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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 Equations Exam Type Questions Radians & Trig Basics More Trigonometric Equations

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www.mathsrevision.com 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 !!

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www.mathsrevision.com 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

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www.mathsrevision.com 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

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www.mathsrevision.com 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 - 1 / 2 2 ) = sin45°cos30° - cos45°sin30° = ( 3 - 1) 2 2 X 2 2 = ( 6 - 2) 4 = ¼( 6 - 2) Trig Identities

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www.mathsrevision.com Higher Outcome 3Example 4 41 40 4 3 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 = 4 2 + 3 2 = 25 So y = 5 sin = 3 / 5 and cos = 4 / 5 Trig Identities NAB type Question

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www.mathsrevision.com Higher Outcome 3 Trig Identities cos( - ) = cos cos + sin sin = ( 40 / 41 X 4 / 5 ) + ( 9 / 41 X 3 / 5 ) = 160 / 205 + 27 / 205 = 187 / 205 Remember this is a NAB type Question sin = 9 / 41 and cos = 40 / 41 sin = 3 / 5 and cos = 4 / 5

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www.mathsrevision.com Higher Outcome 3Example 5 Solve sinx cos30 + cosx sin30 = -0.966 where 0 o < x < 360 o By rule 1a sinx cos30 + cosx sin30 =sin(x + 30) sin(x + 30) = -0.966 AS TC xoxo 180+x o 360-x o 180-x o Quad 3 and Quad 4 sin -1 0.966 = 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

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www.mathsrevision.com 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

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www.mathsrevision.com 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

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www.mathsrevision.com 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

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www.mathsrevision.com 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 = 1 + 1 + 2(sinAcosB - cosAsinB) = 2 + 2sin(A – B) = 2(1 + sin(A – B))= RHS Trig Identities Paper 1 type questions

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www.mathsrevision.com Higher Outcome 3 Double Angle Formulae

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

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www.mathsrevision.com Higher Outcome 3 Double Angle formulae

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www.mathsrevision.com Higher Outcome 3 Double Angle formulae

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www.mathsrevision.com Higher Outcome 3 Double Angle formulae

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 cos2x and sin x, so substitute 1-2sin 2 x Trigonometric Equations

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 Trigonometric Equations

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www.mathsrevision.com Higher Outcome 3 360 o -2 0 2 4 -4 Three problems concerning this graph follow. Trigonometric Equations

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www.mathsrevision.com Higher Outcome 3 360 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

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 From the previous problem we have: Hence Trigonometric Equations

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www.mathsrevision.com Higher Outcome 3 Radian Measurements Reminders i) Radians Converting between degrees and radians:

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www.mathsrevision.com Higher Outcome 3 ii) Exact Values 45 o right-angled triangle: 1 1 45 o Equilateral triangle: 1 2 2 1 60 o 30 o Degree Measurements

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www.mathsrevision.com Higher Outcome 3 degrees0o0o 30 o 45 o 60 o 90 o radians sin cos tan 0 0 1 0 1 0 1 Example: What is the exact value of sin 240 o ? Radians / Degrees

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www.mathsrevision.com Higher Outcome 3 Sine Graph Period = 360 o Amplitude = 1

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www.mathsrevision.com Higher Outcome 3 Cosine Graph Period = 360 o Amplitude = 1

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www.mathsrevision.com Higher Outcome 3 Tan Graph Period = 180 o Amplitude cannot be found for tan function

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 cos 3x is positive so solutions in the first and fourth quadrants x 3 Solving Trigonometric Equations

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www.mathsrevision.com Higher Outcome 3 300 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

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www.mathsrevision.com Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 315 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 112.5 o 157.5 o 172.5 o 585 o 675 o 945 o 1035 o 6t = 225 o

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www.mathsrevision.com Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

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www.mathsrevision.com 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.

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www.mathsrevision.com Higher Outcome 3 210 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

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www.mathsrevision.com Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 120 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

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www.mathsrevision.com Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

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www.mathsrevision.com 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

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www.mathsrevision.com Higher Outcome 3 160.5 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 160.5 o

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www.mathsrevision.com Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

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www.mathsrevision.com 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.

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www.mathsrevision.com Higher Outcome 3 306.9 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

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www.mathsrevision.com Higher Outcome 3 Graphical solution for Solving Trigonometric Equations

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Higher Maths Strategies www.maths4scotland.co.uk Click to start Compound Angles

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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.

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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

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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

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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

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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

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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 6 8 5 12 10 13 Mark up triangles

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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

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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

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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

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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 3 3 4 x Draw in perpendicular 2 h Use fact that sin x = sin ( ½ x + ½ x ) Expand sin ( ½ x + ½ x )

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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

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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

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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

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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 12 13 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

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

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Maths4Scotland Higher Quit Solving Equations Using Compound angle formula for Continue

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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

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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

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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

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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

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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

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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°

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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

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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

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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°

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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

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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°

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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

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

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

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

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www.mathsrevision.com 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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