I Math SL: Trig Practice Problems Circular Functions and Trig - Practice Problems (to 07). In the triangle PQR, PR = 5 cm, QR = 4 cm and PQ = 6 cm. Calculate (a) the size of PQˆ R ; (b) the area of triangle PQR.. The following diagram shows a triangle C, where Ĉ is 90, =, C = and ÂC is. (a) Show that sin = 5. (b) 4 5 Show that sin =. 9 (c) Find the exact value of cos.. The following diagram shows a sector of a circle of radius r cm, and angle at the centre. The perimeter of the sector is 0 cm. (a) 0r Show that =. r (b) The area of the sector is 5 cm. Find the value of r. 4. The following diagram shows the triangle OP, where OP = cm, P = 4 cm and O = cm. diagram not to scale O P (a) Calculate ÔP, giving your answer in radians. The following diagram shows two circles which intersect at the points and. The smaller circle C has centre O and radius cm, the larger circle C has centre P and radius 4 cm, and OP = cm. The point D lies on the circumference of C and E on the circumference of C.Triangle OP is the same as triangle OP in the diagram above. () C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems C C D E O P diagram not to scale (b) Find Ô, giving your answer in radians. (c) Given that Pˆ is.6 radians, calculate the area of (i) sector PE; (ii) sector OD. (d) The area of the quadrilateral OP is 5.8 cm. (i) Find the area of OE. (ii) Hence find the area of the shaded region ED. () (5) (4) (Total 4 marks) ÊD 5. The following diagram shows a pentagon CDE, with = 9. cm, C =. cm, D = 7. cm, =0, Dˆ E = 5 and ˆ D = 60. (a) Find D. (b) Find DE. (c) The area of triangle CD is 5.68 cm. Find Dˆ C. (d) Find C. (e) Find the area of quadrilateral CD. 6. x The diagram below shows the graph of f (x) = + tan for 60 x 60. (4) (4) (4) (4) (5) (Total marks) (a) On the same diagram, draw the asymptotes. () C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems (b) Write down (i) the period of the function; (ii) the value of f (90). (c) Solve f (x) = 0 for 60 x 60. 7. (a) Consider the equation 4x + kx + = 0. For what values of k does this equation have two equal roots? Let f be the function f ( ) = cos + 4 cos +, for 60 60. (b) Show that this function may be written as f ( ) = 4 cos + 4 cos +. (c) Consider the equation f ( ) = 0, for 60 60. (i) How many distinct values of cos satisfy this equation? (ii) Find all values of which satisfy this equation. () () (d) Given that f ( ) = c is satisfied by only three values of, find the value of c. () (Total marks) 8. Ferris wheel with centre O and a radius of 5 metres is represented in the diagram below. Initially seat is at ground level. The next seat is, where Ô =. 6 () () (5) (a) (b) Find the length of the arc. Find the area of the sector O. (c) The wheel turns clockwise through an angle of. Find the height of above the ground. The height, h metres, of seat C above the ground after t minutes, can be modelled by the function h (t) = 5 5 cos t. 4 (d) (i) Find the height of seat C when t =. 4 (ii) Find the initial height of seat C. (iii) Find the time at which seat C first reaches its highest point. (e) Find h (t). (f) For 0 t, (i) sketch the graph of h ; (ii) find the time at which the height is changing most rapidly. 9. Let f (x) = a (x 4) + 8. (a) Write down the coordinates of the vertex of the curve of f. (b) Given that f (7) = 0, find the value of a. (c) Hence find the y-intercept of the curve of f. () () () (8) () (5) (Total marks) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems 0. The following diagram shows a circle with radius r and centre O. The points, and C are on the circle and ÔC =. 4 The area of sector OC is and the length of arc C is. Find the value of r and of.. Let ƒ (x) = a sin b (x c). Part of the graph of ƒ is given below. Given that a, b and c are positive, find the value of a, of b and of c.. The points P(, 4), Q (, ) and R (, 6) are shown in the diagram below. (a) Find the vector PQ. (b) Find a vector equation for the line through R parallel to the line (PQ).. The diagram below shows a circle of radius r and centre O. The angle Ô =. The length of the arc is 4 cm. The area of the sector O is 80 cm. Find the value of r and of. 4. The diagram below shows a quadrilateral CD. = 4, D = 8, CD =, D = 5, ÂD =. Ĉ C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 4 of
I Math SL: Trig Practice Problems (a) Use the cosine rule to show that D = 4 54cos. Let = 40. (b) (i) Find the value of sin Cˆ D. (c) (ii) Find the two possible values for the size of Cˆ D. (iii) Given that Cˆ D is an acute angle, find the perimeter of CD. Find the area of triangle D. 5. (a) Let y = 6x + 60x 56. Given that y has a maximum value, find (i) the value of x giving the maximum value of y; (ii) this maximum value of y. The triangle XYZ has XZ = 6, YZ = x, XY = z as shown below. The perimeter of triangle XYZ is 6. () () () (4) (b) (i) Express z in terms of x. (ii) Using the cosine rule, express z in terms of x and cos Z. (iii) 5x 6 Hence, show that cos Z =. x Let the area of triangle XYZ be. (c) Show that = 9x sin Z. (d) Hence, show that = 6x + 60x 56. (e) (i) Hence, write down the maximum area for triangle XYZ. (ii) What type of triangle is the triangle with maximum area? (7) () (4) () (Total 0 marks) 6. The function f is defined by f : x 0 sin x cos x, 0 x. (a) Write down an expression for f (x) in the form a sin 6x, where a is an integer. (b) Solve f (x) = 0, giving your answers in terms of. 7. The following diagram shows two semi-circles. The larger one has centre O and radius 4 cm. The smaller one has centre P, radius cm, and passes through O. The line (OP) meets the larger semi-circle at S. The semi-circles intersect at Q. C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 5 of
I Math SL: Trig Practice Problems (a) (i) Explain why OPQ is an isosceles triangle. (ii) Use the cosine rule to show that cos OPˆQ =. 9 (iii) Hence show that sin OPˆQ = 80. 9 (iv) Find the area of the triangle OPQ. (b) Consider the smaller semi-circle, with centre P. (i) Write down the size of OPˆQ. (ii) Calculate the area of the sector OPQ. (c) (d) Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS. Hence calculate the area of the shaded region. 8. The graph of a function of the form y = p cos qx is given in the diagram below. (a) Write down the value of p. (b) Calculate the value of q. (7) () () (4) (Total 7 marks) 9. farmer owns a triangular field C. One side of the triangle, [C], is 04 m, a second side, [], is 65 m and the angle between these two sides is 60. (a) Use the cosine rule to calculate the length of the third side of the field. 40 0 0 0 0 0 0 40 y / x (b) Given that sin 60 =, find the area of the field in the form p where p is an integer. Let D be a point on [C] such that [D] bisects the 60 angle. The farmer divides the field into two parts and by constructing a straight fence [D] of length x metres, as shown on the diagram below. C () () 04 m 0 0 x D 65 m (c) (i) 65x Show that the area of l is given by. 4 (ii) Find a similar expression for the area of. C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 6 of
I Math SL: Trig Practice Problems (iii) Hence, find the value of x in the form q, where q is an integer. (d) (i) Explain why sin Dˆ Csin Dˆ. (ii) Use the result of part (i) and the sine rule to show that D 5. DC 8 (7) (5) (Total 8 marks) 0. The following diagram shows a circle of centre O, and radius r. The shaded sector OC has an area of 7 cm. ngle Ô = θ =.5 radians. C O r (a) Find the radius. (b) Calculate the length of the minor arc C. Working: nswers: (a)... (b).... Consider y = sin x. 9 (a) The graph of y intersects the x-axis at point. Find the x-coordinate of, where 0 x. (b) Solve the equation sin x =, for 0 x. 9 Working: nswers: (a)... (b).... The diagram shows a triangular region formed by a hedge [], a part of a river bank [C] and a fence [C]. The hedge is 7 m long and ÂC is 9. The end of the fence, point C, can be positioned anywhere along the river bank. (a) Given that point C is 5 m from, find the length of the fence [C]. 9 5 m C river bank 7 m (b) The farmer has another, longer fence. It is possible for him to enclose two different triangular regions with this fence. He places the fence so that ˆ C is 85. (i) Find the distance from to C. (ii) Find the area of the region C with the fence in this position. (c) To form the second region, he moves the fencing so that point C is closer to point. Find the new distance from to C. (d) Find the minimum length of fence [C] needed to enclose a triangular region C. () (5) (4) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 7 of
I Math SL: Trig Practice Problems. Let f (x) = sin x + cos x for 0 x. (a) (i) Find f (x). One way of writing f (x) is sin x sin x +. (ii) Factorize sin x + sin x. (iii) Hence or otherwise, solve f (x) = 0. The graph of y = f (x) is shown below. y () (Total 4 marks) (6) b 0 a x There is a maximum point at and a minimum point at. (b) Write down the x-coordinate of point. () (c) The region bounded by the graph, the x-axis and the lines x = a and x = b is shaded in the diagram above. (i) Write down an expression that represents the area of this shaded region. (ii) Calculate the area of this shaded region. (5) (Total marks) 4. In triangle PQR, PQ is 0 cm, QR is 8 cm and angle PQR is acute. The area of the triangle is 0 cm. Find the size of angle PQˆ R. 5. Let f (x) = 6 sin x, and g (x) = 6e x, for 0 x. The graph of f is shown on the diagram below. There is a maximum value at (0.5, b). y 0 x (a) Write down the value of b. (b) On the same diagram, sketch the graph of g. (c) Solve f (x) = g (x), 0.5 x.5. 6. Consider the equation cos x + sin x = (a) Write this equation in the form f (x) = 0, where f (x) = p sin x + q sin x + r, and p, q, r. (b) Factorize f (x). (c) Write down the number of solutions of f (x) = 0, for 0 x. 7. The diagram below shows two circles which have the same centre O and radii 6 cm and 0 cm respectively. The two arcs and CD have the same sector angle =.5 radians. C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 8 of
I Math SL: Trig Practice Problems C D O Find the area of the shaded region. 8. Let f (x) = sin (x + ), 0 x. (a) Sketch the curve of y = f (x) on the grid below. y.5 0.5 0 0.5.5.5.5 x 0.5.5 (b) Find the x-coordinates of the maximum and minimum points of f (x), giving your answers correct to one decimal place. 9. In a triangle C, = 4 cm, C = cm and the area of the triangle is 4.5 cm. Find the two possible values of the angle ÂC. Working: nswer:.. 0. Solve the equation cos x = sin x for 0 x, giving your answers in terms of. Working: nswer:... The depth y metres of water in a harbour is given by the equation t y = 0 + 4 sin, where t is the number of hours after midnight. (a) Calculate the depth of the water (i) when t = ; (ii) at 00. The sketch below shows the depth y, of water, at time t, during one day (4 hours). () C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 9 of
I Math SL: Trig Practice Problems 4 5 6 7 8 9 0 4 5 6 7 8 9 0 4 t time (hours) (b) (i) Write down the maximum depth of water in the harbour. (ii) Calculate the value of t when the water is first at its maximum depth during the day. The harbour gates are closed when the depth of the water is less than seven metres. n alarm rings when the gates are opened or closed. (c) (i) How many times does the alarm sound during the day? (ii) (iii) y 5 4 0 9 depth (metres) 8 7 6 5 4 Find the value of t when the alarm sounds first. Use the graph to find the length of time during the day when the harbour gates are closed. Give your answer in hours, to the nearest hour.. The following diagram shows a triangle C, where C = 5 cm, ˆ = 60, Ĉ = 40. () (7) (Total marks) 60 40 5 cm C (a) Calculate. (b) Find the area of the triangle. Working: nswers: (a).. (b)... The diagram below shows a circle of radius 5 cm with centre O. Points and are on the circle, and Ô is 0.8 radians. The point N is on [O] such that [N] is perpendicular to [O]. 5 cm O 0.8 N Find the area of the shaded region. Working: 4. Let f (x) = + cos (x) for 0 x, and x is in radians. nswer:... C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 0 of
I Math SL: Trig Practice Problems (a) (i) Find f (x). (ii) Find the values for x for which f (x) = 0, giving your answers in terms of. The function g (x) is defined as g (x) = f (x), 0 x. (b) (i) The graph of f may be transformed to the graph of g by a stretch in the x-direction with scale factor followed by another transformation. Describe fully this other transformation. (ii) Find the solution to the equation g (x) = f (x) (4) (Total 0 marks) 5. Part of the graph of y = p + q cos x is shown below. The graph passes through the points (0, ) and (, ). y (6) 0 x Find the value of (a) p; (b) q. Working: nswers: (a)... (b)... 6. Find all solutions of the equation cos x = cos (0.5x), for 0 x. Working: nswer:... 7. The diagram below shows a triangle and two arcs of circles. The triangle C is a right-angled isosceles triangle, with = C =. The point P is the midpoint of [C]. The arc DC is part of a circle with centre. The arc EC is part of a circle with centre P. E D P (a) (b) Calculate the area of the segment DCP. Calculate the area of the shaded region ECD. 7 8. The diagram shows a parallelogram OPQR in which OP = 0, OQ =. C C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems y P O R Q x (a) Find the vector OR. 5 (b) Use the scalar product of two vectors to show that cos O PˆQ =. 754 (c) (i) Explain why cos PQˆ R = cos OPˆ Q. (ii) Hence show that sin PQˆ R =. 754 (iii) Calculate the area of the parallelogram OPQR, giving your answer as an integer. (7) (Total 4 marks) 9. The points P, Q, R are three markers on level ground, joined by straight paths PQ, QR, PR as shown in the diagram. QR = 9 km, PQˆ R = 5, PRˆ Q = 5. P (a) (b) (c) 5 5 Q 9 km Find the length PR. Diagram not to scale Tom sets out to walk from Q to P at a steady speed of 8 km h. t the same time, lan sets out to jog from R to P at a steady speed of a km h. They reach P at the same time. Calculate the value of a. The point S is on [PQ], such that RS = QS, as shown in the diagram. P Q Find the length QS. S 40. Consider the function f (x) = cos x + sin x. (a) (i) Show that f ( ) = 0. 4 (ii) Find in terms of, the smallest positive value of x which satisfies f (x) = 0. () (4) () (7) (6) The diagram shows the graph of y = e x (cos x + sin x), x. The graph has a maximum turning point at C(a, b) and a point of inflexion at D. R R () C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems y 6 C(a, b) 4 D x dy (b) Find. dx (c) Find the exact value of a and of b. (d) Show that at D, y = e. (e) Find the area of the shaded region. 4. The graph of the function f (x) = x 4 intersects the x-axis at and the y-axis at. (a) Find the coordinates of (i) ; (ii). (b) Let O denote the origin. Find the area of triangle O. Working: nswers: (a) (i)... (ii)... (b)... 4. (a) Factorize the expression sin x sin x + 6. 4 (b) Consider the equation sin x sin x + 6 = 0. (i) Find the two values of sin x which satisfy this equation, (ii) Solve the equation, for 0 x 80. Working: nswers: (a)... (b) (i)... (ii)... () (4) (5) () (Total 7 marks) 4. The diagram below shows a circle, centre O, with a radius cm. The chord subtends at an angle of 75 at the centre. The tangents to the circle at and at meet at P. C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems cm O 75º P diagram not to scale (a) Using the cosine rule, show that the length of is cos75. (b) (c) Find the length of P. Hence find (i) the area of triangle OP; (ii) the area of triangle P. (d) Find the area of sector O. () (e) Find the area of the shaded region. () (Total marks) 44. Note: Radians are used throughout this question. mass is suspended from the ceiling on a spring. It is pulled down to point P and then released. It oscillates up and down. () () (4) diagram not to scale P Its distance, s cm, from the ceiling, is modelled by the function s = 48 + 0 cos t where t is the time in seconds from release. (a) (i) What is the distance of the point P from the ceiling? (ii) How long is it until the mass is next at P? (b) (i) ds Find. dt (ii) Where is the mass when the velocity is zero? second mass is suspended on another spring. Its distance r cm from the ceiling is modelled by the function r = 60 + 5 cos 4t. The two masses are released at the same instant. (c) Find the value of t when they are first at the same distance below the ceiling. (d) In the first three seconds, how many times are the two masses at the same height? C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 4 of (5) (7) () ()
I Math SL: Trig Practice Problems 45. The following diagram shows a circle of centre O, and radius 5 cm. The arc C subtends an angle of radians at the centre O. C 5 cm rad Diagram not to scale O Ô = radians O = 5 cm Find (a) the length of the arc C; (b) the area of the shaded region. Working: nswers: (a)... (b)... 46. Two boats and start moving from the same point P. oat moves in a straight line at 0 km h and boat moves in a straight line at km h. The angle between their paths is 70. Find the distance between the boats after.5 hours. 47. Let f (x) = sin x and g (x) = sin (0.5x). (a) Write down (i) the minimum value of the function f ; (ii) the period of the function g. (b) Consider the equation f (x) = g (x). Find the number of solutions to this equation, for 0 x. Working: nswers: (a) (i)... (ii)... (b)... 48. Consider the following statements : log0 (0 x ) > 0. : 0.5 cos (0.5x) 0.5. C: arctan x. (a) Determine which statements are true for all real numbers x. Write your answers (yes or no) in the table below. Statement (b) If not true, example C (a) Is the statement true for all real numbers x? (Yes/No) (b) If a statement is not true for all x, complete the last column by giving an example of one value of x for which the statement is false. Working: C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 5 of
I Math SL: Trig Practice Problems 49. The diagram shows a triangle C in which C = 7, C = 6, ˆ C = 45. Diagram not to scale 7 45 6 (a) Use the fact that sin 45 = 6 to show that sin ÂC =. 7 6 The point D is on (), between and, such that sin Dˆ C =. 7 (b) (i) Write down the value of Dˆ C + ÂC. (ii) Calculate the angle CD. (iii) Find the length of [D]. (c) rea of DC D Show that =. rea of C 50. In triangle C, C = 5, C = 7, Â = 48, as shown in the diagram. C C () (6) () (Total 0 marks) 5 7 diagram not to scale 48 Find ˆ, giving your answer correct to the nearest degree. Working: nswer:... 5. Given that sin x =, where x is an acute angle, find the exact value of (a) cos x; (b) cos x. Working: nswers: (a)... (b)... 5. Consider the trigonometric equation sin x = + cos x. (a) Write this equation in the form f (x) = 0, where f (x) = a cos x + b cos x + c, and a, b, c. (b) Factorize f (x). (c) Solve f (x) = 0 for 0 x 60. Working: nswers: (a)... C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 6 of
I Math SL: Trig Practice Problems 5. The following diagram shows a triangle with sides 5 cm, 7 cm, 8 cm. (b)... (c)... 5 7 Diagram not to scale 8 Find (a) the size of the smallest angle, in degrees; (b) the area of the triangle. Working: nswers: (a)... (b)... 54. (a) Write the expression sin x + 4 cos x in the form a cos x + b cos x + c. (b) Hence or otherwise, solve the equation Working: sin x + 4 cos x 4 = 0, 0 x 90. nswers: (a)... (b)... 55. In the following diagram, O is the centre of the circle and (T) is the tangent to the circle at T. T (Total 4 marks) (Total 4 marks) O Diagram not to scale If O = cm, and the circle has a radius of 6 cm, find the area of the shaded region. Working: nswer:... 56. In the diagram below, the points O(0, 0) and (8, 6) are fixed. The angle OPˆ varies as the point P(x, 0) moves along the horizontal line y = 0. y P( x, 0) y=0 (Total 4 marks) (8, 6) O(0, 0) Diagram to scale (a) (i) Show that P x 6x 80. x C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 7 of
I Math SL: Trig Practice Problems (ii) Write down a similar expression for OP in terms of x. () (b) Hence, show that x 8x 40 cos OPˆ, {( x 6x 80)( x 00)} () (c) Find, in degrees, the angle OPˆ when x = 8. () (d) Find the positive value of x such that OPˆ 60. (4) Let the function f be defined by x 8x 40 f ( x) cos OPˆ, 0 x 5. {( x 6x 80)( x 00)} (e) Consider the equation f (x) =. (i) Explain, in terms of the position of the points O,, and P, why this equation has a solution. (ii) Find the exact solution to the equation. (5) 57. The diagram below shows a sector O of a circle of radius 5 cm and centre O. The angle at the centre of the circle is radians. Diagram not to scale (a) Calculate the area of the sector O. (b) Calculate the area of the shaded region. Working: 58. The diagrams below show two triangles both satisfying the conditions = 0 cm, C = 7 cm, ˆ C = 50. Diagrams not to scale O nswers: (a)... (b)... Triangle Triangle (Total 4 marks) (a) Calculate the size of Ĉ in Triangle. (b) Calculate the area of Triangle. Working: C C nswers: (a)... C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 8 of
I Math SL: Trig Practice Problems (b)... (Total 4 marks) 59. The depth, y metres, of sea water in a bay t hours after midnight may be represented by the function y a b cos t, where a, b and k are constants. k The water is at a maximum depth of 4. m at midnight and noon, and is at a minimum depth of 0. m at 06:00 and at 8:00. Write down the value of (a) a; (b) b; (c) k. Working: nswers: (a)... (b)... (c)... (Total 4 marks) 60. Town is 48 km from town and km from town C as shown in the diagram. C km 48km Given that town is 56 km from town C, find the size of angle CÂ 6. (a) Express cos x + sin x in terms of sin x only. to the nearest degree. (Total 4 marks) (b) Solve the equation cos x + sin x = for x in the interval 0 x, giving your answers exactly. (Total 4 marks) 6. Note: Radians are used throughout this question. (a) Draw the graph of y = + x cos x, 0 x 5, on millimetre square graph paper, using a scale of cm per unit. Make clear (i) the integer values of x and y on each axis; (ii) the approximate positions of the x-intercepts and the turning points. (5) (b) Without the use of a calculator, show that is a solution of the equation + x cos x = 0. () (c) Find another solution of the equation + x cos x = 0 for 0 x 5, giving your answer to six significant figures. () (d) Let R be the region enclosed by the graph and the axes for 0 x. Shade R on your diagram, and write down an integral which represents the area of R. () (e) Evaluate the integral in part (d) to an accuracy of six significant figures. (If you consider it necessary, you d can make use of the result ( xsin x cos x ) x cos x.) dx () (Total 5 marks) 6. formula for the depth d metres of water in a harbour at a time t hours after midnight is d P Q cos t, 0 t 4, 6 where P and Q are positive constants. In the following graph the point (6, 8.) is a minimum point and the point (, 4.6) is a maximum point. C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 9 of
I Math SL: Trig Practice Problems d 5 (, 4.6) 0. 5 (6, 8.) 0 6 8 4 t (a) Find the value of (i) Q; (ii) P. () (b) Find the first time in the 4-hour period when the depth of the water is 0 metres. () (c) (i) Use the symmetry of the graph to find the next time when the depth of the water is 0 metres. (ii) Hence find the time intervals in the 4-hour period during which the water is less than 0 metres deep. (4) 64. Solve the equation cos x = 5 sin x, for x in the interval 0 x 60, giving your answers to the nearest degree. (Total 4 marks) 65. 5 If is an obtuse angle in a triangle and sin =, calculate the exact value of sin. (Total 4 marks) 66. (a) Sketch the graph of y = sin x x, x, on millimetre square paper, using a scale of cm per unit on each axis. Label and number both axes and indicate clearly the approximate positions of the x-intercepts and the local maximum and minimum points. (5) (b) Find the solution of the equation sin x x = 0, x > 0. () (c) Find the indefinite integral ( sin x x)dx and hence, or otherwise, calculate the area of the region enclosed by the graph, the x-axis and the line x =. (4) (Total 0 marks) 67. Given that sin θ =, cos θ = and 0 θ 60, (a) find the value of θ; (b) write down the exact value of tan θ. (Total 4 marks) 68. The diagram shows a vertical pole PQ, which is supported by two wires fixed to the horizontal ground at and. P Q Q = 40 m 70 6 0 C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 0 of
I Math SL: Trig Practice Problems Find (a) the height of the pole, PQ; (b) the distance between and. Pˆ Q ÂQ ˆ Q 69. The diagram shows a circle of radius 5 cm. = 6 = 70 = 0 radian (Total 4 marks) Find the perimeter of the shaded region. 70. f (x) = 4 sin x. For what values of k will the equation f (x) = k have no solutions? 7. In this question you should note that radians are used throughout. (a) (i) Sketch the graph of y = x cos x, for 0 x making clear the approximate positions of the positive x-intercept, the maximum point and the end-points. (ii) Write down the approximate coordinates of the positive x-intercept, the maximum point and the end-points. (b) Find the exact value of the positive x-intercept for 0 x. Let R be the region in the first quadrant enclosed by the graph and the x-axis. (c) (i) Shade R on your diagram. (ii) Write down an integral which represents the area of R. (d) Evaluate the integral in part (c)(ii), either by using a graphic display calculator, or by using the following information. d (x sin x + x cos x sin x) = x cos x. dx 7. In this part of the question, radians are used throughout. The function f is given by f (x) = (sin x) cos x. The following diagram shows part of the graph of y = f (x). (Total 4 marks) (Total 4 marks) (7) () () () (Total 5 marks) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems y C O x The point is a maximum point, the point lies on the x-axis, and the point C is a point of inflexion. (a) Give the period of f. (b) From consideration of the graph of y = f (x), find to an accuracy of one significant figure the range of f. (c) (i) Find f (x). (ii) Hence show that at the point, cos x =. (iii) Find the exact maximum value. (d) Find the exact value of the x-coordinate at the point. (e) (i) Find f (x) dx. (ii) Find the area of the shaded region in the diagram. (f) Given that f (x) = 9(cos x) 7 cos x, find the x-coordinate at the point C. (4) (Total 0 marks) 7. triangle has sides of length 4, 5, 7 units. Find, to the nearest tenth of a degree, the size of the largest angle. (Total 4 marks) 74. O is the centre of the circle which has a radius of 5.4 cm. () () (9) () (4) O The area of the shaded sector O is.6 cm. Find the length of the minor arc. (Total 4 marks) 6 75. The circle shown has centre O and radius 6. O is the vector 6, O is the vector and is the 0 OC 0 5 vector. y C O x C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
I Math SL: Trig Practice Problems (a) Verify that, and C lie on the circle. (b) Find the vector C. (c) Using an appropriate scalar product, or otherwise, find the cosine of angle OC ˆ. (d) Find the area of triangle C, giving your answer in the form a, where a. () () () (4) (Total marks) 76. Solve the equation sin x = cos x, for 0 x 80. (Total 4 marks) 77. The diagrams show a circular sector of radius 0 cm and angle θ radians which is formed into a cone of slant height 0 cm. The vertical height h of the cone is equal to the radius r of its base. Find the angle θ radians. 0cm h 0cm r 78. The diagram shows the graph of the function f given by f (x) = sin x +, for 0 x 5, where and are constants, and x is measured in radians. y (,) (5, ) (Total 4 marks) (0, ) 0 4 5 x (, ) The graph includes the points (, ) and (5, ), which are maximum points of the graph. (a) Write down the values of f () and f (5). (b) Show that the period of f is 4. The point (, ) is a minimum point of the graph. (c) Show that =, and find the value of. (d) Show that f (x) = cos x. The line y = k x is a tangent line to the graph for 0 x 5. (e) Find (i) the point where this tangent meets the curve; (ii) the value of k. (f) Solve the equation f (x) = for 0 x 5. () () (5) (4) (6) (5) (Total 4 marks) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of
Circular Functions and Trig - Practice Problems (to 07) MarkScheme. (a) Evidence of using the cosine rule p r q eg cos PQˆ R =, q p r pr cos PQˆ R pr Correct substitution 6 5 eg, 5 46 4 4 6 46 cosq 7 cos PQˆ R = 0. 565 48 PQˆ R = 55.8 (0.97 radians) N (b) rea = pr sin PQˆ R For substituting correctly 46 sin 55.8 = 9.9 (cm ) N. Note: Throughout this question, do not accept methods which involve finding. (a) Evidence of correct approach eg sin =, C 5 sin = 5 G N0 (b) Evidence of using sin = sin cos = 5 4 5 = 9 GN0 (c) Evidence of using an appropriate formula for cos M 4 5 4 5 80 eg,,, 9 9 9 9 8 cos = 9 N. (a) For using perimeter = r + r + arc length 0 = r + r 0r r G (b) 0r Finding = r 0r r r For setting up equation in r M Correct simplified equation, or sketch eg 0r r = 5, r 0r + 5 = 0 r = 5 cm N 4. Notes: Candidates may have differing answers due to using approximate answers from previous parts or using answers from the GDC. Some leeway is provided to accommodate this. N0 C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
(a) METHOD Evidence of using the cosine rule a b c eg cos C =, a b c bccos ab Correct substitution 4 eg cos ÔP =,4 cosôp cos ÔP = 0.5 6 ÔP =.8 (radians) 45 N METHOD rea of OP = 5.8 (from part (d)) rea of triangle OP =.905.9050 = 0.5 sin ÔP ÔP =. or.8 6 ÔP =.8 (radians) 45 N (b) Ô = (.8) (=.64) 8 =.64 (radians) 45 N (c) (i) ppropriate method of finding area eg area = θr rea of sector PE = 4.6 =.0 (cm ) (accept the exact value.04) N (ii) rea of sector OD =.64 =.9 (cm ) N (d) (i) rea OE = rea PE rea OP (=.0 5.8) M = 7.9 (accept 7. from the exact answer for PE) N (ii) rea shaded = rea OD rea OE (=.9 7.9) M = 4.7 (accept answers between 4.6 and 4.7) N 5. (a) Evidence of choosing cosine rule eg a = b + c bc cos Correct substitution eg (D) = 7. + 9. (7.) (9.) cos 60 (D) = 69.7 D = 8.5 (cm) N (b) 80 6 = 8 Evidence of choosing sine rule Correct substitution DE 8.5 eg = sin 8 sin 0 DE =.75 (cm) N (c) Setting up equation [4] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
6. (a) eg ab sin C = 5.68, bh = 5.68 Correct substitution eg 5.68 = (.) (7.) sin Dˆ C,. h = 5.68, (h =.55) sin Dˆ C = 0.5 Dˆ C 0 and/or 50 N (d) Finding ˆ C (60 + D ˆ C) Using appropriate formula eg (C) = () + (C), (C) = () + (C) () (C) cos C Correct substitution (allow FT on their seen ˆ C ) eg (C) = 9. +. C = 9.74 (cm) N (e) For finding area of triangle D Correct substitution rea = 9. 7. sin 60 = 8.8... rea of CD = 8.8... + 5.68 0 = 4.0 (cm ) N y [] 5 60 80 0 80 60 x 5 Correct asymptotes N (b) (i) Period = 60 (accept ) N (ii) f (90) = N (c) 70, 90 NN Notes: Penalize mark for any additional values. Penalize mark for correct answers given in radians,,or4.7,.57. 7. (a) METHOD Using the discriminant = 0 k = 4 4 k = 4, k = 4 N METHOD Factorizing 0 (x ) k = 4, k = 4 N C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
(b) Evidence of using cos = cos M eg ( cos ) + 4 cos + f () = 4 cos + 4 cos + G N0 (c) (i) N (ii) METHOD ttempting to solve for cos M cos = = 40, 0, 40, 0 (correct four values only) N METHOD Sketch of y = 4 cos + 4 cos + y 9 M 60 80 80 60 x Indicating 4 zeros = 40, 0, 40, 0 (correct four values only) N (d) Using sketch c = 9 N 8. Note: ccept exact answers given in terms of. (a) Evidence of using l = r arc = 7.85 (m) N (b) Evidence of using r θ rea of sector O = 58.9 (m ) N (c) METHOD [] 6 angle = 0 6 attempt to find 5 sin 6 height = 5 + 5 sin 6 =.5 (m) N METHOD M C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 4 of 4
angle = attempt to find 5 cos M height = 5 + 5 cos =.5 (m) N (d) (i) h 55cos 4 4 = 5.6 (m) N (ii) h(0) = 5 5 cos 0 4 = 4.9(m) N (iii) METHOD Highest point when h = 0 R 0 = 5 5 cos t 4 M cos t = 4 t =.8 accept 8 N METHOD h 0 Sketch of graph of h M Correct maximum indicated t =.8 N METHOD Evidence of setting h(t) = 0 M sin t 0 4 Justification of maximum R eg reasoning from diagram, first derivative test, second derivative test t =.8 accept 8 N (e) h(t) = 0 sin t (may be seen in part (d)) 4 N (f) (i) h(t) 0 t t 0 C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 5 of 4
N Notes: ward for range 0 to 0, for two zeros. ward for approximate correct sinusoidal shape. (ii) METHOD Maximum on graph of h t = 0.9 N METHOD Minimum on graph of h t =.96 N METHOD Solving h(t) = 0 One or both correct answers t = 0.9, t =.96 N 9. (a) Vertex is (4, 8) N (b) Substituting 0 = a(7 4) + 8 M a = N (c) For y-intercept, x = 0 y = 4 N 0. METHOD Evidence of correctly substituting into = r θ Evidence of correctly substituting into l = r For attempting to eliminate one variable leading to a correct equation in one variable r = 4 = 6 (= 0.54, 0) N METHOD Setting up and equating ratios 4 r r Solving gives r = 4 r = 4 or r = 0.54,0 6 r = 4 = 6 0.54,0 N. a = 4, b =, c = or etc N6. (a) 5 PQ = N (b) Using r = a + tb x 5 t y 6 N4. METHOD Evidence of correctly substituting into l = r C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 6 of 4 []
Evidence of correctly substituting into = r For attempting to solve these equations eliminating one variable correctly r = 5 =.6 (= 9.7) N METHOD Setting up and equating ratios 4 80 r r Solving gives r = 5 r = 4 or r θ 80 =.6 (= 9.7) r = 5 =.6 (= 9.7) N 4. (a) For correct substitution into cosine rule D = 4 8 For factorizing 6, D = 48cosθ = 4 5 4 cos θ GN0 (b) (i) D = 5.565... sin Cˆ D sin 5 5.565 M sin Cˆ D = 0.9 (accept 0.90, subject to P) N (ii) Cˆ D = 65.7 N Or Cˆ D = 80 their acute angle = 4 N (iii) Dˆ C = 89. C 5.565 C or (or cosine rule) sin 89. sin 5 sin 89. sin 65.7 M C =. (accept.7 ) Perimeter = 4 + 8 + +. = 7. N (c) rea = 4 8 sin 40 = 0. N 5. (a) METHOD Note: There are many valid algebraic approaches to this problem (eg completing the square, b using x ). Use the following mark a allocation as a guide. (i) dy Using 0 dx x + 60 = 0 x = 5 N (ii) ymax = 6(5 ) + 60(5) 56 ymax = 44 METHOD 6 5 4 cos θ N C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 7 of 4
(i) Sketch of the correct parabola (may be seen in part (ii)) M x = 5 N (ii) ymax = 44 N (b) (i) z = 0 x (accept x + z = 0) N (c) (ii) z = x + 6 x 6 cos Z N (iii) Substituting for z into the expression in part (ii) Expanding 00 0x + x = x + 6 x cos Z Simplifying x cos Z = 0x 64 0x 64 Isolating cos Z = x 5x 6 cos Z = x Note: Expanding, simplifying and isolating may be done in any order, with the final being awarded for an expression that clearly leads to the required answer. Evidence of using the formula for area of a triangle 6 x sin Z xsin Z 6 x sin Z 4 GN0 = 9x sin Z G N0 (d) Using sin Z = cos Z 5x 6 Substituting for cos Z x 5x 6 5x 60x 56 for expanding to x 9x for simplifying to an expression that clearly leads to the required answer eg = 9x (5x 60x + 56) = 6x + 60x 56 (e) (i) 44 (is maximum value of, from part (a)) max = N (ii) Isosceles N 6. (a) Evidence of choosing the double angle formula f (x) = 5 sin (6x) N (b) Evidence of substituting for f (x) eg 5 sin 6x = 0, sin x = 0 and cos x = 0 6x = 0,, x = 0,, 6 M G N4 7. (a) (i) OP = PQ (= cm) R So OPQ is isosceles GN0 (ii) 4 Using cos rule correctly eg cos OPˆQ = 9 96 cos OPˆQ = 8 8 [0] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 8 of 4
cos OPˆQ = GN0 9 (iii) Evidence of using sin + cos = M sin OPˆQ = 80 8 8 sin OPˆQ = 80 9 GN0 (iv) Evidence of using area triangle OPQ = OPPQ sin P M 80 9 eg, 0.998 9 rea triangle OPQ = (b) (i) OPˆQ =.4594... 80 0 4.47 N OPˆQ =.46 N (ii) Evidence of using formula for area of a sector eg rea sector OPQ =.4594 = 6.57 N (c).4594 QÔP = 0.84 rea sector QOS = 4 0.84 = 6.7 N (d) rea of small semi-circle is 4.5 (= 4.7...) Evidence of correct approach M eg rea = area of semi-circle area sector OPQ area sector QOS + area triangle POQ Correct expression eg 4.5 6.5675... 6.785... + 4.47..., 4.5 (6.785... +.095...), 4.5 (6.5675... +.56...) rea of the shaded region = 5. N 8. (a) p = 0 (b) METHOD Period = q (M) = q = 4 4 METHOD Horizontal stretch of scale factor = q (M) scale factor = 4 q = 4 4 9. (a) using the cosine rule () = b + c bc cos  [7] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 9 of 4
substituting correctly C = 65 +04 (65) (04) cos 60 = 45 + 086 6760 = 88 C = 9 m (b) finding the area, using bc sin  substituting correctly, area = (65) (04) sin 60 = 690 (ccept p = 690) (c) (i) = (65) (x) sin 0 65x = G 4 (ii) = (04) (x) sin 0 M = 6x 65x (iii) starting + = or substituting + 6x = 690 4 69x simplifying = 690 4 4 690 x = 69 x = 40 (ccept q = 40) 4 (d) (i) Recognizing that supplementary angles have equal sines eg Dˆ C = 80 Dˆ sin Dˆ C = sin Dˆ R (ii) using sin rule in ΔD and ΔCD substituting correctly D 65 D sin 0 sin 0 sin Dˆ 65 sin Dˆ and DC 04 DC sin 0 sin 0 sin Dˆ 04 sin Dˆ C M since sin Dˆ = sin Dˆ C D DC D 65 65 04 DC 04 D 5 DC 8 G 5 0. (a) r 7 (.5) r r 6 r 6 cm (C4) (b) rc length r.5 6 rc length = 9 cm (C) Note: Penalize a total of ( mark) for missing units.. (a) when y 0 (may be implied by a sketch) [8] x 8 or.79 9 (C) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 0 of 4
(b) METHOD Sketch of appropriate graph(s) Indicating correct points x. or x5.4 (C)(C) METHOD sinx 9 7 x, x 9 6 9 6 7 x, x 6 9 6 9 9 x, x 8 8 ( x., x5.4) (C)(C). (a) for using cosine rule a b c abcos C C 5 7 5 7 cos 9 C 8.4 m (N0) Notes: Either the first or the second line may be implied, but not both. ward no marks if 8.4 is obtained by assuming a right (angled) triangle (C = 7 sin 9). (i) C 9 7 85 (c) (ii) C 80 (9 85) 66 for using sine rule (may be implied) C 7 sin85 sin 66 7sin85 C sin66 C (8.580 ) 8.5 m rea 7 8.58... sin 9 76.4 m (ccept 76. m ) (N) 5 Ĉ from previous triangle 66 Therefore alternative C ˆ 80 66 4 (or 9 85) C ˆ 80 (9 4) 7 C 9 4 (N) 7 7 C 7 sin 7 sin4 C (.9906 ). m (N) 4 C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
(d) C 9 7. (a) (i) Minimum length for C when showing right triangle C sin 9 7 C 7sin 9 C (8.47 ) 8.4 m (ii) (iii) Ĉ = 90 or diagram (N) f ( x) cosx sin x cosxsin x Note: ward for sin xsin x only if work shown, using product rule on sin xcos x cos x. sin xsin x (sin x)(sin x) or (sin x0.5)(sin x) sin x or sin x sin x 5 x (0.54) x (.6) x (4.7) 6 6 (N) (N)(N) 6 (b) x 0.54 6 (N) (c) (i) EITHER curve crosses axis when x (may be implied) OR 5 6 6 rea f ( x)d x f ( x)dx 5 (N) (N) (N) 6 rea = f ( x) dx () (N) 6 (ii) rea 0.875 0.875.75 (N) 5 4. Using area of a triangle = 0 (0)(8)sin Q ab sin C Note: ccept any letter for Q sin Q = 0.5 PQˆ R = 0 or 6 or 0.54 (C6) 5. (a) b = 6 (C) [4] [] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
(b) y x () (C) (c) x =.05 (accept (.05, 0.896) ) (correct answer only, no additional solutions) () (C) 6. (a) ( sin x) + sin x = 6 sin x sin x = 0 (p = 6, q =, r = ) (C) (b) ( sin x )( sin x + ) (C) (c) 4 solutions () (C) 7. rea of large sector r θ = 6.5 = 9 rea of small sector r θ = 0.5 = 75 Shaded area = large area small area = 9 75 = 7 (C6) 8. (a) y.5 0.5 0 0.5.5.5.5 x 0.5.5 (C) Note: ward for the graph crossing the y-axis between 0.5 and, and for an approximate sine curve crossing the x-axis twice. Do not penalize for x >.4. (b) (Maximum) x = 0.85 4 x = 0. ( dp) (C) (Minimum) x =.856 4 x =.9 ( dp) (C) 9. rea of a triangle = 4 sin 4 sin = 4.5 sin = 0.75 C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
= 48.6 and = (or 0.848,.9 radians) Note: ward (C4) for 48.6 only, (C5) for only. 0. METHOD cos x = sin x cos x () (C6) cos x sin x cos x = 0 cos x(cos x sin x) = 0 cos x = 0, (cos x sin x) = 0 x =, x = (C6) 4 METHOD Graphical solutions EITHER for both graphs y = cos x, y = sin x, OR for the graph of y = cos x sin x. (M) THEN Points representing the solutions clearly indicated.57, 0.785 x =, x = (C6) 4 Notes: If no working shown, award (C4) for one correct answer. ward (C)(C) for each correct decimal answer.57, 0.785. ward (C)(C) for each correct degree answer 90, 45. Penalize a total of [ mark] for any additional answers.. (a) (i) 0 + 4 sin =.4 (ii) t 00, t = 0 + 4 sin 0.5 = 6.48 (N) Note: ward (0) if candidates use t = 00 leading to y =.6. No other ft allowed. (b) (i) 4 metres (ii) t t 4 = 0 + 4 sin sin = t = (.4) (correct answer only) (N) (c) (i) 4 (ii) t 0 + 4 sin = 7 t sin = 0.75 t = 7.98 (N) (iii) depth < 7 from 8 = hours from 00 0 = hours therefore, total = 6 hours (N) 7. (a) ngle 80 5 sin 40 sin80.6 cm (C) (b) rea ac sin (5)(.6)sin 60 7.07 (accept 7.06) cm (C) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 4 of 4 (M) []
Note: Penalize once in this question for absence of units.. METHOD rea sector O (5) (0.8) 0 ON 5cos0.8.48... N 5sin0.8.586... rea of ON ON N 6.49... (cm ) Shaded area 0 6.49...75 (cm ) METHOD (C6) O N rea sector F (5) (.6) 0 rea OF (5) sin.6.5 Twice the shaded area 0.5 ( 7.5) F Shaded area (7.5).75 (cm ) 4. (a) (i) f ( x) 6sinx (ii) EITHER f ( x) sin xcos x 0 sin x 0 or cos x 0 OR sinx 0, for 0 x THEN x 0,, (N4) 6 (b) (i) translation in the y-direction of (ii). (.0 from TRCE is subject to P) () 4 5. = p + q cos 0 = p + q = p + q cos = p q (a) p = (C) (b) q = (C) (C6) [0] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 5 of 4
6. Method y 0.80.5 x 0 (C).80 [ sf] (G) (C).5 [ sf] (G) (C) Method x = ±0.5x + (etc.).5x = 0,, 4 or.5x = 0,, 4 7x = 0, 4, (8) or 5x = 0, 4, (8) 4 4 x = 0, or x = 0, 7 5 4 4 x = 0,, 7 5 (C)(C)(C) 7. (a) area of sector ΑΒDC = () = 4 area of segment DCP = area of C = (C) (b) P = area of semicircle of radius P = ( ) = area of shaded region = ( ) = (C) 8. (a) OR PQ = q p 0 7 = = PO PQ (b) cos OPˆQ PO PQ PO 7 58 PO PQ 5 5 OPˆQ 58 754 OPˆQ PQˆ R =, PQ = = + 6 = 5 cos (G) 4 (c) (i) Since + = 80 (R) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 6 of 4
PQˆ R cos = cos (G) PQˆ R (ii) sin = OR cos = 5 OPˆ Q 754 5 754 59 754 = = (G) 5 754 754 5 754 therefore x = 754 5 = 59 x = sin = 754 (G) Note: ward (0) for the following solution. cos = 5 = 56.89 754 sin = 0.876 = 0.876 sin = 754 754 (iii) rea of OPQR = (area of triangle PQR) = PQ QR sin PQˆ R = 58 754 = sq units. OR rea of OPQR = (area of triangle OPQ) = 7 0 = sq units. 7 Notes: Other valid methods can be used. ward final for the integer answer. 9. (a) Sine rule PR 9 sin5 sin0 9sin 5 PR = sin0 = 5.96 km (b) EITHER x P [4] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 7 of 4
Sine rule to find PQ 9sin 5 PQ = sin0 = 4.9 km OR Cosine rule: PQ = 5.96 + 9 ()(5.96)(9) cos 5 = 9.9 PQ = 4.9 km 4.9 Time for Tom = 8 5.96 Time for lan = a 4.9 5.96 Then = 8 a a = 0.9 7 (c) RS = 4QS 4QS = QS + 8 8 QS cos 5 QS + 4.74QS 8 = 0 (or x + 4.74x 8 = 0) QS = 8.0 or QS =.9 therefore QS =.9 OR QS sinsrˆ Q QS sin5 (G) sin SRˆ Q sin 5 SRˆ Q = 6.7 QŜR Therefore, = 80 (5 + 6.7) = 8. 9 QS SR sin8. sin6.7 sin 5 9sin6.7 9sin 5 QS = sin8. sin8. 4 =.9 6 40. (a) (i) cos, sin 4 therefore cos sin = 0 4 4 (G) (ii) cos x + sin x = 0 + tan x = 0 tan x = l x = 4 Note: ward (0) for.6. OR x = 4 (G) C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 8 of 4
(b) y = e x (cos x + sin x) dy = e x (cos x + sin x) + e x ( sin x + cos x) dx = e x cos x (c) dy = 0 for a turning point e x cos x = 0 dx cos x = 0 x = a = y = e (cos + sin ) = e b = e 4 Note: ward (0)(0) for a =.57, b = 4.8. (d) d y t D, = 0 dx e x cos x e x sin x = 0 e x (cos x sin x) = 0 cos x sin x = 0 x = 4 4 y = e (cos + sin ) 4 4 4 = e (G) 5 4 e x 0 (e) Required area = (cos x + sin x)dx = 7.46 sq units (G) OR Αrea = 7.46 sq units (G) Note: ward (G0) for the answer 9.8 obtained if the calculator is in degree mode. 4. (a) (i) 4 is, 0 (C) (ii) is (0, 4) (C) Note: In each of parts (i) and (ii), award C if and are interchanged, C if intercepts given instead of coordinates. (b) 4 rea = 4 8 = (=.67) (C) 4. (a) ( sin x )(sin x ) (C) Note: ward if x x + 6 correctly factorized to give (x )(x ) (or equivalent with another letter). (b) (i) ( sin x )(sin x ) = 0 [7] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 9 of 4
sin x = sin x = (C) (ii) x = 4.8, 8 (C) Notes: Penalize [ mark] for any extra answers and [ mark] for answers in radians. ie ward 0 for 4.8, 8 and any extra answers. ward 0 for 0.70,.4. ward 0 0 for 0.70,.4 and any extra answers. 4. Note: Do not penalize missing units in this question. (a) = + cos 75 = ( cos 75 ) = ( cos 75 ) = ( cos75) (G) Note: The second is for transforming the initial expression to any simplified expression from which the given result can be clearly seen. (b) PÔ = 7.5 P = tan 7.5 = 9. cm OR Pˆ = 05 ÂP = 7.5 P sin05 sin 7. 5 sin 7.5 P = = 9.(cm) sin 05 (c) (i) rea OP = 9. or tan7. 5 = 55. (cm ) (accept 55. cm ) (ii) rea P = (9.) sin05 = 4.0 (cm ) (accept 40.9 cm ) 4 75 (d) rea of sector = 75 or 80 60 = 94. (cm ) (accept 0 or 94. (cm )) (e) Shaded area = area OP area sector = 6.4 (cm ) (accept 6. cm, 6. cm ) 44. Note: Do not penalize missing units in this question. (a) (i) t release(p), t = 0 s = 48 + 0 cos 0 = 58 cm below ceiling (ii) 58 = 48 +0 cos t cos t = t = sec OR t = sec (G) 5 (b) (i) ds dt = 0 sin t Note: ward for 0, and for sin t. [] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page 0 of 4
(ii) ds v = dt = 0 sin t = 0 sin t = 0 t = 0,... (at least values) s = 48 + 0 cos 0 or s = 48 +0 cos = 58 cm (at P) = 8 cm (0 cm above P) 7 Note: ccept these answers without working for full marks. May be deduced from recognizing that amplitude is 0. (c) 48 +0 cos t = 60 + 5 cos 4t t = 0.6 secs OR t = 0.6 secs (G) (d) times (G) Note: If either of the correct answers to parts (c) and (d) are missing and suitable graphs have been sketched, award (G) for sketch of suitable graph(s); for t = 0.6; for. 45. (a) l = r or C = O = 0 cm (C) (b) Ô (obtuse) = rea = r = ( )(5) 46. = 48 cm ( sf) (C4) d 50 80 70 P () OR.5 0 = 50.5 = 80 d = 50 + 80 50 80 cos 70 d = 78.5 km (C6) 47. (a) (i) (C) (ii) 4 (accept 70 ) () (C) (b) y 48. (G) number of solutions: 4 () (C) Statement (a) Is the statement true for all (b) If not true, example real numbers x? (Yes/No) No x = l (log0 0. = ) (a) () (C) No x = 0 (cos 0 = ) (b) () (C) C Yes N/ C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
Notes: (a) ward for each correct answer. (b) ward () marks for statements and only if NO in column (a). ward () for a correct counter example to statement, for a correct counter example to statement (ignore other incorrect examples). Special Case for statement C: ward if candidates write NO, and give a 5 valid reason (eg arctan = ). 4 49. (a) (b) 7 6 sin sin 45 sin = 6 7 6 = (G) 7 D h (c) OR (i) + = 80 (ii) 6 sin = 7 => = 59.0 or ( sf) (iii) Dˆ C ĈD ÂC => = 80 ( + 45 ) = 4.0 ( sf) D sin 4 7 sin 45 =>D =.69 6 D h rea DC rea C h D = (G) D 6sin 45 rea ΔCD rea ΔC 6sin 45 D = (G) sin sin 48 50. Using sine rule: 5 7 5 sin = sin 48 = 0.508 7 = arcsin (0.508) =.06 C [0] C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4
= (nearest degree) (C6) Note: ward a maximum of [5 marks] if candidates give the answer in radians (0.560). 5. (a) x is an acute angle => cos x is positive. cos x + sin x = => cos x = C:\Users\ob\Documents\Dropbox\Desert\SL\Trig\TestsQuizzesPractice\SLTrigPractice.docx on /8/4 at 8:50 M Page of 4 => cos x = = (= ) (C4) (b) cos x = sin x = 7 = (C) 9 Notes: (a) ward (M0)(0) for cos sin = 0.94. (b) ward (0) for cos sin = 0.778. 5. (a) sin x = ( cos x) = cos x = l + cos x => cos x + cos x l = 0 (C) Note: ward the first for replacing sin x by cos x. sin (b) cos x + cos x = ( cos x )(cos x +) (C) (c) cos x = or cos x = l => x = 60, 80 or 00 (C) Note: ward (0) if the correct answers are given in radians (ie,,, or.05,.4, 5.4) 5. (a) The smallest angle is opposite the smallest side. 8 7 5 cos θ = 8 7 88 = = 0.7857 4 Therefore, θ = 8. (C) (b) rea = 8 7 sin 8. = 7. cm (C) 54. (a) sin x + 4 cos x = ( cos x) + 4cos x 8 9 5 = cos + 4 cos x (C) (b) sin x + 4 cos x 4 = 0 cos x + 4 cos x 4 = 0 x [4]