}\) soi | Award B1 for \(\cos 120 = -\frac{1}{2}\) soi | Award B1 for \(\sin 120 = \frac{\sqrt{3}}{2}\) and \(\sin 30 = \frac{1}{2}\) soi |
23 In the diagram,
Not to scale
Give your answer in the form \(\left(\dfrac{a\sqrt{k}}{b} + \dfrac{8}{3}\pi\right)\) m2. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(AC =\)] \(\dfrac{4}{\cos 30}\) oe | M2 | M1 for \(\frac{4}{AC} = \cos 30\) oe or for \(\frac{\sin 60}{4} = \frac{\sin 90}{AC}\) oe | Accept any variable for AC provided not incorrect length Accept longer methods using 4 tan 30 to find BC then Pythagoras’ or sine rule M2 for AC explicit |
| (\(AC =\)) \(\frac{8}{\sqrt{3}}\) or better | A2 | B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\) oe | oe for B1 e.g. \(\sin 60 = \frac{\sqrt{3}}{2}\) |
| \(\dfrac{45}{360} \times \pi \times \left(\textit{their}\ \dfrac{8}{\sqrt{3}}\right)^2\) | M1 | dep on at least M1 | |
| \(\frac{45}{360} \times \pi \times \frac{64}{3}\) or better or \(\frac{1}{8} \times \pi \times \frac{8}{\sqrt{3}} \times \frac{8}{\sqrt{3}}\) \(= \frac{8}{3}\pi\) | A1 | with no errors seen | Accept \(\frac{1}{8} \times \pi \times \frac{8^2}{3}\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{8\sqrt{3}}{3} + \dfrac{8}{3}\pi\) | 3 | M1 for \(\frac{1}{2} \times 4 \times \frac{8}{\sqrt{3}} \times \sin 30\) FT their AC from (a) or for \(\frac{1}{2} \times 4 \times 4 \times \tan 30\) B1 for \(\sin 30 = \frac{1}{2}\) oe or for \(\tan 30 = \frac{\sqrt{3}}{3}\) oe | Accept e.g. \(\frac{16\sqrt{3}}{6} + \frac{8}{3}\pi\) for 3 marks |
13 The diagram shows a triangle, ABC, with perpendicular height BD.
Not to scale
BC = 12 cm, angle BCD = 30° and angle BAD = 45°.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 6 | 3 | M1 for sin 30 = \(\frac{BD}{12}\) oe B1 for sin 30 = 0.5 oe soi | Accept e.g. sin 30 = \(\frac{\sqrt{1}}{2}\) Do not allow B1 if contradicted |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(6\sqrt{2}\) or \(\frac{12}{\sqrt{2}}\) cao | 3 | B2 for \(\sqrt{72}\) or better OR M1 for \(\frac{their(a)}{\sin 45}\) [= \(AB\)] oe or 2(their (a))\(^2\) oe or \(\frac{12 \times \sin 30}{\sin 45}\) [= \(AB\)] B1 for sin 45 = \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) oe soi | For M1, accept \(\frac{their(a)}{\cos 45}\) [= \(AB\)] oe For M1 , accept e.g. \(\frac{their(a) \times \sin 90}{\sin 45}\) [= \(AB\)] M1 implied by e.g. 72 For B1 accept cos 45 = \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) oe Do not allow B1 if contradicted |
17 The diagram shows triangle ABC.
Not to scale
Find the area of the triangle, in m2.
Give your answer in the form \(a\sqrt{b}\) where \(a\) and \(b\) are integers. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(14\sqrt{2}\) final answer | 3 | M1 for ½ × 8 × 7 × sin 45 oe B1 for sin 45 = \(\frac{1}{\sqrt{2}}\) or better | |
19
[1]
Not to scale
Work out the value of \(x\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{1}{2}\) | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 12 | 2FT | B2FT for final answer \(12\sqrt{3}\), 8, \(8\sqrt{3}\) or 6 if consistent with their (a) or M1 for \(24 \times \dfrac{1}{2}\) or \(24 \times\) their (a) | For 1 or 2 marks, isw attempts to simplify the surds May be implied by their answer Ignore units |
15 ABC is an equilateral triangle of side length 2 cm.
M is the midpoint of AC.
Not to scale
Using this diagram, show that \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
Correctly shows \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) with supporting working | 4 | Using triangle MBC (mark similarly for use of triangle MBA): B1 for MBC = 30° or MC = 1 [cm] soi by use in Pythagoras M2 for [BM =] \(\sqrt{2^2 - 1^{[2]}}\) or \(\sqrt{4 - 1}\) [\(= \sqrt{3}\)] or for [BM2 =] \(2^2 - 1^{[2]}\) oe and \(\sqrt{3}\) or M1 for \(\text{BM}^2 + 1^{[2]} = 2^2\) oe OR B1 for MCB = 60° or MC = 1 [cm] soi by use in cos rule M2 for \(\sqrt{2^2 + 1^{[2]} - 2 \times 2 \times 1 \times \cos 60}\) and \(\cos 60 = \frac{1}{2}\) soi [\(= \sqrt{3}\)] or M1 for \(2^2 + 1^{[2]} - 2 \times 2 \times 1 \times \cos 60\) OR B1 for MCB = 60° soi by use in sine ratio or sine rule M2 for [\(x\) = ] 2sin 60 oe and \(\sin 60 = \frac{\sqrt{3}}{2}\) soi [\(= \sqrt{3}\)] or M1 for \(\frac{\sin 60}{x} = \frac{\sin 90}{2}\) oe AND A1dep for MBC = 30° stated/on diagram and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and with no errors seen leading to the answer | If 0 or 1 scored, SC2 for \(\dfrac{\sin 30}{\cos 30} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\) and \(\tan 30 = \frac{1}{\sqrt{3}}\) May be on diagram For M2 must show the subtraction, do not allow \(\text{BM}^2 + 1^{[2]} = 2^2\) and then \(\sqrt{3}\), this gets M1 only M2 accept [\(x\) = ] \(\dfrac{2\sin 60}{\sin 90}\) and \(\sin 60 = \frac{\sqrt{3}}{2}\) Dep on B1M2 A0 for \(\tan = \frac{1}{\sqrt{3}}\) (without the 30) |
14 Sketch the graph of \(y = \cos x - 1\) for \(0^\circ \leqslant x \leqslant 720^\circ\).
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
 | 3 | B1 for general shape of cosine curve | Starting at its max and completing at least one full cycle; condone incorrect period For full marks, it must be a curve and have correct curvature |
| B1 for max at \(y = 0\), minimum at \(y = -2\) | |||
| B1 for max at \(x = 360, 720\) | Only these two max between \(0 < x \leqslant 720\) | ||
13
Write down the values of \(x\) where this occurs. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct sketch with max at (90, 1) and min at (270, –1) and crossing \(x\)-axis at 0, 180 and 360 | 2 | M1 for correct shape starting at (0, 0) but inaccurate at roots and max/min. Needs at least one cycle, but may have more than one. | Mark intention |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 120 | 1 | ||
| 300 | 1 | FT their 120 + 180 | For FT both must be in range 0 to 360 |
19
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)
Not to scale
Work out the length of BC. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) final answer | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 nfww | 6 | B3 for BD = \(10\sqrt{2}\) oe or M2 for \(10\sqrt{6} \times \tan 30\) oe or M1 for \(\frac{BD}{10\sqrt{6}} = \tan 30\) oe AND M2 for BC = \(\sqrt{\frac{\textit{their}\ BD^2}{2}}\) oe or their BD × their sin 45 oe or M1 for BC2 + CD2 = (their BD)2 or for \(\frac{BC}{\textit{their}\ BD} = \textit{their}\ \sin 45\) | Allow use of other variables for BC and CD (possibly different) |
16 Sketch the graph of \(y = -\sin x\) for \(0^\circ \leqslant x \leqslant 360^\circ\).
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
Correct sketch of \(y = -\sin x\) | 3 | There must be at least one cycle to gain any marks. B1 for a positive or negative sine curve shape starting at (0, 0) and B1 for maximums at (…, 1) and minimum at (…, –1) and B1 for maximum only at (270, …) and minimum only at (90, …) | e.g. B1B1B0 for \(y = \sin x\) drawn B0B1B0 for \(y = \pm\cos x\) drawn B1B1B0 for \(y = \sin 2x\) drawn Before using overlay, check the x-axis line All maximums and minimums within the horizontal zones on overlay Maximum and minimum within the vertical zones on overlay |
15 OAB is a sector of a circle, centre O.
OA = 6 cm and AX is perpendicular to OB.
Not to scale
The area of sector OAB is \(6\pi\) cm\(^2\).
Show that \(\text{AX} = 3\sqrt{3}\) cm. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{x}{360} \times \pi \times 6^2\) or \(\dfrac{[\pi \times]6^2}{[\pi \times]6}\) or shows \(\pi \times 6^2\) and \(\dfrac{1}{6}\) oe | M1 | Accept 36 for \(6^2\) | \(x\) = angle AOX , condone any variable used For M1 may be seen in stages e.g. M1 for \(36\pi \div 6\) |
| \(\dfrac{x}{360}[\times \pi] \times 6^2 = 6[\pi]\) or 360 ÷ 6 | M1dep | Dep on previous M1 | |
| [\(x\) = ] 60 | A1 | Must earn M1M1 before awarding A1 | |
| \(\dfrac{\text{AX}}{6} = \sin\) their 60 oe | M1 | Dep on 0 < their 60 < 90 Accept use of cos 30 or cos 60 and Pythagoras’ or sine rule with 90 | Do not accept assumption that OX = 3 without any evidence |
| \(\text{AX} = 6 \times \dfrac{\sqrt{3}}{2} = 3\sqrt{3}\) or \(\dfrac{3\sqrt{3}}{6} = \dfrac{\text{AX}}{6}\), \(\text{AX} = 3\sqrt{3}\) | M2 | or M1 for \(\sin 60 = \dfrac{\sqrt{3}}{2}\) or \(\cos 30 = \dfrac{\sqrt{3}}{2}\) To award 6 marks, there must be no errors seen | Beware circular methods using \(3\sqrt{3}\) leading to 60, this can only score M1 maximum for \(\sin 60 = \dfrac{\sqrt{3}}{2}\) but ignore circular methods if alongside a correct method |
18
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Cosine-shaped curve with maximum 2 at x = 0°, 360° and 720° and minimum 0 at x = 180° and 540° | 3 | B1 for general shape B1 for max at +2, minimum at 0 B1 for max at x = 0, 360, 720 | Starting at max above the x axis, and completing at least one full cycle For full marks, it must be a curve and have correct curvature |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The maximum value of \(\cos x + 1\) is 2 and 2.7 is greater than 2 oe | 1 | More ‘work’ may be correctly done before an equivalent conclusion, e.g. \(\cos x = 1.7\), and max value of \(\cos x\) is 1 and 1.7 is greater than 1. | |
13 In the diagram, ABC is a right-angled triangle.
P is a point on AB.
BC = 40 m, AP = 20 m and angle ABC = 30°.
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| AC = 40sin30 | M2 | M1 for \(\dfrac{\text{AC}}{40} = \sin 30\) oe | |
| 20 and evidence that sin 30 = 0.5 | A1 | If 0 scored, B1 for sin 30 = 0.5 oe | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(20(\sqrt{3} - 1)\) | 5 | B4 for \(20\sqrt{3} - 20\) or \(\sqrt{1200} - 20\) or B3 for \(\sqrt{1200}\) or \(\dfrac{40\sqrt{3}}{2}\) or M2 for 40cos30 oe or M1 for \(\cos 30 = \dfrac{\text{AB}}{40}\) oe | Other methods are possible e.g. Pythag e.g. M2 for [AB =] \(\sqrt{40^2 - 20^2}\) e.g. M1 for AB² + 20² = 40² |
15
[2]
Give all of the solutions in the range \(0^\circ \leqslant x \leqslant 360^\circ\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct sketch with max at (90, 1) and min at (270, -1) and crossing \(x\)-axis at 0, 180 and 360 | 2 | M1 for correct shape starting at (0, 0) but inaccurate at roots and max/min. Needs at least one cycle, but may have more than one. | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 217° and 323° | 4 | B3 for one correct even if from trials OR M2 for [\(x =\)] −37 to −36.86 OR M1 for \(\sin x = -0.6\) oe If 0 scored SC1 answers summing to 540 to 3sf | Accept answers to greater accuracy 216.8[6…] and 323.1[3…] B3 for grads: [\(x =\)] (−41), 221, 319 OR B2 for grads: [\(x =\)] one of 221, 319 OR M1 implied for grads [\(x =\)] −41 or rads: [\(x =\)] −0.64[…] |
No questions match these filters.