# Higher Unit 2 Trigonometry identities of the form sin(A+B)

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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 !!
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!!

Trig Identities (1) Expand cos(U – V). (use formula 2b )
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 By taking A = 60° and B = 30°,
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 ) + (3/2 X ½) = 3/4 + 3/4 = 3/2 Hence LHS = RHS !!

Trig Identities Prove that sin15° = ¼(6 - 2) sin15° = sin(45 – 30)°
Example 3 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2) = ( ) 22 X  2 = (6 - 2) 4 = ¼(6 - 2)

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

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

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

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

Trig Identities Find the value of x that minimises the expression
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

Trig Identities Paper 1 type questions
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 x cos θ + 1 X sin θ = sin θ

(sinA + cosB)2 + (cosA - sinB)2 = 2(1 + sin(A - B))
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 = (sinAcosB - cosAsinB) = sin(A – B) = 2(1 + sin(A – B)) = RHS

Double Angle Formulae

Double Angle formulae Similarly: 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.

Double Angle formulae

Double Angle formulae

Double Angle formulae

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

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

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

Trigonometric Equations

Trigonometric Equations
-2 2 4 -4 Three problems concerning this graph follow.

Trigonometric Equations
The max & min values of sinbx are 1 and -1 resp. The max & min values of asinbx are 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.

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:

Trigonometric Equations
From the previous problem we have: Hence

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

Radians / Degrees degrees 0o 30o 45o 60o 90o radians sin cos tan 1 1 1
1 1 1 Example: What is the exact value of sin 240o ?

Sine Graph Period = 360o Amplitude = 1

Cosine Graph Period = 360o Amplitude = 1

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

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

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

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

Solving Trigonometric Equations
Graphical solution for

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

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

Solving Trigonometric Equations
Graphical solution for

Solving Trigonometric Equations
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 C A S T 0o 180o 270o 90o (2x – 60o ) = sin-1(1/2) x 2 x 2

Solving Trigonometric Equations
Step 3: Solve the equation 1st quad 2nd quad sin wave repeats every 360o 2x = 90o 210o 450o 570o x = 45o 105o 225o 285o

Solving Trigonometric Equations
Graphical solution for

Solving Trigonometric Equations
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 C A S T 0o 180o 270o 90o 2 solutions 1st and 3rd quads 2 solutions 2nd and 4th quads

Solving Trigonometric Equations
Step 3: Solve the equation 1st quad 2nd quad tan wave repeats every 180o x = 60o 120o 240o 300o

Solving Trigonometric Equations
Graphical solution for

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

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

Solving Trigonometric Equations
Graphical solution for

Solving Trigonometric Equations
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 C A S T 0o 180o 270o 90o Remember this ! Two solutions One solution

Solving Trigonometric Equations
Step 3: Solve the equation Two solutions One solution 1stquad 3rd quad 180o x = 53.1o 306.9o Overall solution in radians x = 0.93 , π and 5.35

Solving Trigonometric Equations
Graphical solution for

Higher Maths Compound Angles Strategies Click to start
Higher Maths Strategies Compound Angles Click to start

Compound Angles The following questions are on
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

This presentation is split into two parts
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

Write down values for cos p and sin p
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. 8 4 p Draw triangle Pythagoras Write down values for cos p and sin p Expand sin (2p) Expand cos (2p) Hint Use m = tan (2p) Previous Quit Quit Next

In triangle ABC show that the exact value of
Maths4Scotland Higher 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 Hint Simplify Previous Quit Quit Next

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

On the co-ordinate diagram shown, A is the point (6, 8) and
Maths4Scotland Higher 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). 10 13 6 8 5 12 Mark up triangles Use Pythagoras Write down values for sin p, cos p, sin q, cos q Expand sin (p + q) Substitute values Hint Simplify Previous Quit Quit Next

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

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

The diagram shows two right angled triangles
Maths4Scotland Higher 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 Use Pythagoras Write down sin x, cos x, sin y, cos y. Expand cos (x + y) Substitute Hint Simplify Previous Quit Quit Next

The framework of a child’s swing has dimensions
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 Draw in perpendicular Use Pythagoras 3 4 x 2 h Use fact that sin x = sin ( ½ x + ½ x) Write down sin ½ x and cos ½ x Expand sin ( ½ x + ½ x) Substitute Hint Simplify Previous Quit Quit Next Table of exact values

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

Find algebraically the exact value of
Maths4Scotland Higher Find algebraically the exact value of Expand sin (q +120) Expand cos (q +150) Use table of exact values Combine and substitute Simplify Hint Previous Quit Quit Next Table of exact values

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

Show that the exact value of
Maths4Scotland Higher For acute angles P and Q Show that the exact value of 12 13 P 5 3 Q 5 4 Draw triangles Use Pythagoras Adjacent sides are 5 and 4 respectively Write down sin P, cos P, sin Q, cos Q Expand sin (P + Q) Substitute Hint Simplify Previous Quit Quit Next

You have completed all 12 questions in this section
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Using Compound angle formula for
Maths4Scotland Higher Using Compound angle formula for Solving Equations Continue Quit Quit

Solve the equation for 0 ≤ x ≤  correct to 2 decimal places
Maths4Scotland Higher Solve the equation for 0 ≤ x ≤  correct to 2 decimal places Replace cos 2x with Determine quadrants Substitute A S C T Simplify Factorise Hence Discard Hint Find acute x Previous Quit Quit Next

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 Determine quadrants Solve simultaneously A S C T Rearrange Check range Find acute 2x Hint Deduce 2x Previous Quit Quit Next Table of exact values

a) Find expressions for: i) f(g(x)) ii) g(f(x))
Maths4Scotland Higher Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x 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° Determine x 1st expression 2nd expression A S C T Determine quadrants Form equation Replace sin 2x Rearrange Hint Common factor Hence Previous Quit Quit Next Table of exact values

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

Solutions for where graphs cross
Maths4Scotland Higher a) 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 Solutions for where graphs cross Common factor Hence By inspection (P) Determine x Find y value A S C T Coords, P Determine quadrants for sin x Hint Previous Quit Quit Next Table of exact values

Solutions are: x= 60°, 132°, 228° and 300°
Maths4Scotland Higher Solve the equation for 0 ≤ x ≤ 360° Replace cos 2x with Determine quadrants Substitute Simplify A S C T A S C T Factorise Hence Find acute x Hint Solutions are: x= 60°, 132°, 228° and 300° Previous Quit Quit Next Table of exact values

Solutions are: Maths4Scotland Higher Solve the equation for 0 ≤ x ≤ 2
Rearrange Find acute x Note range A S C T Determine quadrants Solutions are: Hint Previous Quit Quit Next Table of exact values

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

Solve algebraically, the equation sin 2x + sin x = 0, 0  x  360
Maths4Scotland Higher Solve algebraically, the equation sin 2x + sin x = 0, 0  x  360 Replace sin 2x Determine quadrants for cos x Common factor A S C T Hence Determine x Hint x = 0°, 120°, 240°, 360° Previous Quit Quit Next Table of exact values

Find the exact solutions of 4sin2 x = 1, 0  x  2
Maths4Scotland Higher Find the exact solutions of 4sin2 x = 1, 0  x  2 Rearrange Take square roots Find acute x Determine quadrants for sin x + and – from the square root requires all 4 quadrants A S C T Hint Previous Quit Quit Next Table of exact values

Solutions are: x= 60°, 180° and 300°
Maths4Scotland Higher Solve the equation for 0 ≤ x ≤ 360° Replace cos 2x with Determine quadrants Substitute Simplify A S C T Factorise Hence Find acute x Hint Solutions are: x= 60°, 180° and 300° Previous Quit Quit Next Table of exact values

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

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30° 45° 60° sin cos tan 1 Table of exact values Maths4Scotland Higher
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Make sure you complete and correct ALL of the Trigonometry questions in the past paper booklet.