1. Example 1 – Using a Trigonometric Identity  Solve the equation 1 + sin  = 2 cos 2 .  Solution: We first need to rewrite this equation so that it contains only one trigonometric function. To do this, we use a trigonometric identity:  1 + sin  = 2 cos 2   1 + sin  = 2(1 – sin 2  )  2 sin 2  + sin  – 1 = 0  (2 sin  – 1) (sin  + 1) = 0 Given equation Pythagorean identity Put all terms on one side Factor

2. Example 1 – Solution  2 sin  – 1 = 0 or sin  + 1 = 0  sin  = or sin  = –1   = or  =  Because sine has period 2 , we get all the solutions of the equation by adding integer multiples of 2  to these solutions. cont’d Set each factor equal to 0 Solve for sin  Solve for sin  in interval [0, 2  )

3. Example 1 – Solution  Thus the solutions are  = + 2k   = + 2k   = + 2k  where k is any integer.  cont’d

4. Example 3 – Squaring and Using an Identity  Solve the equation cos  + 1 = sin  in the interval [0, 2  ). Solution:  To get an equation that involves either sine only or cosine only, we square both sides and use a Pythagorean identity:   cos  + 1 = sin   cos 2  + 2 cos  + 1 = sin 2   cos 2  + 2 cos  + 1 = 1 – cos 2   2 cos 2  + 2 cos  = 0 Given equation Pythagorean identity Square both sides Simplify

5. Example 3 – Solution  2 cos  = 0 or cos  + 1 = 0  cos  = 0 or cos  = –1   = or  =   Because we squared both sides, we need to check for extraneous solutions. From Check Your Answers we see that the solutions of the given equation are  /2 and . cont’d Set each factor equal to 0 Solve for cos  Solve for  in [0, 2  )

6. Example 3 – Solution  Check Your Answer:   =  =  =   cos + 1 = sin cos + 1 = sin cos  + 1 = sin   0 + 1 = 1 0 + 1 ≟ –1 –1 + 1 = 0 cont’d

7. Equations with Trigonometric Functions of Multiples of Angles

8.  When solving trigonometric equations that involve functions of multiples of angles, we first solve for the multiple of the angle, then divide to solve for the angle.

9. Example 5 – A Trigonometric Equation Involving Multiple of an Angle  Consider the equation 2 sin 3  – 1 = 0.  (a) Find all solutions of the equation.  (b) Find the solutions in the interval [0, 2  ).  Solution:  (a) We first isolate sin 3  and then solve for the angle 3 .  2 sin 3  – 1 = 0  2 sin 3  = 1  sin 3  = Given equation Add 1 Divide by 2

10. Example 5 – Solution   3  = cont’d Solve for 3  in the interval [0, 2  ) (see Figure 2) Figure 2

11. Example 5 – Solution  To get all solutions, we add integer multiples of 2  to these solutions. So the solutions are of the form  3  = + 2k  3  = + 2k   To solve for , we divide by 3 to get the solutions  where k is any integer. cont’d

12. Example 5 – Solution  (b) The solutions from part (a) that are in the interval [0, 2  ) correspond to k = 0, 1, and 2. For all other values of k the corresponding values of  lie outside this interval.  So the solutions in the interval [0, 2  ) are cont’d

13. Example 6 – A Trigonometric Equation Involving a Half Angle  Consider the equation.  (a) Find all solutions of the equation.  (b) Find the solutions in the interval [0, 4  ).  Solution:  (a) We start by isolating tan :  Given equation Add 1

14. Example 6 – Solution  Since tangent has period , to get all solutions we add integer multiples  of to this solution. So the solutions are of the form  Divide by Solve for in the interval cont’d

15. Example 6 – Solution  Multiplying by 2, we get the solutions   = + 2k   where k is any integer.  (b) The solutions from part (a) that are in the interval [0, 4  ) correspond to k = 0 and k = 1. For all other values of k the corresponding values of x lie outside this interval. Thus the solutions in the interval [0, 4  ) are  cont’d

16. Example 3A: Solving Trigonometric Equations with Trigonometric Identities Use trigonometric identities to solve each equation. tan 2 θ + sec 2 θ = 3 for 0 ≤ θ ≤ 2π. tan 2 θ + (1 + tan 2 θ) – 3 = 0 2tan 2 θ – 2 = 0 tan 2 θ – 1 = 0 (tanθ – 1)(tanθ + 1) = 0 tanθ = 1 or tanθ = – 1 Substitute 1 + tan 2 θ for sec 2 θ by the Pythagorean identity. Simplify. Divide by 2. Factor. Apply the Zero Product Property.

17. Example 3B: Solving Trigonometric Equations with Trigonometric Identities Use trigonometric identities to solve each equation. cos 2 θ = 1 + sin 2 θ for 0° ≤ θ ≤ 360° (1 – sin 2 θ) – 1 – sin 2 θ = 0 – 2sin 2 θ = 0 sin 2 θ = 0 sinθ = 0 θ = 0° or 180° or 360° Substitute 1 – sin 2 θ for cos 2 θ by the Pythagorean identity. Simplify. Divide both sides by – 2. Take the square root of both sides.

18. Check It Out! Example 3a Use trigonometric identities to solve each equation for the given domain. 4sin 2 θ + 4cosθ = 5 4(1 - cos 2 θ) + 4cosθ – 5 = 0 4cos 2 θ – 4cosθ + 1 = 0 (2cos 2 θ – 1) 2 = 0 Substitute 1 – cos 2 θ for sin 2 θ by the Pythagorean identity. Simplify. Factor. Take the square root of both sides and simplify.

19. Check It Out! Example 3b Use trigonometric identities to solve each equation for the given domain. sin2θ = – cosθ for 0 ≤ θ < 2  2cosθsinθ + cosθ = 0 cosθ(2sinθ + 1) = 0 Substitute 2cosθsinθ for sin2θ by the double-angle identity. Factor. Apply the Zero Product Property.

20. Example 4 – Finding Intersection Points  Find the values of x for which the graphs of f (x) = sin x and g (x) = cos x intersect. Solution 1: Graphical The graphs intersect where f (x) = g (x). In Figure 1 we graph y 1 = sin x and y 2 = cos x on the same screen, for x between 0 and 2 . (a) (b) Figure 1

21. Example 4 – Solution 1  Using or the intersect command on the graphing calculator, we see that the two points of intersection in this interval occur where x  0.785 and x  3.927.  Since sine and cosine are periodic with period 2 , the intersection points occur where  x  0.785 + 2k  and x  3.927 + 2k   where k is any integer. cont’d

22. Example 4 – Solution 2  Algebraic  To find the exact solution, we set f (x) = g (x) and solve the resulting equation algebraically:  sin x = cos x  Since the numbers x for which cos x = 0 are not solutions of the equation, we can divide both sides by cos x:  = 1  tan x = 1 cont’d Equate functions Divide by cos x Reciprocal identity

23. Example 4 – Solution 2  The only solution of this equation in the interval ( –  /2,  /2) is x =  /4. Since tangent has period , we get all solutions of the equation by adding integer multiples of  :  x = + k   where k is any integer. The graphs intersect for these values of x.  You should use your calculator to check that, rounded to three decimals, these are the same values that we obtained in Solution 1. cont’d

24. Lesson Quiz Solve cos2θ = 3sinθ + 2 for 0 ≤ 0 ≤ 2π.

25. HW 29 page 396 #17,20, 24, 28, 32, 35- 39odd, 79