1. 1 TF.02.6 - Linear Trigonometric Equations - Graphical Solutions MCR3U - Santowski

2. 2 (A) Review We know what the graphs of the trigonometric functions look like We know what the graphs of the trigonometric functions look like We know that when we algebraically solve an equation in the form of f(x) = 0, then we are trying to find the roots/zeroes/x-intercepts We know that when we algebraically solve an equation in the form of f(x) = 0, then we are trying to find the roots/zeroes/x-intercepts So we should be able to solve trig equations by graphing them and finding the x-intercepts So we should be able to solve trig equations by graphing them and finding the x-intercepts

3. 3 (B) Graphs of sin(x) and cos(x)

4. 4 (B) Graph of tan(x)

5. 5 (C) Examples Solve the equation 3sin(x) – 2 = 0 Solve the equation 3sin(x) – 2 = 0 We can set it up algebraically as sin(x) = 2/3 so x = sin(2/3) giving us 41.8° (and the second angle being 180° - 41.8° = 138.2° We can set it up algebraically as sin(x) = 2/3 so x = sin -1 (2/3) giving us 41.8° (and the second angle being 180° - 41.8° = 138.2° Note that the ratio 2/3 is not one of our standard ratios corresponding to our “standard” angles (30,45,60), so we would use a calculator to actually find the related acute angle of 41.8° Note that the ratio 2/3 is not one of our standard ratios corresponding to our “standard” angles (30,45,60), so we would use a calculator to actually find the related acute angle of 41.8°

6. 6 (C) Examples We can now solve the equation 3sin(x) – 2 = 0 by graphing f(x) = 3sin(x) – 2 and looking for the x-intercepts We can now solve the equation 3sin(x) – 2 = 0 by graphing f(x) = 3sin(x) – 2 and looking for the x-intercepts

7. 7 (C) Examples Notice that there are 2 solutions within the limited domain of 0° <  < 360° Notice that there are 2 solutions within the limited domain of 0° <  < 360° However, if we expand our domain, then we get two new solutions for every additional period we add However, if we expand our domain, then we get two new solutions for every additional period we add The new solutions are related to the original solutions, as they represent the positive and negative co-terminal angles The new solutions are related to the original solutions, as they represent the positive and negative co-terminal angles We can determine their values by simply adding or subtracting multiples of 360° (the period of the given function) We can determine their values by simply adding or subtracting multiples of 360° (the period of the given function)

8. 8 (D) Example #2 Solve 4tan(2x) + 3 = 2 Solve 4tan(2x) + 3 = 2 Again, we can set it up algebraically as tan(2x) = -1/4 and thus (2x) = tan(-1/4) so x = ½ tan(-1/4) Again, we can set it up algebraically as tan(2x) = -1/4 and thus (2x) = tan -1 (-1/4) so x = ½ tan -1 (-1/4) So thus x = ½ of 166° = 84° So thus x = ½ of 166° = 84° and x = ½ of 346° = 173° and x = ½ of 346° = 173° To set it up graphically, we will make one minor change: we have two graphing options  we can graph f(x) = 4tan(2x) + 1 and find the x- intercepts OR we can graph f(x) = 4tan(2x) + 3 and find out where f(x) = 2  to do this, we will simply graph y = 2 as a second equation and find out where the two graphs intersect To set it up graphically, we will make one minor change: we have two graphing options  we can graph f(x) = 4tan(2x) + 1 and find the x- intercepts OR we can graph f(x) = 4tan(2x) + 3 and find out where f(x) = 2  to do this, we will simply graph y = 2 as a second equation and find out where the two graphs intersect

9. 9 (D) Example #2

10. 10 (D) Example #2 Notice that there are 2 solutions within the limited domain of 0° <  < 180° (NOTE: the period of a regular tan function is 180°) Notice that there are 2 solutions within the limited domain of 0° <  < 180° (NOTE: the period of a regular tan function is 180°) However, if we expand our domain, then we get two new solutions for every additional period we add However, if we expand our domain, then we get two new solutions for every additional period we add The new solutions are related to the original solutions, as they represent the positive and negative co-terminal angles The new solutions are related to the original solutions, as they represent the positive and negative co-terminal angles We can determine their values by simply adding or subtracting multiples of 90° (the period of the given function  the effect of the 2x in tan(2x) is a horizontal compression by a factor of 2, so we have reduced the period by a factor of 2) We can determine their values by simply adding or subtracting multiples of 90° (the period of the given function  the effect of the 2x in tan(2x) is a horizontal compression by a factor of 2, so we have reduced the period by a factor of 2)

11. 11 (E) Internet Links Go to the following applet from AnalyzeMath to see the connection between solving equations, triangles, and graphs: Go to the following applet from AnalyzeMath to see the connection between solving equations, triangles, and graphs: Trigonometric Equations and The Unit Circle from AnalyzeMath Trigonometric Equations and The Unit Circle from AnalyzeMath Trigonometric Equations and The Unit Circle from AnalyzeMath Trigonometric Equations and The Unit Circle from AnalyzeMath More on Trigonometric Equations - online math lesson from MathTV More on Trigonometric Equations - online math lesson from MathTV More on Trigonometric Equations - online math lesson from MathTV More on Trigonometric Equations - online math lesson from MathTV

12. 12 (F) Homework Handouts from MHR and Nelson Texts Handouts from MHR and Nelson Texts Nelson text, Chap 5.8, p474, Q6,7,9,12,17,18 Nelson text, Chap 5.8, p474, Q6,7,9,12,17,18