8-3: Algebraic Solutions of Trigonometric Equations Day 1

1. 8-3: Algebraic Solutions of Trigonometric EquationsDay 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 touse 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 allsolutions 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 touse 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 allsolutions 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 tangentfunction, 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: • Use the table of values (restricted sin/cos/tan functions) you should have copied – and started to memorize – by now. • Use the calculator in degree mode, and convert degrees to radians • 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. 8-3: Algebraic Solutions of Trigonometric EquationsDay 2 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 sin2 x – sin x – 2 = 0 in the interval [-, ] • Let u = sin x. This gets us into a quadratic equation • 3u2 – u – 2 = 0 • Factor • (3u + 2)(u – 1) = 0 • Set each parenthesis = 0 and solve for u • u = - 2/3 and u = 1 • Substitute sin x back in for u • sin x = - 2/3 and sin x = 1 • Continued next slide

11. 8-3: Algebraic Solutions to Trig Equations • Factoring Trigonometric Equations (Continued) • Example 7: Solve 3 sin2 x – sin x – 2 = 0 in the interval [-, ] • sin x = - 2/3 and sin x = 1 • Use the inverse sin function to solve for the 1st answer in each equation • x = -.7297 + 2k and x = /2 + 2k • Use the identity rule of sine to find the alternate values • x = 3.1416 – (-.7297) = 3.8713 x =  – /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k x = -.7297 + 2k x = /2 + 2k

12. 8-3: Algebraic Solutions to Trig Equations • Factoring Trigonometric Equations #2 • Example 8: Solve tan x cos2 x = tan x • Write the expression = 0 • tan x cos2 x – tan x = 0 • Take out the GCF • tan x (cos2 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 = 0 cos x = 1 cos x = -1 • Use the inverse trigonometric functions to solve • x = 0 + k x = 0 + 2k x =  + 2k

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

14. 8-3: Algebraic Solutions to Trig Equations • Identities and Factoring • Example 9: -10 cos2 x – 3 sin x + 9 = 0 • Use the Pythagorean identity to write everything in terms of sin • sin2 x + cos2 x = 1 • cos2 x = 1 – sin2 x • Replace the cos2 x • -10(1 – sin2 x) – 3 sin x + 9 = 0 • Distribute • 10 sin2 x – 10 – 3 sin x + 9 = 0 (combine like terms) • 10 sin2 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 cos2 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 (1st values) • x = -0.2014 x = /6 • Use the identity rule of sine to find the alternate values • x = 3.1416 – (-0.2014) = 3.3430 x =  – /6 = 5/6 • Our final solutions: • x = -0.2014 + 2k x = /6 + 2kx = 3.1416 + 2k x = 5/6 + 2k

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