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Occasionally, it might be useful to do as the title here suggests.

Two examples that spring to mind include:

  • solving a\cdot\cos\theta\pm b\cdot\sin\theta=c for \theta (relative velocity example with - below)
  • maximising a\cdot\cos\theta\pm b\cdot\sin\theta without the use of calculus

a\cdot \cos\theta- b\cdot\sin\theta

Note first of all the similarity between:

\displaystyle a\cdot \cos\theta-b\cdot \sin \theta\sim \sin\phi\cos\theta-\cos\phi\sin\theta.

This identity is in the Department of Education formula booklet.

The only problem is that a and b are not necessarily sines and cosines respectively. Consider them, however, as opposites and adjacents to an angle in a right-angled-triangle as shown:

triangle

Using Pythagoras Theorem, the hypotenuse is \sqrt{a^2+b^2} and so if we multiply our expression by \displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} then we have something:

\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\cdot \left(a\cdot \cos\theta- b\cdot\sin\theta\right)

\displaystyle=\sqrt{a^2+b^2}\cdot \left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta-\frac{b}{\sqrt{a^2+b^2}}\sin\theta\right)

=\sqrt{a^2+b^2}\cdot \left(\sin\phi\cos\theta-\cos\phi\sin\theta\right)=\sqrt{a^2+b^2}\sin(\phi-\theta).

Similarly, we have

a\cdot\cos\theta+b\cdot \sin\theta=\sqrt{a^2+b^2}\sin(\phi+\theta),

where \displaystyle\sin\phi=\frac{a}{\sqrt{a^2+b^2}}.

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Quadratics are ubiquitous in mathematics. For the purposes of this piece a quadratic is a real-valued function q:\mathbb{R}\rightarrow \mathbb{R} of the form

q(x)=ax^2+bx+c,

where a,\,b,\,c\in \mathbb{R} such that a\neq 0. There is a little bit more to be said — particularly about the differences between a quadratic and a quadratic function but for those this piece is addressed to (third level: non-maths; all second level), the distinction is unimportant.

Geometry

The basic object we study is the square function, s:\mathbb{R}\rightarrow \mathbb{R}, x\mapsto x^2:

graph1

All quadratics look similar to x^2. If a>0 then the quadratic has this \bigcup geometry. Otherwise it looks like y=-x^2 and has \bigcap geometry

The geometry dictates that quadratics can have either zero, one or two real roots. A root of a function is an input x such that f(x)=0. As the graph of a function is of the form y=f(x), roots are such that y=f(x)=0\Rightarrow y=0, that is where the graph cuts the x-axis. With the geometry of quadratics they can cut the x-axis no times, once (like s(x)=x^2), or twice.

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Here are a number of additional exercises for those who never got them in class:

Class Quiz

Test 1 Sample

Test 1 A 

Test 1 B

The following are all of exam level difficulty:

Test 2 Sample

Test 2

Exam Worksheet 1

Exam Worksheet 2

Exam Worksheet 3

Updated on May 24th 2011.

The results are down the bottom. You are identified by the last three digits of your student number. If there is a zero is means that you did not write down your student number. The results are in alphabetical order but if you are unsure email me and I will tell you what you got.

The scores are itemized as you can see. At the bottom there are some average scores. Finally the last column displays your Continuous Assessment mark for Test 2 (out of 15).

If you would like to see your paper or have it discussed please email me at jippo@campus.ie

St No

One

Two

Three

Four

Five

Six

/40

%

CA/15

390

10

5

10

3

5

2

35

88

13.2

000

10

5

10

3

3

4

35

88

13.2

928

7

4

9

5

5

5

35

88

13.2

136

10

5

10

2

0

4

31

78

11.7

669

10

5

8

3

1

0

27

68

10.2

784

9

5

10

0

2

0

26

65

9.75

051

10

5

5

3

1

2

26

65

9.75

817

9

5

5

3

0

2

24

60

9

417

9

5

4

3

2

1

24

60

9

933

9

2

10

0

0

0

21

53

7.95

175

5

5

10

0

0

0

20

50

7.5

465

6

2

9

0

0

0

17

43

6.45

000

9

2

2

0

0

0

13

33

4.95

917

0

2

4

0

0

0

6

15

2.25

812

3

0

3

0

0

0

6

15

2.25

828

2

1

5

0

0

0

8

20

3

163

5

4

5

3

0

4

21

53

7.95

953

3

2

0

0

0

0

5

13

1.95

513

2

0

2

0

0

0

4

10

1.5

000

2

0

0

0

0

0

2

5

0.75

Ave

6.5

3.2

6.05

1.4

0.95

1

19

49

7.275

Ave %

65

64

60.5

28

19

24

The MATH6014 Test 2 will be held at 5 p.m. Tuesday 03/05/11. The test is worth 15% of your final mark. The test will be 50 minutes long and you must answer all questions. The marks each question carries will be stated on the test and I will attach a set of tables.

Sample to be found here.

The results are down the bottom. You are identified by the last three digits of your student number. If there is a zero is means that you did not write down your student number. The results are in alphabetical order but if you are unsure email me and I will tell you what you got.

The scores are itemized as you can see. At the bottom there are some average scores. Finally the last column displays your Continuous Assessment mark for Test 1 (out of 15).

If you would like to see your paper or have it discussed please email me at jippo@campus.ie

St No 1a 1b 2a 2b 3a 3b % CA
784 5 1 15 3 20 20 64 10
136 15 4 15 15 20 18 87 14
933 10 0 1 15 20 17 63 10
0 0 0 0 0 20 0 20 3
390 13 1 3 8 20 20 65 10
0 12 4 7 6 20 20 69 11
828 0 0 0 0 20 0 20 3
917 0 0 0 0 20 4 24 4
051 8 4 7 3 20 17 59 9
918 0 0 0 1 20 14 35 6
465 2 0 2 3 20 13 40 6
817 8 0 3 2 10 12 35 6
162 0 0 3 0 20 19 42 7
417 15 0 7 12 10 15 59 9
0 0 0 2 0 20 12 34 6
953 2 0 1 0 5 3 11 2
175 0 0 0 0 20 4 24 4
498 0 0 1 0 10 6 17 3
812 6 1 0 0 20 14 41 7
513 5 0 3 0 20 11 39 6
669 4 10 10 15 20 20 79 12
928 9 5 6 0 20 18 58 9
440 0 0 1 0 20 14 35 6
092 3 0 1 0 5 5 14 3
163 4 0 0 8 20 12 44 7
Ave 4.84 1.20 3.52 3.64 17.60 12.32 43.12 6.92
Ave % 32.27 8.00 23.47 24.27 88.00 61.60

The first MATH6014 test will be held at 5 p.m. today. The test is worth 15% of your final mark. The test will be 50 minutes long and you must answer all questions. Question 3 is worth 40 marks; Questions 1 and 2 30 marks each. I will put the formula for the roots of a quadratic equation on the paper. Please find a sample here.

Module Description:

MATH6014

More detailed General Information on this module (* not all one hundred percent accurate at time of publication) may be found after the table of contents in this set of incomplete notes:

MATH6014 Lecture Notes

(last updated 04 May)

February 16

Additional exercises have been added to the section on Basic Algebra (go to the link to the notes above).

Additional exercises are also to be found in the suggested reading.

Past exam papers also comprise additional exercises.

I will strive to include more exercises in future — and will have more algebra exercises in time. I will furnish ye with equation exercises on Friday.

An exam paper from 2007/08:

Technological Mathematics 1

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