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www.mathsrevision.com Higher Outcome 3 Higher Unit 2 www.mathsrevision.com Trigonometry identities of the form sin(A+B) Double Angle formulae Trigonometric.

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Presentation on theme: "www.mathsrevision.com Higher Outcome 3 Higher Unit 2 www.mathsrevision.com Trigonometry identities of the form sin(A+B) Double Angle formulae Trigonometric."— Presentation transcript:

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2 Higher Outcome 3 Higher Unit 2 Trigonometry identities of the form sin(A+B) Double Angle formulae Trigonometric Equations Exam Type Questions Radians & Trig Basics More Trigonometric Equations

3 Higher Outcome 3 Trig Identities 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!! Supplied on a formula sheet !!

4 Higher Outcome 3 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 Trig Identities

5 Higher Outcome 3 = cos30° 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 )°=  3 / 2 RHS = cos60°cos30° + sin60°sin30° = ( ½ X  3 / 2 ) + (  3 / 2 X ½) =  3 / 4 +  3 / 4 =  3 / 2 Hence LHS = RHS !! Trig Identities

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

7 Higher Outcome 3Example 4   Show that cos(  -  ) = 187 / 205 Triangle1 If missing side = x x Then x 2 = 41 2 – 40 2 = 81 So x = 9 sin  = 9 / 41 and cos  = 40 / 41 Triangle2 If missing side = y y Then y 2 = = 25 So y = 5 sin  = 3 / 5 and cos  = 4 / 5 Trig Identities NAB type Question

8 Higher Outcome 3 Trig Identities cos(  -  ) = cos  cos  + sin  sin  = ( 40 / 41 X 4 / 5 ) + ( 9 / 41 X 3 / 5 ) = 160 / / 205 = 187 / 205 Remember this is a NAB type Question sin  = 9 / 41 and cos  = 40 / 41 sin  = 3 / 5 and cos  = 4 / 5

9 Higher Outcome 3Example 5 Solve sinx  cos30  + cosx  sin30  = where 0 o < x < 360 o By rule 1a sinx  cos30  + cosx  sin30  =sin(x + 30)  sin(x + 30)  = AS TC xoxo 180+x o 360-x o 180-x o Quad 3 and Quad 4 sin = 75  Quad 3: angle = 180 o + 75 o x + 30 o = 255 o x = 225 o Quad 4: angle = 360 o – 75 o x + 30 o = 285 o x = 255 o Trig Identities NAB type Question ALWAYS work out Quad 1 first

10 Higher Outcome 3 Trig Identities Example 6 Solve sin5 θ cos3 θ - cos5 θ sin3 θ =  3 / 2 where 0 < θ <  By rule 1b. sin5 θ cos3 θ - cos5 θ sin3 θ = sin(5 θ - 3 θ ) sin2 θ =  3 / 2 AS TC θ  + θ2  - θ  - θ Quad 1 and Quad 2 sin -1  3 / 2 =  / 3 Quad 1: angle =  / 3 2 θ =  / 3 θ =  / 6 Quad 2: angle =  -  / 3 2 θ = 2  / 3 θ =  / 3 = sin2 θ Repeats every  In this example repeats lie out with limits

11 Higher Outcome 3 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 – 32 o = 180 o ie x = 212 o Trig Identities

12 Higher Outcome 3 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 / 2 cos θ +  3 / 2 cos θ – 1 / 2 sin θ + 0 x cos θ + 1 X sin θ = sin θ Trig Identities Paper 1 type questions

13 Higher Outcome 3 Example 9 Prove that (sinA + cosB) 2 + (cosA - sinB) 2 = 2(1 + sin(A - B)) LHS = (sinA + cosB) 2 + (cosA - sinB) 2 = sin 2 A + 2sinAcosB + cos 2 B + cos 2 A – 2cosAsinB + sin 2 B = (sin 2 A + cos 2 A) + (sin 2 B + cos 2 B) + 2sinAcosB - 2cosAsinB = (sinAcosB - cosAsinB) = 2 + 2sin(A – B) = 2(1 + sin(A – B))= RHS Trig Identities Paper 1 type questions

14 Higher Outcome 3 Double Angle Formulae

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

16 Higher Outcome 3 Double Angle formulae

17 Higher Outcome 3 Double Angle formulae

18 Higher Outcome 3 Double Angle formulae

19 Higher Outcome 3 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. cos2A = 2cos 2 A – 1if cosA is also in the equation Rules for solving equations sin2A = 2sinAcosAwhen replacing sin2Aequation cos2A = 1 – 2sin 2 Aif sinA is also in the equation

20 Higher Outcome 3 cos2x and sin x, so substitute 1-2sin 2 x Trigonometric Equations

21 Higher Outcome 3 cos 2x and cos x, so substitute 2cos 2 -1 Trigonometric Equations C A S T 0o0o 180 o 270 o 90 o

22 Higher Outcome 3 Trigonometric Equations

23 Higher Outcome o Three problems concerning this graph follow. Trigonometric Equations

24 Higher Outcome o The max & min values of asinbx are 3 and -3 resp. The max & min values of sinbx are 1 and -1 resp. f(x) goes through 2 complete cycles from 0 – 360 o The max & min values of csinx are 2 and -2 resp. Trigonometric Equations

25 Higher Outcome 3 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: Trigonometric Equations

26 Higher Outcome 3 From the previous problem we have: Hence Trigonometric Equations

27 Higher Outcome 3 Radian Measurements Reminders i) Radians Converting between degrees and radians:

28 Higher Outcome 3 ii) Exact Values 45 o right-angled triangle: o Equilateral triangle: o 30 o Degree Measurements

29 Higher Outcome 3 degrees0o0o 30 o 45 o 60 o 90 o radians sin cos tan Example: What is the exact value of sin 240 o ? Radians / Degrees

30 Higher Outcome 3 Sine Graph Period = 360 o Amplitude = 1

31 Higher Outcome 3 Cosine Graph Period = 360 o Amplitude = 1

32 Higher Outcome 3 Tan Graph Period = 180 o Amplitude cannot be found for tan function

33 Higher Outcome 3 Solving Trigonometric Equations Example: Step 1: Re-Arrange Step 2: consider what solutions are expected C A S T 0o0o 180 o 270 o 90 o

34 Higher Outcome 3 cos 3x is positive so solutions in the first and fourth quadrants x 3 Solving Trigonometric Equations

35 Higher Outcome o Step 3: Solve the equation 1 st quad4 th quad cos wave repeats every 360 o x = 20 o Solving Trigonometric Equations 100 o 140 o 220 o 260 o 340 o 420 o 660 o 780 o 1020 o 3x = 60 o

36 Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

37 Higher Outcome 3 Solving Trigonometric Equations Example: Step 1: Re-Arrange Step 2: consider what solutions are expected C A S T 0o0o 180 o 270 o 90 o sin 6t is negative so solutions in the third and fourth quadrants x 6 x 6

38 Higher Outcome o Step 3: Solve the equation 3 rd quad4 th quad sin wave repeats every 360 o x = 39.1 o Solving Trigonometric Equations 52.5 o 97.5 o o o o 585 o 675 o 945 o 1035 o 6t = 225 o

39 Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

40 Higher Outcome 3 Solving Trigonometric Equations Example: Step 1: Re-Arrange Step 2: consider what solutions are expected C A S T 0o0o 180 o 270 o 90 o (2x – 60 o ) = sin -1 (1/2) x 2 x 2 The solution is to be in radians – but work in degrees and convert at the end.

41 Higher Outcome o Step 3: Solve the equation 1 st quad2 nd quad sin wave repeats every 360 o x = 45 o Solving Trigonometric Equations 105 o 225 o 285 o 450 o 570 o 2x = 90 o

42 Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

43 Higher Outcome 3 Solving Trigonometric Equations Harder Example: Step 1: Re-Arrange Step 2: consider what solutions are expected C A S T 0o0o 180 o 270 o 90 o 2 solutions 1 st and 3 rd quads The solution is to be in radians – but work in degrees and convert at the end. 2 solutions 2 nd and 4 th quads

44 Higher Outcome o Step 3: Solve the equation 1 st quad2 nd quad tan wave repeats every 180 o Solving Trigonometric Equations 240 o 300 o x = 60 o

45 Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

46 Higher Outcome 3 Solving Trigonometric Equations Harder Example: Step 1: Re-Arrange Step 2: Consider what solutions are expected C A S T 0o0o 180 o 270 o 90 o One solution Two solutions

47 Higher Outcome o Step 3: Solve the equation 1 st quad2 nd quad Solving Trigonometric Equations 90 o x = 19.5 o One solution Two solutions Overall solutionx = 19.5 o, 90 o and o

48 Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

49 Higher Outcome 3 Solving Trigonometric Equations Harder Example: Step 1: Re-Arrange Step 2: Consider what solutions are expected C A S T 0o0o 180 o 270 o 90 o One solution Two solutions Remember this ! The solution is to be in radians – but work in degrees and convert at the end.

50 Higher Outcome o Step 3: Solve the equation 1 st quad3 rd quad Solving Trigonometric Equations 180 o x = 53.1 o One solution Two solutions Overall solution in radians x = 0.93, π and 5.35

51 Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

52 Higher Maths Strategies Click to start Compound Angles

53 Maths4Scotland Higher Compound Angles The following questions are on Non-calculator questions will be indicated Click to continue You will need a pencil, paper, ruler and rubber.

54 Maths4Scotland Higher Quit This presentation is split into two parts Using Compound angle formula for Exact values Solving equations Choose by clicking on the appropriate button

55 Maths4Scotland Higher Hint PreviousNext Quit 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 (2 p ) ii) cos (2 p ) The line OB is inclined at an angle 2 p radians to the x -axis. b) Write down the exact value of the gradient of OB. Draw trianglePythagoras Write down values for cos p and sin p Expand sin (2p) Expand cos (2p) Use m = tan (2p) 8 4 p

56 Maths4Scotland Higher Hint PreviousNext Quit In triangle ABC show that the exact value of Use Pythagoras Write down values for sin a, cos a, sin b, cos b Expand sin (a + b) Substitute values Simplify

57 Maths4Scotland Higher Hint PreviousNext Quit Using triangle PQR, as shown, find the exact value of cos 2 x Use Pythagoras Write down values for cos x and sin x Expand cos 2 x Substitute values Simplify

58 Maths4Scotland Higher Hint PreviousNext Quit On the co-ordinate diagram shown, A is the point (6, 8) and B is the point (12, -5). Angle AOC = p and angle COB = q Find the exact value of sin ( p + q ). Use Pythagoras Write down values for sin p, cos p, sin q, cos q Expand sin ( p + q ) Substitute values Simplify Mark up triangles

59 Maths4Scotland Higher Hint PreviousNext Quit Draw triangles Use Pythagoras Expand sin 2A A and B are acute angles such that and. Find the exact value of a) b) c) 4 3 A 12 5 B Hypotenuses are 5 and 13 respectively 5 13 Write down sin A, cos A, sin B, cos B Expand cos 2A Expand sin (2A + B) Substitute

60 Maths4Scotland Higher Hint PreviousNext Quit Draw triangle Use Pythagoras Expand sin (x + 30) If x° is an acute angle such that show that the exact value of 3 4 x Hypotenuse is 5 5 Write down sin x and cos x Substitute Simplify Table of exact values

61 Maths4Scotland Higher Hint PreviousNext Quit Use Pythagoras Expand cos (x + y) Write down sin x, cos x, sin y, cos y. Substitute Simplify The diagram shows two right angled triangles ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm. Angle DBC = x ° and angle ABD is y °. Show that the exact value of 5

62 Maths4Scotland Higher Hint PreviousNext Quit Draw triangle Use Pythagoras The framework of a child’s swing has dimensions as shown in the diagram. Find the exact value of sin x° Write down sin ½ x and cos ½ x Substitute Simplify Table of exact values x Draw in perpendicular 2 h Use fact that sin x = sin ( ½ x + ½ x ) Expand sin ( ½ x + ½ x )

63 Maths4Scotland Higher Hint PreviousNext Quit Given that find the exact value of Write down values for cos a and sin a Expand sin 2 a Substitute values Simplify 3 a Draw triangle Use Pythagoras

64 Maths4Scotland Higher Hint PreviousNext Quit Find algebraically the exact value of Expand sin (  +120) Use table of exact values Combine and substitute Table of exact values Expand cos (  +150) Simplify

65 Maths4Scotland Higher Hint PreviousNext Quit If find the exact value of a)b) Write down values for cos  and sin  Expand sin 2  Draw triangle Use Pythagoras 5  4 3 Opposite side = 3 Expand sin 4  (4  = 2  + 2  ) Expand cos 2  Find sin 4 

66 Maths4Scotland Higher Hint PreviousNext Quit Draw triangles Use Pythagoras Expand sin (P + Q) For acute angles P and Q Show that the exact value of P 5 3 Q Adjacent sides are 5 and 4 respectively 5 4 Write down sin P, cos P, sin Q, cos Q Substitute Simplify

67 Maths4Scotland Higher Previous Quit You have completed all 12 questions in this section Back to start

68 Maths4Scotland Higher Quit Solving Equations Using Compound angle formula for Continue

69 Maths4Scotland Higher Hint PreviousNext Quit Solve the equation for 0 ≤ x ≤  correct to 2 decimal places Replace cos 2x with Substitute Simplify Factorise Hence Discard Find acute x Determine quadrants AS CT

70 Maths4Scotland Higher Hint PreviousNext Quit Table of exact values Solve simultaneously Rearrange Find acute 2 x Determine quadrants AS CT The diagram shows the graph of a cosine function from 0 to . 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 Check range Deduce 2x

71 Functions f and g are defined on suitable domains by f ( x ) = sin ( x ) and g ( x ) = 2 x a)Find expressions for: i) f ( g ( x ))ii) g ( f ( x )) b)Solve 2 f ( g ( x )) = g ( f ( x )) for 0  x  360° Maths4Scotland Higher Hint PreviousNext Quit Table of exact values 2 nd expression Form equation Rearrange Determine quadrants AS CT 1 st expression Common factor Replace sin 2x Hence Determine x

72 Functions are defined on a suitable set of real numbers a)Find expressions for i) f ( h ( x ))ii) g ( h ( x )) b)i) Show that ii) Find a similar expression for g ( h ( x )) iii) Hence solve the equation Maths4Scotland Higher Hint PreviousNext Quit Table of exact values 2 nd expression Simplify 1 st expr. Similarly for 2 nd expr. Determine quadrants AS CT 1 st expression Use exact values Form Eqn. Simplifies to Rearrange: acute x

73 a)Solve the equation sin 2 x - cos x = 0 in the interval 0  x  180° b)The diagram shows parts of two trigonometric graphs, y = sin 2 x and y = cos x. Use your solutions in (a) to write down the co-ordinates of the point P. Maths4Scotland Higher Hint PreviousNext Quit Table of exact values Determine quadrants for sin x AS CT Common factor Replace sin 2 x Hence Determine x Solutions for where graphs cross By inspection (P) Find y value Coords, P

74 Maths4Scotland Higher Hint PreviousNext Quit Solve the equation for 0 ≤ x ≤ 360° Replace cos 2x with Substitute Simplify Factorise Hence Find acute x Determine quadrants AS CT Table of exact values AS CT Solutions are: x= 60°, 132°, 228° and 300°

75 Maths4Scotland Higher Hint PreviousNext Quit Solve the equation for 0 ≤ x ≤ 2  Rearrange Find acute x Determine quadrants AS CT Table of exact values Solutions are: Note range

76 Maths4Scotland Higher Hint PreviousNext Quit a) Write the equation cos 2  + 8 cos  + 9 = 0 in terms of cos  and show that for cos  it has equal roots. b) Show that there are no real roots for  Rearrange Divide by 2 Deduction Factorise Replace cos 2  with Equal roots for cos  Try to solve: Hence there are no real solutions for  No solution

77 Solve algebraically, the equation sin 2 x + sin x = 0, 0  x  360 Maths4Scotland Higher Hint PreviousNext Quit Table of exact values Determine quadrants for cos x AS CT Common factor Replace sin 2 x Hence Determine x x = 0°, 120°, 240°, 360°

78 Find the exact solutions of 4sin 2 x = 1, 0  x  2  Maths4Scotland Higher Hint PreviousNext Quit Table of exact values Determine quadrants for sin x AS CT Take square roots Rearrange Find acute x + and – from the square root requires all 4 quadrants

79 Maths4Scotland Higher Hint PreviousNext Quit Solve the equation for 0 ≤ x ≤ 360° Replace cos 2x with Substitute Simplify Factorise Hence Find acute x Determine quadrants AS CT Table of exact values Solutions are: x= 60°, 180° and 300°

80 Maths4Scotland Higher Hint PreviousNext Quit Solve algebraically, the equation for 0 ≤ x ≤ 360° Replace cos 2x with Substitute Simplify Factorise Hence Find acute x Determine quadrants Table of exact values AS CT Solutions are: x= 60° and 300° Discard above

81 Maths4Scotland Higher Previous Quit You have completed all 12 questions in this presentation Back to start

82 Maths4Scotland Higher Return 30°45°60° sin cos tan1 Table of exact values

83 Maths4Scotland Higher Previous Quit You have completed all 12 questions in this presentation Back to start

84 Higher Outcome 3 Are you on Target ! Update you log book Make sure you complete and correct ALL of the Trigonometry questions in the past paper booklet.Trigonometry


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