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Slide 6-1 Equations 6.1 Solving Trigonometric Equations 6.2 More on Trigonometric Equations 6.3 Trigonometric Equations Involving Multiples Angles 6.4.

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Presentation on theme: "Slide 6-1 Equations 6.1 Solving Trigonometric Equations 6.2 More on Trigonometric Equations 6.3 Trigonometric Equations Involving Multiples Angles 6.4."— Presentation transcript:

1 Slide 6-1 Equations 6.1 Solving Trigonometric Equations 6.2 More on Trigonometric Equations 6.3 Trigonometric Equations Involving Multiples Angles 6.4 Parametric Equations and Further Graphing Chapter 6

2 Slide 6-2  Decide whether the equation is linear or quadratic in form, so you can determine the solution method.  If only one trigonometric function is present, first solve the equation for that function.  If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve. 6.1 Solving Trigonometric Equations

3 Slide 6-3  If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval.  Try using identities to change the form of the equation. If may be helpful to square both sides of the equation first. If this is done, check for extraneous solutions.

4 Slide 6-4 Example: Linear Method  Solve 2 cos 2 x  1 = 0  Solution: First, solve for cos x on the unit circle.

5 Slide 6-5 Example: Factoring  Solve 2 cos x + sec x = 0  Solution Since neither factor of the equation can equal zero, the equation has no solution.

6 Slide 6-6 Example: Factoring  Solve 2 sin 2 x + 3sin x + 1 = 0 (2 Sin x + 1)(Sin x +1) = 0  Set each factor equal to 0. a) 2 Sin x + 1 = 0 b) Sin x + 1 = 0   continued

7 Slide 6-7 Example: Factoring continued 

8 Slide 6-8 Example: Squaring  Solve cos x + 1 = sin x [0, 2  ] Check the solutions in the original equation. The only solutions are  /2 and .

9 Slide 6-9   a) over the interval [0, 2  ), and  b) give all solutions  Solution: Write the interval as the inequality 6.2 More on Trigonometric Equations

10 Slide 6-10 Example: Using a Half-Angle Identity continued  The corresponding interval for x/2 is  Solve  Sine values that corresponds to 1/2 are

11 Slide 6-11 Example: Using a Half-Angle Identity continued  b) Sine function with a period of 4 , all solutions are given by the expressions where n is any integer.

12 Slide 6-12 Example: Double Angle  Solve cos 2x = cos x over the interval [0, 2  ).  First, change cos 2x to a trigonometric function of x. Use the identity

13 Slide 6-13 Example: Double Angle continued  Over the interval

14 Slide 6-14 Example: Multiple-Angle Identity  Solve over the interval [0, 360  ).

15 Slide 6-15  List all solutions in the interval.  The final two solutions were found by adding 360  to 60  and 120 , respectively, giving the solution set 6.3 Trigonometric Equations Involving Multiples Angles

16 Slide 6-16 Example: Multiple Angle  Solve tan 3x + sec 3x = 2 over the interval [0, 2  ).  Tangent and secant are related so use the identity

17 Slide 6-17 Example: Multiple Angle continued 

18 Slide 6-18 Example: Multiple Angle continued  Use a calculator and the fact that cosine is positive in quadrants I and IV,  Since both sides of the equation were squared, each proposed solution must be checked. The solution set is {.2145, , }.

19 Slide 6-19 Solving for x in Terms of y Using Inverse Function Example: y = 3 cos 2x for x. Solution: We want 2x alone on one side of the equation so we can solve for 2x, and then for x.

20 Slide 6-20 Solving an Equation Involving an Inverse Trigonometric Function Example: Solve 2 arcsin Solution: First solve for arcsin x, and then for x. The solution set is {1}.

21 Slide 6-21 Solving an Equation Involving Inverse Trigonometric Functions Example: Solve Solution: Let Then sin and for u in quadrant I, the equation becomes

22 Slide 6-22 Solving an Equation Involving Inverse Trigonometric Functions continued Sketch a triangle and label it using the facts that u is in quadrant I and Since x = cos u, x = and the solution set is { }.

23 Slide 6-23 Solving an Inverse Trigonometric Equation Using an Identity Example: Solve Solution: Isolate one inverse function on one side of the equation. (1)

24 Slide 6-24 Solving an Inverse Trigonometric Equation Using an Identity continued Let u = arccos x, so 0  u  by definition. (2) Substitute this result into equation (2) to get (3)

25 Slide 6-25 Solving an Inverse Trigonometric Equation Using an Identity continued From equation (1) and by the definition of the arcsine function, Since we must have Thus x > 0. From this triangle we find that

26 Slide 6-26 Solving an Inverse Trigonometric Equation Using an Identity continued Now substituting into equation (3) using The solution set is { }.

27 Slide 6-27  A plane curve is a set of points (x, y) such that x = f(t), y = g(t), and f and g are both defined on an interval I. The equations x = f(t) and y = g(t) are parametric equations with parameter t. 6.4 Parametric Equations and Further Graphing

28 Slide 6-28 Graphing a Plane Curve Defined Parametrically Example: Let x = t 2 and y = 2t + 3, for t in [  3,3]. Graph the set of ordered pairs (x, y). Solution: Make a table of corresponding values of t, x, and y over the domain of t 11 11 4 22 33 9 33 yxt

29 Slide 6-29 Graphing a Plane Curve Defined Parametrically continued Plotting the points shows a graph of a portion of a parabola with horizontal axis y = 3. The arrowheads indicate the direction the curve traces as t increases.

30 Slide 6-30 Finding an Equivalent Rectangular Equation Example: Find a rectangular equation for the plane curve of the previous example defined as follows. x = t 2, y = 2t + 3, for t in [  3, 3] Solution: Solve either equation for t.

31 Slide 6-31 Finding an Equivalent Rectangular Equation continued Now substitute this result into the first equation to get  This is the equation of a horizontal parabola opening to the right. Because t is in [  3, 3], x is in [0, 9] and y is in [  3, 9]. This rectangular equation must be given with its restricted domain as 4x = (y  3) 2, for x in [0, 9].

32 Slide 6-32 Graphing a Plane Curve Defined Parametrically Example: Graph the plane curve defined by x = 2 sin t, y = 3cos t, for t in [0,2 ]. Solution: Use the fact that sin 2 t + cos 2 t = 1. Square both sides of each equation; solve one for sin 2 t, the other for cos 2 t.

33 Slide 6-33 Graphing a Plane Curve Defined Parametrically continued Now add corresponding sides of the two equations. This is the equation of an ellipse.

34 Slide 6-34 Finding Alternative Parametric Equation Forms Give two parametric representations for the equation of the parabola y = (x + 5) Solution: The simplest choice is to let x = t, y = (t + 5) for t in ( ,  ) Another choice, which leads to a simpler equation for y, is x = t + 5, y = t for t in ( ,  ).

35 Slide 6-35 Application A small rocket is launched from a table that is 3.36 ft above the ground. Its initial velocity is 64 ft per sec, and it is launched at an angle of 30° with respect to the ground. Find the rectangular equation that models its path. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t  16t Or equivalently,

36 Slide 6-36 Application continued From we obtain Substituting into the other parametric equations for t yields Simplifying, we find that the rectangular equation is Because the equation defines a parabola, the rocket follows a parabolic path.


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