1. Essential Question: How do we find the non-calculator solution to inverse sin and cosine functions?

2. 8-3: Algebraic Solutions to Trig Equations Solving Basic Cosine Equations Example 1: Solve cos x = 0.6 Using the graphing calculator gets us the first solution cos -1 (0.6) = 0.9273 To find the second solution, we need to use the identity rule of cosine cos(-x) = cos(x) This tells us that our second solution is at x = -0.9273 Remember, cosine is a cyclical wave, so all solutions are given by: x = 0.9273 + 2  k and x = -0.9273 + 2  k

3. 8-3: Algebraic Solutions to Trig Equations Solving Basic Sine Equations Example 2: Solve sin x = -0.75 Using the graphing calculator gets us the first solution sin -1 (-0.75) = -0.8481 To find the second solution, we need to use the identity rule of sine sin(  - x) = sin(x) This tells us that our second solution is at x = 3.1416 – (-0.8481) = 3.9897 Remember, sine is a cyclical wave, so all solutions are given by: x = -0.8481 + 2  k and x = 3.9897 + 2  k (because the cycle is 2 , 3.9897  -2.2935)

4. 8-3: Algebraic Solutions to Trig Equations Solving Basic Tangent Equations Example 3: Solve tan x = 3 Using the graphing calculator gets us the solution tan -1 (3) = 1.2490 There is no second solution on a tangent function, but remember, tangent is a cyclical wave, so all solutions are given by: x = 1.2490 +  k

5. 8-3: Algebraic Solutions to Trig Equations Using the Solution Algorithm Example 4: Solve 8 cos x – 1 = 0 Isolate the trigonometric equation 8 cos x = 1 cos x = 1 / 8 Use the inverse cosine to find the first solution cos -1 ( 1 / 8 ) = x x = 1.4455 Use the identity rule of cosine [cos(-x) = cos(x)] for the other x value x = -1.4455 All solutions are given by: x = 1.4455 + 2  k and x = -1.4455 + 2  k

6. 8-3: Algebraic Solutions to Trig Equations Solving Basic Equations with Special Values Example 5: Solve sin u = exactly, without using a calculator. Find the first value by either: a) Use the table of values (restricted sin/cos/tan functions) you should have copied – and started to memorize – by now. b) Use the calculator in degree mode, and convert degrees to radians c) Use the calculator in radian mode, and divide your answer by  x =  / 4 Use the identity rule of sine [sin(  - x) = sin(x)] for the other x value x =  -  / 4 = 3  / 4 All solutions are given by: x =  / 4 + 2  k and x = 3  / 4 + 2  k

7. 8-3: Algebraic Solutions to Trig Equations Assignment Page 545 Problems 1 – 21, odd problems

8. Essential Question: How do we find the non-calculator solution to inverse sin and cosine functions?

9. 8-3: Algebraic Solutions to Trig Equations Solving basic equations with substitution Example 6: Solve sin 2x = exactly, without using a calculator. Let u = 2x, this gets us into a basic equation (we actually solved it yesterday) sin u = u =  / 4 + 2  k and u = 3  / 4 + 2  k Substitute 2x back in for u, then solve for x 2x =  / 4 + 2  k and 2x = 3  / 4 + 2  k x =  / 8 +  k and x = 3  / 8 +  k

10. 8-3: Algebraic Solutions to Trig Equations Factoring Trigonometric Equations Example 7: Solve 3 sin 2 x – sin x – 2 = 0 in the interval [- ,  ] Let u = sin x. This gets us into a quadratic equation 3u 2 – u – 2 = 0 Factor (3u + 2)(u – 1) = 0 Set each parenthesis = 0 and solve for u u = - 2 / 3 andu = 1 Substitute sin x back in for u sin x = - 2 / 3 andsin x = 1 Continued next slide

11. 8-3: Algebraic Solutions to Trig Equations Factoring Trigonometric Equations (Continued) Example 7: Solve 3 sin 2 x – sin x – 2 = 0 in the interval [- ,  ] sin x = - 2 / 3 andsin x = 1 Use the inverse sin function to solve for the 1 st answer in each equation x = -.7297 + 2  kandx =  / 2 + 2  k Use the identity rule of sine to find the alternate values x = 3.1416 – (-.7297) = 3.8713x =  –  / 2 =  / 2 3.8713 is outside the defined interval ([- ,  ]), but if you subtract a revolution (3.8713 – 6.2832), -2.4119 is within the interval. This gives us our final solutions x = -2.4119 + 2  kx = -.7297 + 2  kx =  / 2 + 2  k

12. 8-3: Algebraic Solutions to Trig Equations Factoring Trigonometric Equations #2 Example 8: Solve tan x cos 2 x = tan x Write the expression = 0 tan x cos 2 x – tan x = 0 Take out the GCF tan x (cos 2 x – 1) = 0 The right-part of the equation can be factored tan x (cos x – 1)(cos x + 1) = 0 Set each parenthesis = 0 and solve tan x = 0cos x = 1cos x = -1 Use the inverse trigonometric functions to solve x = 0 +  kx = 0 + 2  kx =  + 2  k

13. 8-3: Algebraic Solutions to Trig Equations Factoring Trigonometric Equations #2 Solutions from the previous slide: x = 0 +  kx = 0 + 2  kx =  + 2  k So where are those solutions? Meaning all the solutions can be summed up as: x = 0 +  k 00   22 33 22 44 55 33 66 77

14. 8-3: Algebraic Solutions to Trig Equations Identities and Factoring Example 9: -10 cos 2 x – 3 sin x + 9 = 0 Use the Pythagorean identity to write everything in terms of sin sin 2 x + cos 2 x = 1 cos 2 x = 1 – sin 2 x Replace the cos 2 x -10(1 – sin 2 x) – 3 sin x + 9 = 0 Distribute 10 sin 2 x – 10 – 3 sin x + 9 = 0(combine like terms) 10 sin 2 x – 3 sin x – 1 = 0 Factor (5 sin x + 1)(2 sin x – 1) = 0 Continued next slide

15. 8-3: Algebraic Solutions to Trig Equations Identities and Factoring (Continued) Example 9: -10 cos 2 x – 3 sin x + 9 = 0 (5 sin x + 1)(2 sin x – 1) = 0 Set each parenthesis equal to 0 and solve sin x = – 1 / 5 sin x = 1 / 2 Use the inverse sine function to solve for x (1 st values) x = -0.2014x =  / 6 Use the identity rule of sine to find the alternate values x = 3.1416 – (-0.2014) = 3.3430x =  –  / 6 = 5  / 6 Our final solutions: x = -0.2014 + 2  kx =  / 6 + 2  k x = 3.1416 + 2  kx = 5  / 6 + 2  k

16. 8-3: Algebraic Solutions to Trig Equations Assignment Page 546 Problems 23 – 53, odd problems