Unit 3 Outcome 4

Unit 3 Outcome 4

Starter

Higher

Friday, April 21, 2023

Lesson 1

Q1Solution

Answer D

∫-2

0

x dx +

2

x0

dx∫

2

∫0

x dx2

The wave function and acos α + asin α

Unit 3 Outcome 4Higher



Example Write 4 cosx + 3sinx in the form K cos(x-α) 0 ≤ α ≥ 360

4 cosx + 3 sinx = K cos(x - α) 0 ≤ α ≥ 360

= K [cosx cosα + sinxsin α]

= K cosx cosα + K sinxsin α

= K cosα cosx + K sin α sinx4 cosx + 3 sinx

K cosα = 4K sin α = 3

C

AS

T

Cosα is pos

√

√sin α is pos

√√α is therefore in the 1st quadrant since cos and sin both positive

tan =cos αsin α

tan α = sin αcos α

K

K

34

α = 36.9

Find K K = 42 + 32

K = √ (42 + 32)

K = √ (16 + 9)

K = √ 25K = 5

Find α Reminder

tan α =

4 cosx + 3 sinx = K cos(x - α) 4 cosx + 3 sinx = 5 cos(x – 36.9) y = 5 cos (x – 36.9)




Write the following in the form K cos (x - α) 0 ≤ α ≥ 360

a) 6 cosx + 8 sinx

b) 5 cosx + 12 sinx

c) 8 cosx + 15 sinx

a) y = 10 cos(x – 53.13)

b) y = 13 cos(x – 67.38)

c) y = 17 cos(x – 61.927)



Wave Function

Page 253 Exercise 1A Q2, Q3, Q4


Extension Wave Function

Page 253 Exercise 1B Q1, Q2, Q4




Write the following in the form K cos (x - α) 0 ≤ α ≥ 360

Exam Standard Question

Maths4Scotland Higher

Hint

Expand ksin(x - a): sin( ) sin cos cos sink x a k x a k x a

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Equate coefficients: cos 2 sin 5k a k a

Square and add2 2 22 5 29k k

Dividing:

Put together: 2cos 5sin 29 sin 68x x x

5

2tan a acute 68a a is in 1st quadrant

(sin and cos are both +) 68a

Express in the form2sin 5cosx x sin( ) , 0 360 and 0k x k


Hint

Max for sine occurs ,2

(...)


Max value of sine function:

Max value of function:

The diagram shows an incomplete graph of

3sin , for 0 23

y x x

Find the coordinates of the maximum stationary point.

5

6x

Sine takes values between 1 and -1

3

Coordinates of max s.p. 5,

63


Hint

Expand kcos(x - a): cos( ) cos cos sin sink x a k x a k x a


Equate coefficients: cos 2 sin 3k a k a

Square and add2 2 22 3 13k k

Dividing:

Put together: 2cos 3sin 13 cos 56x x x

3

2tan a acute 56a a is in 1st quadrant

(sin and cos are both + )56a

( ) 2 cos 3sinf x x x a) Express f (x) in the form where andcos( ) 0 0 360k x k

for( ) 0.5 0 360f x x b) Hence solve algebraically

Solve equation. 13 cos 56 0.5x 0.5

13cos 56x

56 82acute x Cosine +, so 1st & 4th quadrants 138 334x or x


Hint

Use tan A = sin A / cos A

5

2tan x


Divide

acute 68x

Sine and cosine are both + in original equations

68x

Solve the simultaneous equations

where k > 0 and 0 x 360

sin 5

cos 2

k x

k x

Find acute angle

Determine quadrant(s)

Solution must be in 1st quadrant

State solution


Hint

Use Rcos(x - a): cos( ) cos cos sin sinR x a R x a R x a


Equate coefficients: cos 3 sin 2R a R a

Square and add 22 22 3 13R R

Dividing:

Put together: 2sin 3cos 13 cos 146x x x

2

3tan a acute 34a a is in 2nd quadrant

(sin + and cos - ) 146a

Solve equation. 13 cos 146 2.5x 2.5

13cos 146x

146 46acute x Cosine +, so 1st & 4th quadrants

or (out of range, so subtract 360°)192 460x x

Solve the equation in the interval 0 x 360. 2sin 3cos 2.5x x

or100 192x x


Return

30° 45° 60°

sin

cos

tan 1

6

4

3

1

2

1

23

2

3

2

1

21

21

3 3

Table of exact values

Unit 2 Outcome 4Circle

Higher


Intersection of a Line and a circle

Unit 3 Outcome 4

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