1 / 82

Higher Unit 2

Higher Unit 2. Trigonometry identities of the form sin(A+B). Double Angle formulae. Trigonometric Equations. Radians & Trig Basics. More Trigonometric Equations. Exam Type Questions. Trig Identities. Supplied on a formula sheet !!.

louise
Download Presentation

Higher Unit 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Higher Unit 2 Trigonometry identities of the form sin(A+B) Double Angle formulae Trigonometric Equations Radians & Trig Basics More Trigonometric Equations Exam Type Questions www.mathsrevision.com

  2. Trig Identities Supplied on a formula sheet !! The following relationships are always true for two angles A and B. 1a. sin(A + B) = sinAcosB + cosAsinB 1b. sin(A - B) = sinAcosB - cosAsinB 2a. cos(A + B) = cosAcosB – sinAsinB 2b. cos(A - B) = cosAcosB + sinAsinB Quite tricky to prove but some of following examples should show that they do work!!

  3. Trig Identities Examples 1 (1) Expand cos(U – V). (use formula 2b ) cos(U – V) = cosUcosV + sinUsinV (2) Simplify sinf°cosg° - cosf°sing° (use formula 1b ) sinf°cosg° - cosf°sing° = sin(f – g)° (3) Simplify cos8 θ sinθ + sin8 θ cos θ (use formula 1a ) cos8 θ sin θ + sin8 θ cos θ = sin(8 θ + θ) = sin9 θ

  4. Trig Identities Example 2 By taking A = 60° and B = 30°, prove the identity for cos(A – B). NB: cos(A – B) = cosAcosB + sinAsinB LHS = cos(60 – 30 )° = cos30° = 3/2 RHS = cos60°cos30° + sin60°sin30° = ( ½ X3/2 ) + (3/2X ½) = 3/4 + 3/4 = 3/2 Hence LHS = RHS !!

  5. Trig Identities Example 3 Prove that sin15° = ¼(6 - 2) sin15° = sin(45 – 30)° = sin45°cos30° - cos45°sin30° = (1/2X3/2 ) - (1/2X ½) = (3/22 - 1/22) = (3 - 1) 22 X2 2 = (6 - 2) 4 = ¼(6 - 2)

  6. Trig Identities NAB type Question Example 4 y 41 x 3   4 40 Show that cos( - ) = 187/205 Triangle2 Triangle1 If missing side = y If missing side = x Then x2 = 412 – 402 = 81 Then y2 = 42 + 32 = 25 So x = 9 So y = 5 sin = 9/41 and cos = 40/41 sin  = 3/5 and cos = 4/5

  7. Trig Identities sin = 9/41 and cos = 40/41 sin  = 3/5 and cos = 4/5 cos( - ) = coscos + sinsin = (40/41X4/5) + (9/41X3/5) = 160/205 + 27/205 = 187/205 Remember this is a NAB type Question

  8. S A 180-xo xo 360-xo 180+xo T C NAB type Question Trig Identities Example 5 Solve sinxcos30 + cosxsin30 = -0.966 where 0o< x < 360o ALWAYS work out Quad 1 first By rule 1a sinxcos30 + cosxsin30 = sin(x + 30) sin(x + 30) = -0.966 Quad 3 and Quad 4 sin-1 0.966 = 75 Quad 3: angle = 180o+ 75o Quad 4: angle = 360o – 75o x + 30o= 285o x + 30o= 255o x = 225o x = 255o

  9. S A  -θ θ  +θ 2  -θ T C Trig Identities Example 6 Solve sin5 θ cos3 θ - cos5 θ sin3 θ = 3/2 where 0 < θ <  sin(5θ - 3θ) = sin2θ By rule 1b. sin5θ cos3θ - cos5θ sin3θ = sin2θ = 3/2 Quad 1 and Quad 2 sin-13/2 = /3 Repeats every  Quad 1: angle = /3 Quad 2: angle =  - /3 In this example repeats lie out with limits 2 θ = /3 2 θ = 2/3 θ = /3 θ = /6

  10. Trig Identities Example 7 Find the value of x that minimises the expression cosxcos32 + sinxsin32 Using rule 2(b) we get cosxcos32 + sinxsin32 = cos(x – 32) cos graph is roller-coaster min value is -1 when angle = 180 ie x – 32o = 180o ie x = 212o

  11. Paper 1 type questions Trig Identities Example 8 Simplify sin(θ - /3) + cos(θ + /6) + cos(/2 - θ) sin(θ - /3) + cos(θ + /6) + cos(/2 - θ) = sin θ cos/3 – cos θ sin/3 + cos θ cos/6 – sin θ sin/6 + cos/2 cos θ + sin/2 sin θ = 1/2 sin θ– 3/2cos θ + 3/2 cos θ – 1/2sin θ + 0 xcos θ + 1 X sin θ = sin θ

  12. Paper 1 type questions Trig Identities Example 9 Prove that (sinA + cosB)2 + (cosA - sinB)2 = 2(1 + sin(A - B)) LHS = (sinA + cosB)2 + (cosA - sinB)2 = sin2A + 2sinAcosB + cos2B + cos2A – 2cosAsinB + sin2B = (sin2A + cos2A) + (sin2B + cos2B) + 2sinAcosB - 2cosAsinB = 1 + 1 + 2(sinAcosB - cosAsinB) = 2 + 2sin(A – B) = 2(1 + sin(A – B)) = RHS

  13. Double Angle Formulae

  14. Double Angle formulae Mixed Examples: Substitute form the tan (sin/cos) equation +ve because A is acute 3-4-5 triangle ! Similarly: A is greater than 45 degrees – hence 2A is greater than 90 degrees.

  15. Double Angle formulae

  16. Double Angle formulae

  17. Double Angle formulae

  18. Trigonometric Equations Double angle formulae (like cos2A or sin2A) often occur in trig equations. We can solve these equations by substituting the expressions derived in the previous sections. Rules for solving equations sin2A = 2sinAcosA when replacing sin2Aequation cos2A = 2cos2A – 1 if cosA is also in the equation cos2A = 1 – 2sin2A if sinA is also in the equation

  19. Trigonometric Equations cos2x and sin x, so substitute 1-2sin2x

  20. 90o A S 180o 0o C T 270o Trigonometric Equations cos 2x and cos x, so substitute 2cos2-1

  21. Trigonometric Equations

  22. 4 2 360o 0 -2 -4 Trigonometric Equations Three problems concerning this graph follow.

  23. 360o Trigonometric Equations The max & min values of sinbxare 1 and -1 resp. The max & min values of asinbxare 3 and -3 resp. f(x) goes through 2 complete cycles from 0 – 360o The max & min values of csinx are 2 and -2 resp.

  24. Trigonometric Equations From the previous problem we now have: Hence, the equation to solve is: Expand sin 2x Divide both sides by 2 Spot the common factor in the terms? Is satisfied by all values of x for which:

  25. Trigonometric Equations From the previous problem we have: Hence

  26. Radian Measurements Reminders i) Radians Converting between degrees and radians:

  27. 30o 2 2 1 60o 45o 1 1 1 Degree Measurements Equilateral triangle: ii) Exact Values 45o right-angled triangle:

  28. Radians / Degrees 0 0 1 1 0 0 1 What is the exact value of sin 240o ? Example:

  29. Sine Graph Period = 360o Amplitude = 1

  30. Cosine Graph Period = 360o Amplitude = 1

  31. Tan Graph Period = 180o Amplitude cannot be found for tan function

  32. 90o A S 180o 0o C T 270o Solving Trigonometric Equations Example: Step 2: consider what solutions are expected Step 1: Re-Arrange

  33. Solving Trigonometric Equations cos 3x is positive so solutions in the first and fourth quadrants x 3 x 3

  34. Solving Trigonometric Equations Step 3: Solve the equation cos wave repeats every 360o 1st quad 4th quad 3x = 60o 420o 660o 780o 1020o 300o x = 20o 100o 140o 220o 260o 340o

  35. Solving Trigonometric Equations Graphical solution for

  36. 90o A S 180o 0o C T 270o Solving Trigonometric Equations Example: Step 2: consider what solutions are expected Step 1: Re-Arrange sin 6t is negative so solutions in the third and fourth quadrants x 6 x 6

  37. Solving Trigonometric Equations Step 3: Solve the equation sin wave repeats every 360o 3rd quad 4th quad 6t = 225o 585o 675o 945o 1035o 315o x = 39.1o 52.5o 97.5o 112.5o 157.5o 172.5o

  38. Solving Trigonometric Equations Graphical solution for

  39. 90o A S 180o 0o C T 270o The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Example: Step 2: consider what solutions are expected Step 1: Re-Arrange (2x – 60o ) = sin-1(1/2) x 2 x 2

  40. Solving Trigonometric Equations Step 3: Solve the equation sin wave repeats every 360o 1stquad 2ndquad 2x = 90o 450o570o 210o x = 45o 105o 225o 285o

  41. Solving Trigonometric Equations Graphical solution for

  42. 90o A S 180o 0o C T 270o The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Harder Example: Step 2: consider what solutions are expected Step 1: Re-Arrange 2 solutions 1st and 3rdquads 2 solutions 2nd and 4thquads

  43. Solving Trigonometric Equations Step 3: Solve the equation tan wave repeats every 180o 1stquad 2ndquad x = 60o 240o300o 120o

  44. Solving Trigonometric Equations Graphical solution for

  45. 90o A S 180o 0o C T 270o Solving Trigonometric Equations Harder Example: Step 2: Consider what solutions are expected Step 1: Re-Arrange Two solutions One solution

  46. Solving Trigonometric Equations Step 3: Solve the equation Two solutions One solution 1stquad 2ndquad 90o x= 19.5o 160.5o Overall solution x = 19.5o , 90o and 160.5o

  47. Solving Trigonometric Equations Graphical solution for

  48. 90o A S 180o 0o C T 270o The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Harder Example: Step 2: Consider what solutions are expected Step 1: Re-Arrange Remember this ! Two solutions One solution

  49. Solving Trigonometric Equations Step 3: Solve the equation Two solutions One solution 1stquad 3rdquad 180o x= 53.1o 306.9o Overall solution in radians x = 0.93, πand 5.35

  50. Solving Trigonometric Equations Graphical solution for

More Related