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# Trigonometry

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### Trigonometry

1. 1. Higher Unit 2 www.mathsrevision.com Higher Outcome 3 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. 2. Trig Identities Outcome 3 www.mathsrevision.com Higher Supplied on a formula sheet !! The following relationships are always true for two angles A and B. 1a. 1b. sin(A + B) = sinAcosB + cosAsinB 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. 3. Trig Identities Higher Examples 1 Outcome 3 www.mathsrevision.com (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. 4. Trig Identities Higher Example 2 Outcome 3 www.mathsrevision.com 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 = Hence LHS = RHS !! √3 /4 + = √3 /2 /2 ) + (√3/2 X ½) √3 /4
5. 5. Trig Identities Higher Example 3 Outcome 3 www.mathsrevision.com Prove that sin15° = ¼(√6 - √2) sin15° = sin(45 – 30)° = sin45°cos30° - cos45°sin30° = (1/√2 = X √3 /2 ) - (1/√2 X ½) (√3/2√2 - 1 = (√3 - 1) 2√2 = (√6 - √2) 4 = ¼(√6 - √2) /2√2) X √2 √2
6. 6. Trig Identities www.mathsrevision.com Higher Example 4 NAB type Question Outcome 3 y β 41 x α 40 Show that 3 4 cos(α - β) = 187 /205 Triangle2 Triangle1 If missing side = x If missing side = y Then x2 = 412 – 402 = 81 Then y2 = 42 + 32 = 25 So So x=9 y=5 sinα = 9/41 and cosα = 40/41 sin β = 3/5 and cosβ = 4/5
7. 7. Higher Trig Identities Outcome 3 www.mathsrevision.com sinα = 9/41 and cosα = 40/41 sin β = 3/5 and cosβ = 4/5 cos(α - β) = cosαcosβ + sinαsinβ = (40/41 X /5) + (9/41 4 = 160 /205 + = 187 X /5 ) 3 /205 /205 27 Remember this is a NAB type Question
8. 8. Trig Identities Higher Example 5 Solve Outcome 3 sinx°cos30° + cosx°sin30° = -0.966 ALWAYS work out By Quad 1 rule 1a first sin(x www.mathsrevision.com NAB type Question where 0o < x < 360o sinx°cos30° + cosx°sin30° = sin(x + 30)° S 180-xo + 30)° = -0.966 Quad 3 and Quad 4 sin-1 0.966 = 75° Quad 3: angle = 180o + 75o x + 30o = 255o x = 225o A xo 180+x T 360-xo C o Quad 4: angle = 360o – 75o x + 30o = 285o x = 255o
9. 9. Trig Identities Higher Solve www.mathsrevision.com Outcome 3 Example 6 sin5 θ cos3 θ - cos5 θ sin3 θ = √3/2 By rule 1b. where 0 < θ < π sin5θ cos3θ - cos5θ sin3θ =sin(5θ - 3θ) = sin2θ sin2θ = √3/2 Repeats every π Quad 1 and Quad 2 sin-1 √3/2 = π/3 S π-θ A θ π+θ 2π-θ T C Quad 1: angle = π/3 Quad 2: angle = π - π/3 In this example repeats lie out 2 θ = π/ 3 2 θ = 2π/3 θ = π /6 θ = π/ 3 with limits
10. 10. Trig Identities Higher Example 7 Outcome 3 www.mathsrevision.com 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. 11. Paper 1 type questions Trig Identities www.mathsrevision.com Higher Example 8 Outcome 3 Simplify sin(θ - π/3) + cos(θ + π/6) + cos(π/2 - θ) sin(θ - π/3) + cos(θ + π/6) + cos(π/2 - θ) = sin θ cosπ/3 – cos θ sinπ/3 + cos θ cosπ/6 – sin θ sinπ/6 + cosπ/2 cos θ + sinπ/2 sin θ = 1/2 sin θ – √3/2cos θ + √3/2 cos θ – 1/2sin θ + 0 x cos θ + 1 X sin θ = sin θ
12. 12. Paper 1 type questions Trig Identities Higher Example 9 Outcome 3 www.mathsrevision.com 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. 13. Double Angle Formulae Outcome 3 www.mathsrevision.com Higher sin 2 A = 2sin A cos A cos 2 A = cos 2 A − sin 2 A = 2cos 2 A − 1 2 = 1 − 2sin A Two further formulae derived from the cos 2 A formulae. cos A = 1 (1 + cos 2 A) 2 2 sin A = 1 (1 − cos 2 A) 2 2
14. 14. Double Angle formulae www.mathsrevision.com Higher Mixed Examples: Outcome 3 4 Given that A is an acute angle and tan A = , calculate sin 2 A and cos 2 A. 3 sin A 4 = cos A 3 2 sin 2 A + ( 3 sin A) 2 = 1 4 sin A = ± Similarly: sin 2 A = 2sin A cos A = Substitute form the tan (sin/cos) equation sin A + cos A = 1 2 cos A = 16 = 25 3 5 9 − 16 = 25 +ve because A is acute 3-4-5 triangle ! 24 25 cos 2 A = cos 2 A − sin 2 A = 4 5 −7 25 A is greater than 45 degrees – hence 2A is greater than 90 degrees.
15. 15. Double Angle formulae Outcome 3 Higher www.mathsrevision.com Find the exact value of sin 75o. 2 sin(75o ) = sin(45 + 30) o sin(75 ) = sin 45cos30 + cos 45sin 30 45 1 o 1 3 1 1 1+ 3 = + = 2 2 22 2 2 Prove that 1 2 30 o 1 sin(α + β ) = tan α + tan β cos α cos β sin(α + β ) sin α cos β + cos α sin β = cosα cos β cosα cos β = sin α sin β + cosα cos β = tan α + tan β 3
16. 16. Double Angle formulae Outcome 3 Higher www.mathsrevision.com For the diagram opposite show that cos LMN = 5 . 5 cos LMN = cos(α + β ) Length of LM = Length of MN = M 18 = 3 2 10 3 2 cos(α + β ) = cosα cos β − sin α sin β = = = 1 3 1 1 − 2 10 2 10 2 2 = = 20 4 5 5 5 1 5 α 3 L 3 β 10 1 N
17. 17. Double Angle formulae Outcome 3 Higher www.mathsrevision.com Prove that, cos 4 α − sin 4 α = cos 2α . cos 4 α − sin 4 α = (cos 2 α ) 2 − (sin 2 α ) 2 Using x 2 − y 2 = ( x − y )( x + y ) = (cos 2 α − sin 2 α )(cos 2 α + sin 2 α ) cos 2 α + sin 2 α = 1 = cos 2 α − sin 2 α cos 2α = cos 2 α − sin 2 α = cos 2α
18. 18. Trigonometric Equations www.mathsrevision.com Higher Outcome 3 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. 19. Trigonometric Equations Outcome 3 Higher cos 2 x o − 4sin x o + 5 = 0 for 0 ≤ x ≤ 360o. www.mathsrevision.com Solve: cos2x and sin x, (1 − 2sin x) − 4sin x + 5 = 0 2 6 − 4sin x − 2sin x = 0 2 so substitute 1-2sin2x compare with 6 − 4 z − 2 z 2 = 0 (6 + 2sin x)(1 − sin x) = 0 sin x = 1 or sin x = −3 x = 90 o 0 ≤ sin x ≤ 1 for all real angles
20. 20. Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Solve: 5cos 2 x o = cos x o − 2 for 0 ≤ x ≤ 360o cos 2x and cos x, 5(2cos 2 x − 1) = cos x − 2 so substitute 2cos2 -1 π 2 10cos 2 x − cos x − 3 = 0 (5cos x − 3)(2cos x + 1) = 0 3 1 cos x = or cos x = − 5 2 π x = 90 + 30 = 120 x = 51.3o and x = 360 − 51.3 = 308.7 90o o o x = 270 − 30 = 240o 180 o and S A T C 270o 3π 2 0o
21. 21. Trigonometric Equations www.mathsrevision.com Higher Outcome 3
22. 22. Trigonometric Equations Outcome 3 Higher www.mathsrevision.com The diagram shows the graphs of f ( x ) = a sin bx o and g ( x ) = c sin x o for 0 ≤ x ≤ 360o. 4 y y y = f ( x) 2 360 0 -2 o x x y = g ( x) -4 Three problems concerning this graph follow.
23. 23. Trigonometric Equations Outcome 3 Higher www.mathsrevision.com i) State the values of a, b and c. y y y = f ( x) f ( x) = a sin bx o g ( x) = c sin x o The max & min values of sinbx are 1 and -1 resp. The max & min values of asinbx are 3 and -3 resp. 360o y = g ( x) a=3 f(x) goes through 2 complete cycles from 0 – 360o b = 2 g ( x) = c sin x o The max & min values of csinx are 2 and -2 resp. c=2 x x
24. 24. Trigonometric Equations Outcome 3 Higher f ( x ) = g ( x) algebraically. www.mathsrevision.com ii) Solve the equation From the previous problem we now have: f ( x) = 3sin 2 x g ( x) = 2sin x and Hence, the equation to solve is: 3(2sin x cos x) = 2sin x Expand sin 2x 6sin x cos x − 2sin x = 0 3sin x cos x − sin x = 0 sin x(3cos x − 1) = 0 sin x = 0 or 1 cos x = 3 3sin 2 x = 2sin x Divide both sides by 2 Spot the common factor in the terms? Is satisfied by all values of x for which:
25. 25. Trigonometric Equations Outcome 3 Higher www.mathsrevision.com iii) find the coordinates of the points of intersection of the graphs for 0 ≤ x ≤ 360o. From the previous problem we have: sin x = 0 and sin x = 0 Hence x = 0o x = 180o x = 360o 1 cos x = 3 1 cos x = 3 x = 70.5o x = (360 − 70.5)o = 289.5o
26. 26. Radian Measurements Outcome 3 www.mathsrevision.com Higher Reminders i) Radians π radians = 180 Converting between degrees and radians: 120o = 120. π 18 0 5π 5π 180 = . 6 6 π 2π = radians 3 = 150o o
27. 27. Degree Measurements Outcome 3 Higher Equilateral triangle: www.mathsrevision.com ii) Exact Values 45o right-angled triangle: 1 2 60o 1 2 45o 1 cos 45o = sin 45o = tan 45 = 1 o 30o 2 3 1 1 2 sin 60o = 1 2 3 2 sin 30o = cos 60o = 1 2 cos30o = 3 2 tan 30o = 1 3 tan 60o = 3
28. 28. Radians / Degrees Outcome 3 www.mathsrevision.com Higher degrees 0o 30o 45o radians 0 π π π π 0 cos 1 4 1 2 1 2 3 3 2 1 2 2 sin 0 tan 0 6 1 2 3 2 1 3 1 3 ∞ Example: 60o 90o 1 What is the exact value of sin 240o ? 240 = 180 + 60 sin(180 + α ) = − sin α sin 240o = − 3 2
29. 29. Sine Graph www.mathsrevision.com Higher Outcome 3 Period = 360o Amplitude = 1
30. 30. Cosine Graph Outcome 3 www.mathsrevision.com Higher Period = 360o Amplitude = 1
31. 31. Tan Graph www.mathsrevision.com Higher Outcome 3 Period = 180o Amplitude cannot be found for tan function
32. 32. Solving Trigonometric Equations Outcome 3 www.mathsrevision.com Higher Solve 2cos3 x − 1 = 0 (0 ≤ x ≤ 360o ) Example: Step 2: consider what solutions Step 1: Re-Arrange are expected π 2 2cos3 x − 1 = 0 2cos3 x = 1 1 cos3 x = 2 90o π 180 o S A T C 270o 3π 2 0o
33. 33. Solving Trigonometric Equations Outcome 3 www.mathsrevision.com Higher cos 3x is positive so solutions 1 in the first and fourth quadrants cos3 x = 2 0 ≤ x ≤ 360o Since x3 Then has 2 solutions x3 0 ≤ 3 x ≤ 1080o has 6 solutions
34. 34. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Step 3: Solve the equation 1 cos3 x = 2 1 3 x = cos  ÷ = 60o 2 −1 1st quad 4th quad 3x = 60o 300o x = 20o 100o cos wave repeats every 360o 420o 660o 780o 1020o 140o 220o 260o 340o
35. 35. Solving Trigonometric Equations Higher Outcome 3 www.mathsrevision.com Graphical solution for 1 cos3 x = 2
36. 36. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Example: Solve 1 + 2 sin 6t = 0 Step 1: Re-Arrange (0 ≤ t ≤ 180o ) Step 2: consider what solutions are expected −1 sin 6t = 2 π 2 90o sin 6t is negative so solutions in the third and fourth quadrants Since 0 ≤ t ≤ 180o x6 x6 Then has 2 solutions 0 ≤ 6t ≤ 1080o has 12 solutions π 180 o S A T C 270o 3π 2 0o
37. 37. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Step 3: Solve the equation −1 sin 6t = 2  −1  6t = sin −1  ÷ 2  3rd quad 4th quad 6t = 225o x = 39.1o 315o  1  o st sin −1  ÷ = 45 always 1 Quad first  2 sin wave repeats every 360o 585o 675o 945o 1035o 52.5o 97.5o 112.5o 157.5o 172.5o
38. 38. Solving Trigonometric Equations www.mathsrevision.com Higher Outcome 3 −1 Graphical solution for sin 6t = 2
39. 39. The solution is to be in radians – but work in Trigonometric at the Solvingdegrees and convertEquations end. Outcome 3 Higher www.mathsrevision.com Example: π Solve 2sin(2 x − ) = 1 3 (0 ≤ x ≤ 2π ) Step 2: consider what solutions are expected Step 1: Re-Arrange 1 sin(2 x − 60 ) = 2 π 2 o 90o (2x – 60o ) = sin-1(1/2) Since 0 ≤ x ≤ 360 has 2 solutions o x2 Then π x2 0 ≤ 2 x ≤ 720 o has 4 solutions 180 o S A T C 270o 3π 2 0o
40. 40. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Step 3: Solve the equation sin(2 x − 60o ) = 1 2 1 2 x − 60o = sin −1  ÷ 2 1 sin −1  ÷ = 30o 2 (1st quadrant) 2 x − 60o = 30o and 150o 1st quad 2nd quad 2x = 90o x = 45o π 4 210o sin wave repeats every 360o 450o 570o 105o 225o 7π 5π 12 4 285o 19π 12
41. 41. Solving Trigonometric Equations www.mathsrevision.com Higher Outcome 3 Graphical solution for 1 sin(2 x − 60 ) = 2 o
42. 42. The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Harder Example: Solve tan 2 x = 3 (0 ≤ x ≤ 2π ) Step 2: consider what solutions are expected Step 1: Re-Arrange π 2 tan x = ± 3 90o tan x = + 3 tan x = − 3 2 solutions 1 and 3 st rd quads 2 solutions 2 nd and 4 th quads π 180 o S A T C 270o 3π 2 0o
43. 43. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Step 3: Solve the equation tan x = + 3 tan x = − 3 π 3 π 3 (60o in the 1st quadrant) 120o tan wave repeats every 180o 240o 300o 2π 3 4π 3 1st quad 2nd quad x = 60o tan −1 3 = 5π 3
44. 44. Solving Trigonometric Equations www.mathsrevision.com Higher Outcome 3 Graphical solution for tan x = 3 2
45. 45. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Harder Example: Solve 3sin 2 x − 4sin x + 1 = 0 (0 ≤ x ≤ 360o ) Step 1: Re-Arrange Step 2: Consider what solutions (3sin x − 1)(sin x − 1) = 0 1 sin x = 3 are expected π 2 90o sin x = 1 π Two solutions One solution 180 o S A T C 270o 3π 2 0o
46. 46. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Step 3: Solve the equation sin x = 1 3 sin x = 1 Two solutions 1stquad 2nd quad x = 19.5o 160.5o Overall solution One solution 90o x = 19.5o , 90o and 160.5o
47. 47. Solving Trigonometric Equations Higher Outcome 3 www.mathsrevision.com Graphical solution for 3sin 2 x − 4sin x + 1 = 0
48. 48. The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Harder Example: Solve 5sin 2 x − 2 = 2cos x Step 1: Re-Arrange Step 2: Consider what solutions 5(1 − cos x) − 2 = 2cos x 2 are expected π 2 Remember o this ! 90 2 2 3 − 2cos x − 5cos 2 x = 0 (3 − 5cos x)(1 + cos x) = 0 3 cos x = 5 (0 ≤ x ≤ 2π ) cos x = −1 sin α + cos α = 1 cos 2 α = 1 − sin 2 α A2 o 2 S 180 α = 1 − cos α sin π C T 270o Two solutions One solution 3π 2 0o
49. 49. Solving Trigonometric Equations Outcome 3 Higher www.mathsrevision.com Step 3: Solve the equation cos x = 3 5 Two solutions 1stquad 3rd quad x = 53.1o 306.9o Overall solution in radians cos x = −1 One solution 180o x = 0.93 , π and 5.35
50. 50. Solving Trigonometric Equations Higher Outcome 3 www.mathsrevision.com Graphical solution for 5 sin 2 x − 2 = 2cos x
51. 51. www.maths4scotland.co.uk Higher Maths Strategies Compound Angles Click to start
52. 52. Maths4Scotland Higher The following questions are on Compound Angles Non-calculator questions will be indicated You will need a pencil, paper, ruler and rubber. Click to continue
53. 53. Maths4Scotland Higher This presentation is split into two parts Using Compound angle formula for Exact values Solving equations Choose by clicking on the appropriate button Quit Quit
54. 54. Maths4Scotland Higher A is the point (8, 4). The line OA is inclined at an angle p radians to the x-axis a) Find the exact values of: i) sin (2p) ii) cos (2p) The line OB is inclined at an angle 2p radians to the x-axis. b) Write down the exact value of the gradient of OB. Draw triangle 80 Pythagoras 4 p 8 8 sin p = 80 4 8 ⇒ 2× × ⇒ 80 80 cos p = Write down values for cos p and sin p Expand sin (2p) sin 2 p = 2sin p cos p Expand cos (2p) cos 2 p = cos p − sin p ⇒ Use m = tan (2p) tan 2 p = Previous 2 2 sin 2 p cos 2 p Quit ⇒ ( ) ( ) 8 80 2 − 4 80 2 ⇒ 4 80 64 4 ⇒ 80 5 64 − 16 3 ⇒ 80 5 4 3 4 ÷ ⇒ 5 5 3 Quit Hint Next
55. 55. Maths4Scotland Higher In triangle ABC show that the exact value of sin(a + b) is Use Pythagoras sin a = sin a, cos a, sin b, cos b Substitute values Simplify Previous 10 2 AC = 2 CB = 10 Write down values for Expand sin (a + b) 2 5 1 2 cos a = 1 2 sin b = 1 10 cos b = 3 10 sin( a + b) = sin a cos b + cos a sin b sin( a + b) = sin(a + b) = 3 20 + 1 20 Quit 1 2 × → 3 10 + 4 20 Quit 1 2 → × 1 10 4 4 → → 4×5 2 5 2 5 Hint Next
56. 56. Maths4Scotland Higher Using triangle PQR, as shown, find the 11 exact value of cos 2x Use Pythagoras PR = 11 Write down values for cos x and sin x 2 cos x = 11 7 sin x = 11 Expand cos 2x cos 2 x = cos 2 x − sin 2 x Substitute values ( ) −( ) Simplify Previous 2 11 cos 2x = 4 7 cos 2 x = − 11 11 Quit 2 7 11 → − 2 3 11 Quit Hint Next
57. 57. Maths4Scotland Higher On the co-ordinate diagram shown, A is the point (6, 8) and 10 B is the point (12, -5). Angle AOC = p and angle COB = q Mark up triangles Find the exact value of sin (p + q). OA = 10 OB = 13 Use Pythagoras Write down values for sin p = sin p, cos p, sin q, cos q Expand sin (p + q) Substitute values Simplify Previous 8 , 10 cos p = 6 , 10 sin q = 8 6 12 5 13 5 , 13 cos q = 12 13 sin ( p + q ) = sin p cos q + cos p sin q sin ( p + q) = sin ( p + q) = 96 130 + 30 130 Quit 8 12 × 10 13 → + 6 5 × 10 13 126 130 Quit → 63 65 Hint Next
58. 58. Maths4Scotland Higher A and B are acute angles such that tan A = 3 4 Draw triangles Use Pythagoras Write down sin A, cos A, sin B, cos B sin A = sin 2 A = 2sin A cos A Expand cos 2A cos 2 A = cos A − sin A 2 Expand sin (2A + B) 13 3 A 5 B 4 12 Hypotenuses are 5 and 13 respectively Expand sin 2A Previous 5 Find the exact value of cos 2A b) sin(2 A + B) c) sin 2A a) Substitute 5 and tan B = 12 . 2 3 , 5 cos A = 4 , 5 sin B = 3 5 sin 2 A = 2 × × cos 2A = 2 4 5 5 , 13 ⇒ 2 4  3  ÷ − ÷ 5 5 ⇒ cos B = 12 13 24 25 16 9 − 25 25 ⇒ 7 25 sin ( 2 A + B ) = sin 2 A cos B + cos 2 A sin B sin ( 2 A + B ) = 24 12 7 5 323 × + × = 25 13 25 13 325 Quit Quit Hint Next
59. 59. Maths4Scotland Higher tan x° If x° is an acute angle such that = 4 3 5 4 3 +3 sin( x + 30)° is show that the exact value of 10 4 x 3 Draw triangle Use Pythagoras Write down sin x and cos x Expand sin (x + 30) sin x = Hypotenuse is 5 4 , 5 cos x = 3 5 sin( x + 30) = sin x cos 30 + cos x sin 30 Substitute sin( x + 30) = 4 3 3 1 × + × 5 2 5 2 Simplify sin( x + 30) = 4 3 3 + 10 10 ⇒ 4 3 +3 10 Hint Previous Table of exact values Quit Quit Next
60. 60. Maths4Scotland Higher The diagram shows two right angled triangles 24 ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm. Angle DBC = x° and angleyABD is 20 − 6 6 cos( x + )° is y°. 35 Show that the exact value of BD = 5, AD = 24 Use Pythagoras Write down sin x, cos x, sin y, cos y. Expand cos (x + y) sin x = 3 , 5 cos x = 4 , 5 sin y = 5 24 , 7 cos y = 5 7 cos( x + y ) = cos x cos y − sin x sin y Substitute cos( x + y ) = 4 5 3 24 × − × 5 7 5 7 Simplify cos( x + y ) = 20 − 3 4 × 6 20 − 6 6 20 3 24 ⇒ ⇒ − 35 35 35 35 Previous Quit Quit Hint Next
61. 61. Maths4Scotland Higher The framework of a child’s swing has dimensions as shown in the diagram. Find the exact value of sin x° Draw triangle Use Pythagoras Draw in perpendicular h= 5 3 Use fact that sin x = sin ( ½ x + ½ x) sin Write down sin ½ x and cos ½ x sin Expand sin ( ½ x + ½ x) Substitute Simplify sin ( x x + 2 2 sin x = )= ( x x 2 h5 x x + 2 2 2 3 ( ) = , cos ( ) = ) = sin cos + sin 2× × x 2 2 3 x 2 x 2 x 2 5 3 x x cos = 2 2 3 2 4 x 2 2sin cos x 2 5 3 4 5 9 Previous Table of exact values Hint Quit Quit Next
62. 62. Maths4Scotland Higher 11 π , 0<α < 3 2 sin 2α find the exact value of tan Given that α = Draw triangle 11 a Use Pythagoras Write down values for cos a and sin a 20 hypotenuse 3 cos a = 20 = 20 3 11 sin a = 20 Expand sin 2a sin 2a = 2 sin a cos a Substitute values sin 2a = 2 × Simplify Previous 11 3 × 20 20 6 11 sin 2a = 20 Quit ⇒ Quit 3 11 10 Hint Next
63. 63. Maths4Scotland Higher Find algebraically the exact value of sin θ ° + sin ( θ + 120 ) ° + cos(θ + 150)° Expand sin (θ +120) sin ( θ + 120 ) = sin θ cos120 + cos θ sin120 Expand cos (θ +150) cos ( θ + 150 ) = cos θ cos150 − sin θ sin150 Use table of exact values cos 120 = − cos 60 = − sin 120 = Combine and substitute Simplify sin 60 = sin θ + sin θ . 1 2 sin θ − sin θ + 3 2 1 2 3 2 cos 150 = − cos 30 = − sin 150 = sin 30 = 3 2 1 2 ( ) + cosθ .( ) + cos θ .( ) − sin θ . ( ) cos θ − − 1 2 3 2 3 2 − 3 2 1 2 1 2 cos θ − sin θ =0 Previous Table of exact values Quit Quit Hint Next
64. 64. Maths4Scotland If cos θ = Higher 4 π , 0 ≤θ ≤ 5 2 sin 2θ a) find the exact value of sin 4θb) Draw triangle cos θ = cos θ and sin θ Expand cos 2θ Find sin 4θ Previous Opposite side = 3 4 5 sin θ = sin 2θ = 2 sin θ cos θ Expand sin 4θ (4θ = 2θ + 2θ) 4 3 5 3 4 24 ⇒ 2× × ⇒ 5 5 25 sin 4θ = 2 sin 2θ cos 2θ cos 2θ = cos θ − sin θ 2 sin 4θ = 2 × 3 θ Use Pythagoras Write down values for Expand sin 2θ 5 24 7 × 25 25 Quit 2 ⇒ 16 9 7 ⇒ − ⇒ 25 25 25 336 625 Quit Hint Next
65. 65. Maths4Scotland Higher For acute angles sin P = P and Q 12 and 13 sin Q = 3 5 13 63 sin ( P +Q) = 65 Show that the exact value of Draw triangles Use Pythagoras Write down sin P, cos P, sin Q, cos Q Expand sin (P + Q) 5 12 P 3 Q 5 4 Adjacent sides are 5 and 4 respectively sin P = 12 , 13 cos P = 5 , 13 sin Q = 3 , 5 cos Q = 4 5 sin ( P + Q ) = sin P cos Q + cos P sin Q Substitute sin ( P + Q ) = 12 4 5 3 × + × 13 5 13 5 Simplify sin ( P + Q ) = 48 15 + 65 65 Previous Quit ⇒ Quit 63 65 Hint Next
66. 66. Maths4Scotland Higher You have completed all Previous Quit 12 questions in this section Quit Back to start
67. 67. Maths4Scotland Higher Using Compound angle formula for Solving Equations Continue Quit Quit
68. 68. Maths4Scotland Higher Solve the equation 3cos(2 x) + 10 cos( x) − 1 = 0 for 0 ≤ x ≤ π correct to 2 decimal places Replace cos 2x with Substitute Simplify cos 2 x = 2 cos 2 x − 1 3 ( 2 cos x − 1) + 10 cos x − 1 = 0 2 Determine quadrants S A T 6 cos x + 10 cos x − 4 = 0 2 C 3cos 2 x + 5cos x − 2 = 0 Factorise Hence ( 3cos x − 1) ( cos x + 2 ) = 0 cos x = x = 1.23 Find acute x Previous acute x = 1.23 rad Quit 2π − 1.23 rads x = 5.05 cos x = −2 Discard or x = 1.23 1 3 rads rads Hint Quit Next
69. 69. Maths4Scotland Higher The diagram shows the graph of a cosine function from 0 to π. a) State the equation of the graph. b) The line with equation y = -√3 intersects this graph at points A and B. Find the co-ordinates of B. Equation y = 2 cos 2 x Determine quadrants 2 cos 2 x = − 3 Solve simultaneously cos 2x = − Rearrange Check range 0≤ x ≤π Find acute 2x Deduce 2x acute 2x = 2x = S T x = Table of exact values 6π π + rads 6 6 Quit Quit 5π 7π or 12 12 B π 6 Previous C 3 2 ⇒ 0 ≤ 2 x ≤ 2π 6π π − or 6 6 A ( is B 7π 12 ,− 3 ) Next Hint
70. 70. Maths4Scotland Higher Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x a) i) 1 expression st 2nd expression Form equation Find expressions for: f(g(x)) g(f(x)) Determine x ii) for 0 ≤ x sin360° 0 ⇒ x = 0°, 360° ≤ x= f b)g ( xSolve f (2f(g(x))sing(f(x)) ( )) = 2 x) = = 2 x cos x = g ( f ( x )) = g (sin x) = 2sin x 2sin 2 x = 2sin x → sin 2 x = sin x Replace sin 2x 2sin x cos x − sin x = 0 Common factor Hence or 2 cos x − 1 = 0 ⇒ cos x = Previous Table of exact values Quit acute x = 60° S A T C Determine x = 0°, 60°, 300°, 360° sin x ( 2 cos x − 1) = 0 sin x = 0 ⇒ quadrants x = 60°, 300° 2sin x cos x = sin x Rearrange 1 2 1 2 Quit Hint Next
71. 71. Maths4Scotland Higher Functions f ( x) = sin x, g ( x) = cos x a) b) i) 2 expression nd h( x ) = x + Find expressions for 1 1 f (h( x)) = Show that 1st expression and sin x + 2 2 π 4 are defined on a suitable set of real numbers i) f(h(x)) cos x ii) Find a similar expression for g(h(x)) f (h( x )) − g (h( x )) = 1 for 0 ≤ x ≤ 2π iii) Hence solve the equation 2 π π sin x = 1 f (h( x)) = f x + = sin x + Simplifies to 2 4 4 ( ) ( ) g (h( x)) = g ( x + ) = cos ( x + ) π 4 f Simplify 1 expr. st π 4 π (h( x)) = sin x cos 4 1 2 Rearrange: sin x = π + cos x sin 4 1 2 Use exact values f ( h( x)) = Similarly for 2nd expr. g (h( x )) = cos x cos − sin x sin sin x + g (h( x)) = Form Eqn. ii) g(h(x)) 1 2 acute x cos x π 4 cos x − 1 2 Determine π 4 sin x f ( h( x)) − g ( h( x)) = 1 Previous Table of exact values Quit Quit acute 2 = 2 x= 2 1 = 2 2 2 π 4 S A T C quadrants x= π 3π , 4 4 Hint Next
72. 72. Maths4Scotland a) Higher Solve the equation sin 2x - cos x = 0 in the interval 0 ≤ x ≤ 180° b) The diagram shows parts of two trigonometric graphs, y = sin 2x and y = cos x. Use your solutions in (a) to write down the co-ordinates of the point P. Replace sin 2x 2sin x cos x − cos x = 0 Common factor cos x ( 2sin x − 1) = 0 Hence cos x = 0 Determine x or Solutions for where graphs cross 2sin x − 1 = 0 ⇒ sin x = 1 2 cos x = 0 ⇒ x = 90°, ( 270° out of range) sin x = 1 2 ⇒ acute x = 30° S A Determine quadrants T Previous Table of exact values x =150° y = cos150° Find y value Coords, P C Quit By inspection (P) y=− x = 30°, 150° for sin x x = 30°, 90°, 150° Quit ( P 150°, − 3 2 ) 3 2 Hint Next
73. 73. Maths4Scotland Higher Solve the 3cos(2 x) + cos( x) = − 1 equation for 0 ≤ x ≤ 360° cos 2 x = 2 cos 2 x − 1 Replace cos 2x with Determine quadrants 3 ( 2 cos x − 1) + cos x = −1 2 Substitute cos x = − 2 3 cos x = 1 2 6 cos 2 x + cos x − 2 = 0 Factorise x = 48° acute x = 60° S Simplify acute A S A ( 3cos x + 2 ) ( 2 cos x − 1) = 0 cos x = − Hence Find acute x acute 2 3 x = 48° cos x = acute 1 2 x = 60° T C x = 132° x = 228° T C x = 60° x = 300° Solutions are: x= 60°, 132°, 228° and 300° Previous Table of exact values Quit Quit Hint Next
74. 74. Maths4Scotland Higher ( π 6 ) 2sin x − Solve the2equation= 1 sin Rearrange ) ( ) ( π = 6 Determine quadrants T 2x − ) π 6 = and for range ( 0 ≤ x ≤ 2π ⇒ 0 ≤ 2 x ≤ 4π S 0 ≤ 2 x ≤ 2π for range 1 = 2 π acute 2 x − 6 Find acute x Note range ( π 2x − 6 for 0 ≤ x ≤ 2π 2x − A π 6 ) = π 6 ( 2x − π 6 π 6 ) = 5π 6 ) = 17π 6 2π ≤ 2 x ≤ 4π 13π 6 ( 2x − Solutions are: x= C π π 7π 3π , , , 6 2 6 2 Hint Previous Table of exact values Quit Quit Next
75. 75. Maths4Scotland Higher a) Write the equation cos 2θ + 8 cos θ + 9 = 0 in terms of cos θ and show that for cos θ it has equal roots. b) Show that there are no real roots for θ Replace cos 2θ with Rearrange Divide by 2 Factorise Deduction cos 2θ = 2 cos 2 θ − 1 Try to solve: ( cos θ + 2 ) = 0 2 cos 2 θ + 8cos θ + 8 = 0 cos θ = −2 cos 2 θ + 4 cos θ + 4 = 0 No solution Hence there are no real solutions for θ ( cos θ + 2 ) ( cos θ + 2 ) = 0 Equal roots for cos θ Hint Previous Quit Quit Next
76. 76. Maths4Scotland Higher Solve algebraically, the equation sin 2x + sin x = 0, 0 ≤ x ≤ 360 Replace sin 2x 2sin x cos x + sin x = 0 Determine quadrants sin x ( 2 cos x + 1) = 0 for cos x S A Common factor Hence sin x = 0 or Determine x 2 cos x + 1 = 0 ⇒ cos x = − 1 2 1 2 acute x = 60° x = 0°, Previous Table of exact values C x = 120°, 240° sin x = 0 ⇒ x = 0°, 360° cos x = − ⇒ T Quit 120°, 240°, 360° Quit Hint Next
77. 77. Maths4Scotland Higher Find the exact solutions of 4sin2 x = 1, 0 ≤ x ≤ 2π sin 2 x = Rearrange Take square roots Find acute x 1 4 sin x = ± 1 2 x = π 6 acute Determine quadrants for sin x S T + and – from the square root requires all 4 quadrants A C π 5π 7π 11π x = , , , 6 6 6 6 Hint Previous Table of exact values Quit Quit Next
78. 78. Maths4Scotland Higher Solve the equation x + cos x = 0 cos 2 Replace cos 2x with for 0 ≤ x ≤ 360° cos 2 x = 2 cos 2 x − 1 Substitute 2 cos 2 x + cos x − 1 = 0 Factorise cos x = 2 cos 2 x − 1 + cos x = 0 Simplify Determine quadrants 1 2 ( 2 cos x − 1) ( cos x + 1) = 0 cos x = Hence Find acute x acute 1 2 x = 60° acute S cos x = −1 x = 60° A T C x = 60° x = 300° x = 180° Solutions are: x= 60°, 180° and 300° Previous Table of exact values Quit Quit Hint Next
79. 79. Maths4Scotland Higher cos Solve algebraically, the equation 2 x + 5cos x − 2 = 0 Replace cos 2x with Substitute cos 2 x = 2 cos 2 x − 1 for 0 ≤ x ≤ 360° Determine quadrants 2 cos x − 1 + 5cos x − 2 = 0 2 cos x = acute Simplify 2 cos 2 x + 5cos x − 3 = 0 Factorise 1 2 x = 60° ( 2 cos x − 1) ( cos x + 3) = 0 Hence Find acute x cos x = acute 1 2 x = 60° S cos x = −3 Discard above A T C x = 60° x = 300° Solutions are: x= 60° and 300° Previous Table of exact values Quit Quit Hint Next
80. 80. Maths4Scotland Higher You have completed all Previous 12 questions in this presentation Quit Quit Back to start
81. 81. Maths4Scotland Higher Table of exact values 30° sin cos tan Return 45° 60° π 6 1 2 π 4 π 3 1 2 1 2 3 2 3 2 1 3 1 1 2 3
82. 82. Maths4Scotland Higher You have completed all Previous 12 questions in this presentation Quit Quit Back to start