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5.8 Trigonometric Equations To calculate the angle at which a curved section of highway should be banked, an engineer uses the v2 , where x is the angle of the equation tan x = 224 000 bank and v is the speed limit on the curve, in kilometres per hour. 2 v The equation tan x = is an example of a 224 000 trigonometric equation. It will be used in Example 6. A trigonometric equation is an equation that contains one or more trigonometric functions. Other examples of trigonometric equations include the following. 2 2 sin x + cos x = 1 2cos x − 1 = 0 The equation sin2 x + cos2 x = 1 is an identity. Recall that a trigonometric identity is an equation that is true for all values of the variable for which the expressions on both sides of the equation are defined. The equation 2cos x − 1 = 0 is not an identity and is only true for certain values of x. To solve a trigonometric equation that is not an identity means to find all values of the variable, x, that make the equation true. I NVESTIGATE & I NQUIRE To graph the system y = 2x + 3 and y = 9 − x, use the Y= editor to enter Y1 = 2X + 3 and Y2 = 9 − X. Graph the equations in the standard viewing window. 1. Use the intersect operation to find the coordinates of the point of intersection. 2. Equate the right sides of the two equations from question 1 and solve the resulting equation for x algebraically. 3. Compare the value of x you found in question 3 to the value of x you found in question 2. Explain your findings. 4. 402 MHR • Chapter 5 graph the equations y = 4cos x and y = 1 + 2cos x, use the Y= editor to enter Y1 = 4cos(X) and Y2 = 1 + 2cos(X). Using the mode settings, select the degree mode. Using the window variables, adjust the window to 0° ≤ x ≤ 360° and −5 ≤ y ≤ 5 by setting Xmin = 0, Xmax = 360, Ymin = −5, and Ymax = 5. Display the graph. 5. To 6. Use the intersect operation to find the coordinates of the points of intersection. 7. Equate the right sides of the two equations from question 5, and solve the resulting equation for cos x algebraically. 8. If 0° ≤ x ≤ 360°, what values of x, in degrees, give the value of cos x from question 7? Compare the values of x you found in question 8 to the values of x you found in question 6. Explain your findings. 9. EXAMPLE 1 Solving a Trigonometric Equation sin θ − 1 = 0 on the interval 0° ≤ θ ≤ 360°. Solve 2 SOLUTION 1 Paper-and-Pencil Method 2sin θ − 1 = 0 Add 1 to both sides: 2sin θ = 1 1 Divide both sides by 2: sin θ = 2 The sine of an angle is positive in the first and second quadrants. y y r 0 θ y x y x r θ –x 0 x y sin θ = – r The solutions are 45° and 135°. π 3π In radians, the solutions are and . 4 4 Use the memory aid CAST to remember which trigonometric ratios are positive in each quadrant. 5.8 Trigonometric Equations • MHR 403 SOLUTION 2 Graphing-Calculator Method Graph the related trigonometric function y = 2sin x − 1 for 0° ≤ x ≤ 360°. Using the mode settings, select the degree mode. Find the x-intercepts using the zero operation. The graph intersects the x-axis at 45° and 135°. The solutions are 45° and 135°. π 3π In radians, the solutions are θ = and . 4 4 The window variables include Xmin = 0, Xmax = 360, Ymin = –3, Ymax = 1. EXAMPLE 2 Solving a Trigonometric Equation Find the exact solutions for 3cos θ = cos θ + 1, if 0° ≤ θ ≤ 360°. SOLUTION 1 Paper-and-Pencil Method 3cos θ = cos θ + 1 Subtract cos θ from both sides: 2cos θ = 1 Divide both sides by 2: cos θ = 1 2 The cosine of an angle is positive in the first and fourth quadrants. y y r 0 θ y x x θ r cos θ = x–r The exact solutions are 60° and 300°. 5π . In radians, the exact solutions are π and 3 3 404 MHR • Chapter 5 x x 0 –y SOLUTION 2 Graphing-Calculator Method Graph y = 3cos x and y = cos x + 1 in the same viewing window for 0° ≤ x ≤ 360°. Use the intersect operation to determine the coordinates of the points of intersection. The window variables include Xmin = 0, Xmax = 360, Ymin = –4, Ymax = 4. The graphs intersect at x = 60° and x = 300°. The solutions are 60° and 300°. 5π . In radians, the solutions are π and 3 3 Note that an alternative graphing-calculator method for solving Example 2 involves rewriting 3cos x = cos x + 1 as 2cos x − 1 = 0. The equation 2cos x − 1 = 0 can be solved by graphing y = 2cos x − 1 and using the zero operation to find the x-intercepts. The window variables include Xmin = 0, Xmax = 360, Ymin = –4, Ymax = 2. EXAMPLE 3 Solving by Factoring 2 Solve the equation 2cos x − cos x − 1 = 0 on the interval 0 ≤ x ≤ 2π. SOLUTION 2cos2 x − cos x − 1 = 0 Factor the left side: (2cos x + 1)(cos x − 1) = 0 Use the zero product property: 2cos x + 1 = 0 or cos x − 1 = 0 2cos x = −1 cos x = 1 cos x = – 1 x = 0 or 2π 2 The solution can be modelled 2π or 4π x= graphically. The window variables 3 3 include Xmin = 0, Xmax = 360, 2π , 4π , and 2π. The solutions are 0, 3 3 Ymin = –3, Ymax = 3. 5.8 Trigonometric Equations • MHR 405 EXAMPLE 4 Solving by Factoring 2 Solve the equation 2sin x − 7sin x + 3 = 0 for 0 ≤ x ≤ 2π. Express answers as exact solutions and as approximate solutions, to the nearest hundredth of a radian. SOLUTION 2sin2 x − 7sin x + 3 = 0 Factor the left side: (sin x − 3)(2sin x − 1) = 0 Use the zero product property: sin x − 3 = 0 or 2sin x − 1 = 0 sin x = 3 2sin x = 1 sin x = 1 2 5π x = π or 6 6 There is no solution to sin x = 3, since all values of sin x are ≥ −1 or ≤ 1. 5π . The exact solutions are π and 6 6 The approximate solutions are 0.52 and 2.62. EXAMPLE 5 Using a Trigonometric Identity Solve 6cos2 x − sin x − 5 = 0 for 0° ≤ x ≤ 360°. Round approximate solutions to the nearest tenth of a degree. SOLUTION Use the Pythagorean identity cos2 x = 1 − sin2 x to write an equivalent equation that involves only the sine function. 6cos2 x − sin x − 5 = 0 2 2 Substitute 1 − sin x for cos x: 6(1 − sin2 x) − sin x − 5 = 0 Expand: 6 − 6sin2 x − sin x − 5 = 0 Simplify: −6sin2 x − sin x + 1 = 0 Multiply both sides by –1: 6sin2 x + sin x − 1 = 0 Factor the left side: (2sin x + 1)(3sin x − 1) = 0 Use the zero product property: 2sin x + 1 = 0 or 3sin x − 1 = 0 2sin x = −1 or 3sin x = 1 sin x = – 1 or sin x = 1 2 3 x = 210° or 330° x =⋅ 19.5° or 160.5° The solutions are 19.5°, 160.5°, 210°, and 330°. 406 MHR • Chapter 5 EXAMPLE 6 Bank Angles v2 Engineers use the equation tan x = to calculate the angle at which 224 000 a curved section of highway should be banked. In the equation, x is the angle of the bank and v is the speed limit on the curve, in kilometres per hour. a) Calculate the angle of the bank, to the nearest tenth of a degree, if the speed limit is 100 km/h. b) The four turns at the Indianapolis Motor Speedway are banked at an angle of 9.2°. What is the maximum speed through these turns, to the nearest kilometre per hour? SOLUTION 2 v tan x = 224 000 2 100 = 224 000 ⋅ x = 2.6° The angle of the bank is 2.6°, to the nearest tenth of a degree. a) 2 b) v = tan x 224 000 v2 = 224 000tan x v = 224 0tan 00x v = 224 0tan 002° 9. ⋅ v = 190 The maximum speed is 190 km/h, to the nearest kilometre per hour. Key Concepts • To solve a trigonometric equation that is not an identity, find all values of the variable that make the equation true. • Trigonometric equations can be solved a) with paper and pencil using the methods used to solve algebraic equations b) graphically using a graphing calculator • Answers can be expressed in degrees or radians. 5.8 Trigonometric Equations • MHR 407 Communicate Yo u r Understanding Describe how you would solve 2sin x – 3 = 0, 0° ≤ x ≤ 360°. Describe how you would solve 2cos2 x − 3cos x + 1 = 0, 0 ≤ x ≤ 2π. Justify your method. 3. Explain why the equation cos x − 2 = 0 has no solutions. 1. 2. Practise A 1. a) b) c) d) e) f) Solve each equation for 0 ≤ x ≤ 2π. sin x = 0 2cos x + 1 = 0 tan x = 1 2 sin x + 1 = 0 2cos x − 3 = 0 2sin x + 3 = 0 2. a) b) c) d) e) f) Solve each equation for 0 ≤ x ≤ 360°. sin x + 1 = 0 2 cos x − 1 = 0 2sin x − 3 = 0 2 cos x + 1 = 0 2sin x + 1 = 0 tan x = –1 3. a) b) c) d) e) f) g) Solve each equation for 0° ≤ x ≤ 360°. 2cos2 x − 7cos x + 3 = 0 3sin x = 2cos2 x 2sin2 x − 3sin x − 2 = 0 sin2 x − 1 = cos2 x tan2 x − 1 = 0 2sin2 x + 3sin x + 1 = 0 2cos2 x + 3sin x − 3 = 0 408 MHR • Chapter 5 Solve each equation for 0 ≤ x ≤ 2π. Express answers as exact solutions and as approximate solutions, to the nearest hundredth of a radian. 2 a) sin x − 2sin x − 3 = 0 2 b) 2cos x = sin x + 1 c) 2sin x cos x + sin x = 0 2 d) sin x = 6sin x − 9 2 e) sin x + sin x = 0 f) cos x = 2sin x cos x 4. Solve for x on the interval 0° ≤ x ≤ 360°. Round approximate solutions to the nearest tenth of a degree. a) 4cos x − 3 = 0 b) 1 + sin x = 4sin x 2 c) 6cos x − cos x − 1 = 0 2 d) 9sin x − 6sin x + 1 = 0 2 e) 16cos x − 4cos x + 1 = 0 2 f) 6cos x + sin x − 4 = 0 5. Solve for x on the interval 0 ≤ x ≤ 2π. Give the exact solution, where possible. Otherwise, round to the nearest hundredth of a radian. a) sin x − 3sin x cos x = 0 2 b) 6cos x + cos x − 1 = 0 2 c) 8cos x + 14cos x = −3 2 d) 8sin x − 10cos x − 11 = 0 6. Apply, Solve, Communicate 7. Refraction of light The diagram shows how a light ray bends as it travels from air into water. The bending of the ray is a process known as refraction. In the diagram, i is the angle of incidence and r is the angle of refraction. For water, the index of refraction, n, is defined as follows. sin i n= sin r a) The index of refraction of water has a value of 1.33. For a ray with an angle of incidence of 40°, find the angle of refraction, to the nearest tenth of a degree. b) State any restrictions on the value of r in the above equation. Explain. i r B Show that the area, A, of an isosceles triangle with base b and equal angles x, can be found using the equation 2 b tan x . A= 4 2 b) For an isosceles triangle with A = 40 cm and b = 8 cm, use the equation to find x, to the nearest tenth of a degree. 8. Isosceles triangle a) In right triangle ABC, BD ⊥ AC, AC = 4, and BD = 1. 1 a) Show that sin A cos A = . 4 b) Describe how you would use a graphing calculator to find the measure of ∠A, in degrees. c) Find the exact measure of ∠A, in degrees. x x 9. Application A B D C Solve for x on the interval 0 ≤ x ≤ 2π. Give the exact solution, if possible. Otherwise, round to the nearest hundredth of a radian. 2 2 a) tan x − 4tan x = 0 b) tan x − 5tan x + 6 = 0 2 2 c) tan x − 4tan x + 4 = 0 d) 6tan x − 7tan x + 2 = 0 10. 11. Solve for x on the interval 0° ≤ x ≤ 360°. a) 2sin x tan x − tan x − 2sin x + 1 = 0 b) cos x tan x − 1 + tan x − cos x = 0 12. a) Graph y = sin x and y = 0.5 for b) For what values of x is sin x ≥ 0.5? c) For what values of x is sin x ≤ 0.5? Check your solutions. 0° ≤ x ≤ 360°. 5.8 Trigonometric Equations • MHR 409 Write a short paragraph, including examples, to distinguish the terms trigonometric identity and trigonometric equation. 2 2 b) Inquiry/Problem Solving If (cos x + sin x) + (cos x − sin x) = k, for what value(s) of k is the equation an identity? 13. a) Communication C Solve each equation on the interval 0° ≤ x ≤ 360°. a) sin 2x = 1 b) cos 2x = −1 c) 2sin 2x = 1 d) 2cos 2x = 1 e) 2 cos 2x = 1 f) 2sin 2x = 3 g) 2cos 2x = −3 h) sin 2x = 0 i) 2sin 0.5x = 1 14. Use a graphing calculator to find the exact solutions, in degrees, for 0° ≤ x ≤ 360°. a) cos 2x = cos x b) sin 2x = 2cos x c) tan x = sin 2x 15. Technology Describe how you would use a graphing calculator to solve the equation sin (cos x) = 0, 0° ≤ x ≤ 360°. b) Find the exact solutions to this equation. 16. Technology a) A C H I E V E M E N T Check Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application A storm drain has a cross section in the shape of an isosceles trapezoid. x The shorter base and each of the two equal sides measure 2 m, and x is 2m the angle formed by the longer base and each of the equal sides. a) Write an expression for the area, A, of the cross section in terms of sin x and cos x. b) Describe how you would use a graphing calculator to find the value of x, in degrees, if the area of the cross section is 5 m2. c) Find the value of x, to the nearest degree. NUMBER Power Place the digits from 1 to 9 in the boxes to make the statements true. Use the order of operations. ■×■−■=2 (■ + ■) ÷ ■ = 2 ■+■−■=2 410 MHR • Chapter 5 x 2m 2m