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MAT01A1
Appendix D: Trigonometry
Dr Craig
12 February 2019
Introduction
Who:
Dr Craig
What:
Lecturer & course coordinator for MAT01A1
Where:
C-Ring 508 [email protected]
Web:
http://andrewcraigmaths.wordpress.com
Important information
Course code: MAT01A1
NOT: MAT1A1E, MAT1A3E, MATE0A1,
MAEB0A1, MAA00A1, MAT00A1,
MAFT0A1
Learning Guide: available on Blackboard.
Please check Blackboard twice a week.
Student email: check this email account
twice per week or set up forwarding to an
address that you check frequently.
Important information
Lecture times: Tuesday 08h50 – 10h25
Wednesdays 17h10 – 18h45
Lecture venues: C-LES 102, C-LES 103
Tutorials: Tuesday afternoons
13h50 – 15h25: D-LES 104 or B-LES 102
OR
15h30 – 17h05: C-LES 203
Other announcements
I No tuts for MAT01A1 on Wednesdays. If
you see this on your timetable, it is an
error. (To move your Chem. prac., email
Mr Kgatshe [email protected])
I CSC02A2 students. Email Dr Craig
regarding tutorial clash.
I Maths Learning Centre in C-Ring 512:
10h30 – 15h25 Mondays
08h00 – 15h25 Tuesday to Thursday
08h00 – 12h05 Fridays
Lecturers’ Consultation Hours
Monday:
14h40 – 15h25 Dr Craig (C-508)
Tuesday:
11h20 – 13h45 Dr Robinson (C-514)
Wednesday:
15h30 – 17h05 Dr Robinson (C-514)
Thursday:
11h20 – 12h55 Dr Craig (C-508)
Friday:
11h20 – 12h55 Dr Craig (C-508)
Warm up
Let a > 0. Then
4. |f (x)| = a if and only if f (x) = a or
f (x) = −a.5. |f (x)| < a if and only if −a < f (x) < a.
6. |f (x)| > a if and only if f (x) > a or
f (x) < −a.Now solve:
1 < |x + 1| < 4
The solution is the values of x that satisfy:
1 < |x + 1| AND |x + 1| < 4
Therefore x ∈ (−5,−2) ∪ (0, 3). This
algebraic solution agrees with the sketch:
The triangle inequality
If a, b ∈ R, then |a + b| 6 |a| + |b|.
How do we prove this?
First observe that −|a| 6 a and a 6 |a|.
Similarly, −|b| 6 b and b 6 |b|.
Applications:
Example: If |x− 4| < 0.1 and |y − 7| < 0.2,
use the Triangle Inequality to estimate
|(x + y)− 11|.
Exercise: show that if |x + 3| < 12, then
|4x + 13| < 3.
How big is a degree?
If you didn’t have a protractor, how would
you draw an angle of size 30◦?
You could do it using the 1, 2,√3 special
triangle. But what about 32◦?
Would it be possible to draw this angle with
only a ruler, compass, and a piece of string?
Radian measure
From now on in your mathematical life,
angles will be measured in radians.
Radians measure the ratio between the arc
length and the radius. When a =arc length,
we have:
θ =a
rand a = rθ.
Note: these formulas are only valid when θ is
measured in radians.
Example: 90◦ = π/2 radians.
How do we get this?
Consider any circle. Let the radius of the
circle be r. The circumference of the circle is
equal to 2πr. The arc length of 90◦ is a
quarter of the total circumference, so
a =2πr
4=πr
2.
∴ 90◦ =πr
2× 1
r=π
2radians.
Using the fact that 360◦ will be 2π radians,
we can use the following formulas to convert
between degrees and radians:
1 radian =180
π1◦ =
π
180radians.
Example: convert
(a) −72◦ to radians,
(b) 5π/2 to degrees.
More examples:
(a) If the radius of a circle is 5cm, what
angle is subtended by an arc of
6cm?
(b) If a circle has radius 3cm, what is
the length of an arc subtended by a
central angle of 3π/8?
30◦ =π
645◦ =
π
460◦ =
π
3
90◦ =π
2120◦ =
2π
3135◦ =
3π
4
180◦ = π 270◦ =3π
2360◦ = 2π
Trigonometric functions
Trig functions take as input an angle
(measured in radians) and output the ratio
between two distances.
θ
hypotenuse
adjacent
opposite
sin θ =opp
hypcos θ =
adj
hyptan θ =
opp
adj
csc θ =hyp
oppsec θ =
hyp
adjcot θ =
adj
opp
More generally, angles can be measured in a
coordinate system:
θ > 0
θ < 0
A positive and negative angle drawn in the
standard position.
Trig functions in a coordinate system
P (x, y)
rθ
sin θ =y
rcos θ =
x
rtan θ =
y
x
csc θ =r
ysec θ =
r
xcot θ =
x
y
Signs of trig functions
AS
T C
All > 0sin θ > 0
tan θ > 0 cos θ > 0
Special angles
π/4
π/4
1
1
√2
π6
π/3
√3
1
2
Exercise: calculate all of the trig ratios for
θ = 5π/3.
Graphs of trig functions: f (x) = sinx
Note: −1 6 sinx 6 1.
Question: why is sin(3π/2) = −1? Think of
how 3π/2 is sketched in xy-plane.
Graph of f (x) = cos(x). Again, | cosx| 6 1.
Note: cos 0 = 1. Also sin(x + 2π) = sinx
and cos(x + 2π) = cosx.
Now that we are using radians we should
label our axes using radian intervals like π/2
or maybe π/4, depending what is most
suitable.
Graph of f(x) = tan(x).
Note the vertical asymptotes at π/2 and −π/2.
Why does tan(x) get very big as x
approaches π2 from below?
Graph of f (x) = csc(x).
Note: | csc(x)| > 1, x /∈ { z.π | z ∈ Z }.
Graph of f (x) = csc(x) and g(x) = sin(x).
Note: | csc(x)| > 1, x /∈ { z.π | z ∈ Z }.
Graph of f (x) = sec(x).
Graph of f (x) = cot(x).
Note: cot(x) = 0 wherever tan(x) has an
asymptote.
Graph of f (x) = cot(x) and g(x) = tan(x).
Note: cot(x) = 0 wherever tan(x) has an
asymptote.
Trig identities
Trig identities are useful relationships
between trig functions. Some of the basic
identities are:
csc θ =1
sin θsec θ =
1
cos θ
tan θ =sin θ
cos θcot θ =
1
tan θ=
cos θ
sin θ
More trig identities
sin2 θ + cos2 θ = 1
Divide both sides by cos2 θ to get
sin2 θ
cos2 θ+
cos2 θ
cos2 θ=
1
cos2 θ
∴ tan2 θ + 1 = sec2 θ
Or, divide both sides of the original by sin2 θ:
sin2 θ
sin2 θ+
cos2 θ
sin2 θ=
1
sin2 θ
∴ 1 + cot2 θ = csc2 θ
Example
sin2 θ + cos2 θ = 1
Prove the following trig identities:
cot2 θ + sec2 θ = tan2 θ + csc2 θ
tan2 α− sin2 α = tan2 α sin2 α
Addition formulas
sin(x + y) = sinx. cos y + cosx. sin y
cos(x + y) = cosx. cos y − sinx. sin y
How can we use
sin(x + y) = sinx. cos y + cosx. sin y
to get a formula for
sin(x− y)?Result:
sin(x− y) = sinx. cos y − cosx. sin y
We can also use the addition formulas to get
the double-angle formulas:
sin(2x) = 2 sinx. cosx
cos(2x) = cos2 x− sin2 x
cos(2x) = 2 cos2 x− 1
cos(2x) = 1− 2 sin2 x
Example: Find all of the values of x in the
interval [0, 2π] such that sinx = sin 2x.
