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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 !!.

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Higher Unit 2

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## Higher Unit 2

Trigonometry identities of the form sin(A+B)

Double Angle formulae

Trigonometric Equations

More Trigonometric Equations

Exam Type Questions

www.mathsrevision.com

## 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

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

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 !!

## 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)

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

## 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

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

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

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

sin-13/2 = /3

Repeats every 

Quad 2: angle =  - /3

In this example repeats lie out with limits

2 θ = /3

2 θ = 2/3

θ = /3

θ = /6

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

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

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

## 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.

## 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 = 2sinAcosAwhen replacing sin2Aequation

cos2A = 2cos2A – 1 if cosA is also in the equation

cos2A = 1 – 2sin2Aif sinA is also in the equation

Trigonometric Equations

cos2x and sin x,

so substitute 1-2sin2x

90o

A

S

180o

0o

C

T

270o

Trigonometric Equations

cos 2x and cos x,

so substitute 2cos2-1

Trigonometric Equations

4

2

360o

0

-2

-4

Trigonometric Equations

Three problems concerning this graph follow.

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.

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

Reminders

30o

2

2

1

60o

45o

1

1

1

Degree Measurements

Equilateral triangle:

ii) Exact Values

45o right-angled triangle:

0

0

1

1

0

0

1

What is the exact value of sin 240o ?

Example:

Sine Graph

Period = 360o

Amplitude = 1

Cosine Graph

Period = 360o

Amplitude = 1

Tan Graph

Period = 180o

Amplitude cannot be found for tan function

90o

A

S

180o

0o

C

T

270o

Solving Trigonometric Equations

Example:

Step 2: consider what solutions

are expected

Step 1: Re-Arrange

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

3x = 60o

420o 660o 780o 1020o

300o

x = 20o

100o

140o

220o

260o

340o

Solving Trigonometric Equations

Graphical solution for

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

Solving Trigonometric Equations

Step 3: Solve the equation

sin wave repeats every 360o

6t = 225o

585o 675o 945o 1035o

315o

x = 39.1o

52.5o

97.5o

112.5o

157.5o

172.5o

Solving Trigonometric Equations

Graphical solution for

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

Solving Trigonometric Equations

Step 3: Solve the equation

sin wave repeats every 360o

2x = 90o

450o570o

210o

x = 45o

105o

225o

285o

Solving Trigonometric Equations

Graphical solution for

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

2 solutions

Solving Trigonometric Equations

Step 3: Solve the equation

tan wave repeats every 180o

x = 60o

240o300o

120o

Solving Trigonometric Equations

Graphical solution for

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

Solving Trigonometric Equations

Step 3: Solve the equation

Two solutions

One solution

90o

x= 19.5o

160.5o

Overall solution x = 19.5o , 90o and 160.5o

Solving Trigonometric Equations

Graphical solution for

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

Solving Trigonometric Equations

Step 3: Solve the equation

Two solutions

One solution

180o

x= 53.1o

306.9o

Overall solution in radians x = 0.93, πand 5.35

Solving Trigonometric Equations

Graphical solution for

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Higher Maths

Strategies

Compound Angles

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Compound Angles

Non-calculator questions will be indicated

You will need a pencil, paper, ruler and rubber.

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This presentation is split into two parts

Using Compound angle formula for

Exact values

Solving equations

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4

8

Hint

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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

Pythagoras

Write down values for cos p and sin p

Expand sin (2p)

Expand cos (2p)

Use m = tan (2p)

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Hint

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

Simplify

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Hint

Maths4Scotland Higher

Using triangle PQR, as shown, find the

exact value of cos 2x

Use Pythagoras

Write down values for

cosxandsinx

Expand cos 2x

Substitute values

Simplify

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10

8

6

12

5

13

Hint

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).

Mark up triangles

Use Pythagoras

Write down values for

sin p, cos p, sin q, cos q

Expand sin (p + q)

Substitute values

Simplify

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A and B are acute angles such that and .

Find the exact value of

a) b) c)

13

5

B

A

3

5

12

4

Hint

Maths4Scotland Higher

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)

Substitute

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x

4

3

Hint

Maths4Scotland Higher

If x° is an acute angle such that

show that the exact value of

5

Draw triangle

Use Pythagoras

Hypotenuse is 5

Write down sin x and cos x

Expand sin (x + 30)

Substitute

Simplify

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Table of exact values

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

Hint

Maths4Scotland Higher

Use Pythagoras

Write down

sin x, cos x, sin y, cos y.

Expand cos (x + y)

Substitute

Simplify

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x

3

3

h

2

4

Hint

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

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

Simplify

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Table of exact values

a

3

Hint

Maths4Scotland Higher

Given that

find the exact value of

Draw triangle

Use Pythagoras

Write down values for

cosaandsina

Expand sin 2a

Substitute values

Simplify

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Hint

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

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Table of exact values

5

q

4

Hint

Maths4Scotland Higher

If find the exact value of

a)b)

3

Opposite side = 3

Draw triangle

Use Pythagoras

Write down values for

cosqandsinq

Expand sin 2q

Expand sin 4q (4q= 2q + 2q)

Expand cos 2q

Find sin 4q

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13

5

3

12

P

Q

5

4

Hint

Maths4Scotland Higher

For acute angles P and Q

Show that the exact value of

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

Simplify

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Using Compound angle formula for

Solving Equations

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Solve the equation for 0 ≤ x ≤  correct to 2 decimal places

S

A

T

C

Hint

Maths4Scotland Higher

Replace cos 2x with

Substitute

Simplify

Factorise

Hence

Find acute x

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S

A

T

C

Hint

Maths4Scotland Higher

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

Solve simultaneously

Rearrange

Check range

Find acute 2x

Deduce 2x

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Table of exact values

S

A

T

C

Hint

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

Form equation

Determine

Replace sin 2x

Rearrange

Common factor

Hence

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Table of exact values

S

A

T

C

Hint

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

Simplifies to

1st expression

Rearrange:

2nd expression

acute x

Simplify 1st expr.

Use exact values

Determine

Similarly for 2nd expr.

Form Eqn.

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Table of exact values

S

A

T

C

Hint

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

Coords, P

for sin x

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Table of exact values

S

S

A

A

T

T

C

C

Hint

Maths4Scotland Higher

Solve the equation for 0 ≤ x ≤ 360°

Replace cos 2x with

Substitute

Simplify

Factorise

Hence

Find acute x

Solutions are: x= 60°, 132°, 228° and 300°

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Table of exact values

S

A

T

C

Hint

Maths4Scotland Higher

Solve the equation for 0 ≤ x ≤ 2

Rearrange

Find acute x

Note range

Solutions are:

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Table of exact values

Hint

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

Try to solve:

Replace cos 2q with

Rearrange

No solution

Divide by 2

Hence there are no real solutions for q

Factorise

Equal roots for cos q

Deduction

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S

A

T

C

Hint

Maths4Scotland Higher

Solve algebraically, the equation sin 2x + sin x = 0, 0  x  360

for cos x

Replace sin 2x

Common factor

Hence

Determine x

x= 0°, 120°, 240°, 360°

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Table of exact values

S

A

T

C

Hint

Maths4Scotland Higher

Find the exact solutions of 4sin2x = 1, 0  x  2

Rearrange

Take square roots

Find acute x

+ and – from the square root requires all 4 quadrants

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Table of exact values

S

A

T

C

Hint

Maths4Scotland Higher

Solve the equation for 0 ≤ x ≤ 360°

Replace cos 2x with

Substitute

Simplify

Factorise

Hence

Find acute x

Solutions are: x= 60°, 180° and 300°

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Table of exact values

S

A

T

C

Hint

Maths4Scotland Higher

Solve algebraically, the equation for 0 ≤ x ≤ 360°

Replace cos 2x with

Substitute

Simplify

Factorise

Hence

Find acute x

Solutions are: x= 60° and 300°

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Table of exact values

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Table of exact values

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