Trigonometric functions — Paper 4 · IGCSE Mathematics 0580

10 B 8.5 cm 12.5 cm NOT TO 60° x cm A C SCALE 46° 76° 58° D The diagram shows a quadrilateral ABCD. (a) The length of AC is x cm. Use the cosine rule in triangle ABC to show that 2x2 – 17x – 168 = 0. [4] (b) Solve the equation 2x2 – 17x – 168 = 0. Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] (c) Use the sine rule to calculate the length of CD. CD = … cm [3] (d) Calculate the area of the quadrilateral ABCD. … cm2 [3]

14 marks

Mark scheme: 10(a) M2 x 2 + 8.5 2 − 12.5 2 12.52 = x2 + 8.52 – 2 × x × 8.5cos60 oe isw M1 for cos60 = 2 × x × 8.5 156.25 = x2 + 72.25 – 8.5x A1 or better 2x2 – 17x – 168 = 0 A1 with no errors or omissions 10(b) 2 2 2 [ −− ]17 ± ([ − ]17) − 4 ( 2 )( −168 ) B1 for ([ − ]17) − 4(2)( −168) or better seen 2 × 2 p + or − q and if in form r B1 for p = [− −] 17 and r = 2 × 2 14.35, –5.85 final answers 1, 1 SC1 for 14.352 to 14.353 and –5.853 to –5.852 seen or 14.3 or 14.4 and –5.8 or –5.9 as final answers or −14.35 and 5.85 as final answers or 14.35 and –5.85 seen in working 10(c) 12.2 or 12.17… nfww 3 their 14.35 × sin46 M2 for sin58 sin 46 sin58 or M1 for = CD their14.35 10(d) 138 or 137.5 to 137.8 nfww 3 M1 for 0.5 × their 14.35 × 8.5sin60 M1 for 0.5 × their 14.35 × their12.2 × sin76

This question in 0580/41 Oct/Nov 2017

1 (a) The angles of a triangle are in the ratio 2 : 3 : 5. (i) Show that the triangle is right-angled. [1] (ii) The length of the hypotenuse of the triangle is 12 cm. Use trigonometry to calculate the length of the shortest side of this triangle. … cm [3] (b) The sides of a different right-angled triangle are in the ratio 3 : 4 : 5. (i) The length of the shortest side is 7.8 cm. Calculate the length of the longest side. … cm [2] (ii) Calculate the smallest angle in this triangle. … [3]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 180 ÷ (2 + 3 + 5) × 5 [= 90] 1 with no errors seen 1(a)(ii) 7.05 or 7.053…. 3 x M2 for = sin36 oe or better 12 or B1 for 36 or 54 seen 1(b)(i) 13 2 M1 for 7.8 ÷ 3 soi 1(b)(ii) 36.9 or 36.86 to 36.87 3 B1 for smallest angle identified 3 M1 for sin[ ] = oe 5 7.8 or sin[ ] = oe their ( b )(i) If zero scored, SC1 for calculation of 53.1

This question in 0580/43 Oct/Nov 2017

8 (a) (i) On the axes, sketch the graph of y = sin x for 0° G x G 360 ° . y 1 0 90° 180° 270° 360° x – 1 [2] (ii) Describe fully the symmetry of the graph of y = sin x for 0° G x G 360 ° . … … [2] (b) Solve 4 sinx - 1 = 2 for 0° G x G 360° . x = … and x = … [3] (c) (i) Write x 2 + 10 x + 14 in the form ( x + a ) 2 + b . … [2] (ii) On the axes, sketch the graph of y = x 2 + 10x + 14 , indicating the coordinates of the turning point. y O x [3]

12 marks

Mark scheme: 8(a)(i) Correct sketch 2 B1 for correct shape but inaccurate 8(a)(ii) Rotational [symmetry] order 2 2 B1 for rotational [symmetry] [centre] (180, 0) 8(b) 48.6 or 48.59 to 48.60 3 B2 for 48.6 or 48.59 to 48.60 or 131.4 or 131.40 and 131.4 or 131.40 to 131.41 to 131.41 or M1 for sin x = 0.75 or better If 0 scored, SC1 for two answers adding to 180 8(c)(i) (x + 5)2 – 11 2 M1 for (x + 5)2 + k or (x + their 5)2 + 14 – (their 5)2 or a = 5 8(c)(ii) Sketch of U-shaped parabola with a 3 FT their (x + 5)2 – 11 provided in that form minimum indicated at (–5, –11) B1 for U shape curve with no part of graph in 4th quadrant B1FT for turning point at (–5, k) or (k, –11)

This question in 0580/41 May/June 2020

10 (a) y NOT TO SCALE B A O x C The diagram shows a sketch of the curve y = x 2 + 3x - 4 . (i) Find the coordinates of the points A, B and C. A ( … , … ) B ( … , … ) C ( … , … ) [4] (ii) Differentiate x 2 + 3x - 4 . … [2] (iii) Find the equation of the tangent to the curve at the point (2, 6). … [3] (b) y 0 x 90° 180° 270° 360° (i) On the diagram, sketch the graph of y = tan x for 0° G x G 360° . [2] (ii) Solve the equation 5 tanx =- 7 for 0° G x G 360° . x = … or x = … [3]

14 marks

Mark scheme: 10(a)(i) A(–4, 0) 4 B3 for A and B correct B(1, 0) Or B2 for B (–4, 0) and A (1, 0) C(0, –4) Or B1 for (x + 4)(x – 1) or for −±3 32 −×4 1 ×−4 oe 2 and B1 for A or B correct B1 for C(0, –4) OR SC2 for –4, 1 and –4 in correct positions on the graph 10(a)(ii) 2x + 3 [ ± 0] final answer 2 B1 for answer 2x +c or for ax + 3, a ≠ 0 or for correct answer seen 10(a)(iii) y = 7x – 8 oe 3 B2 for answer 7x – 8 OR M1 for [gradient =] 2(2) + 3 FT their part (a)(ii) of the form ax + b M1dep for substitution of (2, 6) into y = their mx + c oe 10(b)(i) Correct sketch 2 B1 for one correct section out of 4 OR B1 for two properties correct from • Crosses x-axis at (0, 0) (180, 0) and (360, 0) only • Correct curvature in each section of 90o • Asymptotes at x = 90 and x = 270 0 90 180 270 360 10(b)(ii) 125.5 or 125.53 to 125.54 3 B2 for one correct angle and or B1 for –54.5 or –54.46… or for 2 angles 305.5 or 305.53 to 305.54 with a difference of 180.

This question in 0580/41 Oct/Nov 2020

8 (a) O 53° NOT TO 9.5 cm A B SCALE X Y The diagram shows a sector OXY of a circle with centre O and radius 9.5 cm. The sector angle is 53°. A lies on OX, B lies on OY and OA = OB . (i) Show that the area of the sector is 41.7 cm 2, correct to 1 decimal place. [2] 1 (ii) The area of triangle OAB is of the area of sector OXY. 3 Calculate OA. OA = … cm [4] (b) O 60° NOT TO 24 cm SCALE P Q The diagram shows a sector OPQ of a circle with centre O and radius 24 cm. The sector angle is 60°. A cone is made from this sector by joining OP to OQ. O NOT TO SCALE P Q Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [6]

12 marks

Mark scheme: 8(a)(i) 53 2 M1 × π× 9.5 360 41.74 to 41.75 A1 8(a)(ii) 5.9[0] or 5.899 to 5.903.. 4 2 13 × 41.7 M3 for  OA =  oe   1 2 sin53 1 2 1 M2 for × OA × sin53 = × 41.7 oe 2 3 1 1 M1 for × OA × OB × sin53 = × 41.7 seen or 2 3 better 8(b) 396 or 397 or 396.4 to 396.6 6  60  M2 for [ r = ]  ×2× π×24  ÷ 2π oe or better  360  60 or M1 for 2πr = × 2 × π× 24 oe 360 M2 for 24 2 −a 2 or M1 for h2 + a2 = 242 1 2 M1 for 3π× their r × their h

This question in 0580/42 Feb/March 2021

12 (a) Solve the equation tan x = 11.43 for 0° G x G 360 ° . x = … or x = … [2] (b) Sketch the curve y = x 3 - 4x . y x O [3] (c) A curve has equation y = x 3 + ax + b . The stationary points of the curve have coordinates (2, k) and (-2, 10 - k). Work out the value of a, the value of b and the value of k. a = … , b = … , k = … [6]

11 marks

Mark scheme: 12(a) 85[.0], 265[.0] and no others 2 B1 for each If 0 scored SC1 for two values in the range with a difference of 180 but not multiples of 90 12(b) correct shape and passes through 3 B1 for any positive cubic shape origin B1 for sketch with one max and one min and with 3 roots including zero If 0 scored, SC1 for x(x + 2)(x – 2) soi 12(c) a = –12 6 B5 for 2 correct b = 5 OR k = –11 B2 for 3x2 + a or B1 for 3x2 isw d y M1dep on at least B1 for their = 0 d x M1dep on at least B1M1 for x = 2 or x = – 2 d y substituted in their = 0 equation d x M1 for k = 23 + 2 × their a + b and 3 10 − k = ( −2 ) + ( −2 ) × their a + b

This question in 0580/42 Feb/March 2022

9 y 1 0 x 360° – 1 (a) On the diagram, sketch the graph of y = sin x for 0° G x G 360° . [2] (b) Solve the equation 5 sinx + 4 = 0 for 0° G x G 360 ° . x = … or x = … [3]

5 marks

Mark scheme: 9(a) Correct sketch to go through (0, 0), and (360, 0) 2 y M1 for correct sine curve shape through the origin or for almost correct sketch fitting all tramlines but with an omission at either end or incorrect curvature in one place only 0 360º x 9(b) 233.1 or 233.13… 3 B2 for one correct angle and or M1 for sin x = –0.8 oe 306.9 or 306.86 to 306.87 If 0 scored SC1 for 2 reflex angles that add to 540 or two non- reflex angles that add to 180

This question in 0580/42 Oct/Nov 2022

12 (a) Sketch the graph of y = tan x for 0 ° G x G 360° . y 0 x 90° 180° 270° 360° [2] 1 (b) Find x when tanx = and 0° G x G 360° . 3 … [2]

4 marks

Mark scheme: 12(a) Correct4444 sketch 2 Condone curve touching asymptotes but not crossing 2222 B1 for one section correct 0000 0000 50505050 100100100100 150150150150 200200200200 250250250250 300300300300 350350350350 -2-2-2-2 or for 3 sections in correct part of graph but with incorrect curvature and no other sections in incorrect part of graph -4-4-4-4 12(b) 30 and 210 final answer 2 B1 for each If 0 scored SC1 for two answers (one acute and one reflex) with a difference of 180

This question in 0580/42 Feb/March 2023

4 (a) P NOT TO SCALE 8 cm Q R 24 cm (i) Calculate the area of triangle PQR. … cm2 [2] (ii) Calculate angle PRQ. Angle PRQ = … [2] (b) NOT TO SCALE 11 cm 6 cm The diagram shows a half-cylinder of radius 6 cm and length 11 cm. Calculate the volume of the half-cylinder. … cm3 [2] (c) T T D C C 44 cmcm S S O NOT TO 15 cm X X SCALE A B A B 20 cm (i) ABCD is a rectangle with AB = 20 cm and BC = 15 cm. S, X and T are points on a circle centre O, such that DSA and DTC are tangents to the circle. The radius of the circle is 4 cm and TX is a diameter of the circle. The shape DSXT is removed from the corner of the rectangle, leaving the shaded shape shown in the second diagram. Calculate the area of the shaded shape. … cm2 [5] (ii) Calculate the perimeter of the shaded shape. … cm [3]

14 marks

Mark scheme: 4(a)(i) 96 2 1 M1 for  24  8 2 4(a)(ii) 18.4 or 18.43... 2 8 M1 for tan x  oe 24 4(b) 622 or 622.0 to 622.1 … 2 1 2 1 2 M1 for [ 2] 6  11 or 2  6 [ 11] 4(c)(i) 246 or 246.2 to 246.3... 5 270 2 M4 for 15  20 4 4   4 oe 360 OR 270 2 M2 for  4 oe 360 or M1 for k   4 2 , where k  1 M1 for15  20 or 4  4 oe 4(c)(ii) 80.8 or 80.9 or 80.84 to 80.85... 3 M1 for 15  20  11  16 oe 3 M1 for 2 4 oe 4

This question in 0580/41 May/June 2023

7 (a) The diagram shows the graph of a function. y x O Put a ring around the word which correctly identifies the type of function. reciprocal quadratic cubic exponential linear [1] (b) (i) y x O 1 On the diagram, sketch the graph of y = , x ! 0 . [2] 2 x 1 (ii) Solve the equation = 2x . 2 x x = … and x = … [2] (c) (i) y 1 0 180° 360° x – 1 On the diagram, sketch the graph of y = sin x for 0° G x G 360 ° . [2] (ii) Solve the equation 3 sinx + 1 = 0 for 0° G x G 360° . x = … and x = … [3]

10 marks

Mark scheme: 7(a) Cubic 1 7(b)(i) Correct sketch 2 B1 for one branch correct or an attempt at the correct shape y Maximum 1 mark if sketch crosses x- axis or y-axis O x 7(b)(ii) 1 2 M1 for 4 x 2  1 oe  nfww 2 1 1 or B1 for or  nfww 2 2 7(c)(i) Correct sketch through (0, 0) (180, 0) and 2 B1 for correct sine curve shape, starting (360, 0) with max and min at 1 and –1 resp. at the origin, with minimum of 1 cycle. 180 360 7(c)(ii) 199.5 or 199.47... 3 B2 for one correct 1 and or M1 for sin x =  oe 3 340.5... If 0 scored, SC1 for two reflex angles with a sum of 540 or 2 non-reflex angles with a sum of 180

This question in 0580/43 May/June 2023

6 NOT TO SCALE (2t + 3) cm t cm w° 5 cm The diagram shows a right-angled triangle. Find the value of w. w = … [7]

7 marks

Mark scheme: 6 11.9 or 11.91 to 11.92 7 B5 for t = 1.055 or 1.0550... their t M1 for tan w = oe 5 OR 2 2 2 M1 for ( 2t + 3) = t + 5 oe seen isw M2 for 3t 2 + 12t − 16  = 0 oe seen isw or B1 for 4t 2 + 6t + 6t + 9 −12  12 2 − 4(3)( −16) M1FT for oe 2(3) their t M1 for tan w = oe 5

This question in 0580/42 Oct/Nov 2023

7 (a) Complete the table of values for y = 3 cos 2x° . Values are given correct to 1 decimal place. x 0 10 20 30 40 45 50 60 70 80 90 y 3.0 2.8 2.3 1.5 0.5 - .05 - .23 - .30 [3] (b) Draw the graph of y = 3 cos 2x° for 0 G x G 90 . y 3 2 1 0 x 10 20 30 40 50 60 70 80 90 - 1 - 2 - 3 [4] (c) Use your graph to solve the equation 3 cos 2x° =- 2 for 0 G x G 90 . x = … [1] (d) By drawing a suitable straight line, solve the equation 120 cos 2x° = 80 - x for 0 G x G 90 . x = … [3]

11 marks

Mark scheme: 7(a) 0, –1.5 oe, –2.8 3 B1 for each 7(b) Correct graph 4 B3 FT for 10 or 11 correct points FT their table or B2 FT for 8 or 9 correct points FT their table or B1 FT for 6 or 7 correct points FT their table 7(c) 65 to 67 1 FT intersection of their graph with y = – 2 7(d) M2 y = 2 −x oe ruled M1 for [ y =] 2 −x oe soi 40 40 or for 3 cos 2x = 2 −x oe soi 40 32 to 36 B1

This question in 0580/43 Oct/Nov 2023

4 In this question all the measurements are in centimetres. NOT TO r + 2 SCALE 30° r + 5 r + 1 The area of the triangle is equal to the area of the square. (a) Show that 3r 2 + r - 6 = 0 . [4] (b) Solve the equation 3r 2 + r - 6 = 0 . Give your answer to 2 decimal places. You must show all your working. r = … or r = … [3] (c) Find the perimeter of the square. … cm [2]

9 marks

Mark scheme: 4(a) 1 2 M2 1 ( r  5)( r  2)sin30  ( r  1) M1 for ( r  5)( r  2)sin30 oe 2 2 r 2  5r  2 r  10 or r 2  r  r  1 soi B1 Leading to 3r 2  r  6  0 with no errors A1 Dependent on both expansions seen or omissions 4(b) 2 B2 1 1  4(3)( 6) 1 p B1 for 21  4(3)( 6) or for 2(3) 2(3) Or 1 p or 2(3) or 1  1  2   2   oe 2 6  6   1   r   or  6   2  or 1 1  1     18    oe 3  2  2   2    1   3r    2  –1.59 and 1.26 B1 4(c) 9.028 to 9.040 2 M1 for (their root (greater than –1) + 1) × 4

This question in 0580/43 May/June 2024

11 (a) Q P NOT TO SCALE x° O 3 In the circle, centre O, the length of the minor arc PQ is of the length of the major arc PQ. 7 Show that x = 108 . [3] (b) A NOT TO SCALE r y° B O The diagram shows a sector, OAB, of a circle with centre O and radius r. The area of triangle OAB is half the area of the sector. Angle AOB = y° and is obtuse. (i) Show that 360 siny = r y . [2] (ii) Complete the table, giving your answers correct to two decimal places. y 360 siny ry 108.4 341.60 340.55 108.5 341.40 340.86 108.6 341.20 108.7 [3] (iii) Complete the statement. The value of y, correct to one decimal place, that satisfies the equation 360 siny = r y is … . [1]

9 marks

Mark scheme: 11(a) 3 M2  360 oe 3 x 10 M1 for = 3  7 360 x 3 360  x or for [ 2r] =  [ 360 7 360 2r] oe or better or 10 360  x 1 [ 2r] =  [ 2r] oe or 7 360 better 360 or  k (k = 1 or 7) 7  3 108 A1 11(b)(i) 1 1 y 2 y 1 r2 siny =   r2 M1 for  r2 or for r2 siny 2 2 360 360 2 y 1 or  r2 = [2  ] r2 siny 360 2 and one further step leading to 360siny = y with no errors 11(b)(ii) 341.18 or 341.22 3 B1 for each 341.00 341.49 or 341.54 11(b)(iii) 108.6 cao 1

This question in 0580/43 May/June 2024

19 Solve the equation 2 + 5 cos x = 0 for 0° # x # 360°. x = … or x = … [3]

3 marks

Mark scheme: 19 113.6 and 246.4 3 B2 for one correct angle as answer or M1 for cos x = –0.4 oe If M1 or 0 scored, SC1 for 2 angles that add and round to 360.0

This question in 0580/42 Feb/March 2025