Carto Wong
Notes for HKCEE Mathematics
Trigonometry
Definition.
Suppose the point P(x, y) makes angle  with positive x-axis. Define
y

sin



r

x

 cos  
r

y

 tan   x

P(x, y)
r
y

x
x2  y 2 .
where r
Important relationship.
tan  
sin 
cos 
and
sin 2   cos 2   1 .
Special Values.

 in degree
sin 
cos 
tan 
0
0
1
0
30
1
2
3
2
1
45
2
2
2
2
1
60
3
2
1
2
3
90
1
0

3
You are suggested to memorize all these values!
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Carto Wong
The Graphs of Trigonometric Functions.
The Graph of Sine Function
- 360

o
- 180
o
180 o
T h eG r a p oh f C o s i nFeu n c t i o n
360 o
o
- 3 6 0
1 8 o0
3 6 o0
The functions sin  and cos are periodic with period 360 .
The Graph of Tangent
- 180

o
- 1 8 0
o
- 90
Function
o
90 o
180 o
The function tan  is periodic with period 180 .
General Solution for Trigonometric Equation.
Equation
General Solution
sin  k
  n  180  (1)n sin 1 k
cos  k
  n  360  cos 1 k
tan  k
  n  180  tan 1 k
Example.
The general solution for sin  
1
is   n  180  (1)n  30 (n is integer), i.e. the
2
solutions are
,  330,  210, 30, 150, 390, 510,
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Reduction Principle.
tan(  )   tan 
sin(  )   sin 
cos(  )  cos
sin(90   )  cos 
cos(90   )  sin 
tan(90   ) 
sin(90   )  cos 
cos(90   )   sin 
tan(90   )  
sin(180   )  sin 
cos(180   )   cos
tan(180   )   tan 
sin(180   )   sin 
cos(180   )   cos 
tan(180   )  tan 
sin(270   )   cos
cos(270   )   sin 
tan(270   ) 
sin(270   )   cos
cos(270   )  sin 
tan(270   )  
sin(360   )   sin 
cos(360   )  cos
tan(360   )   tan 
1
tan 
1
tan 
1
tan 
1
tan 
The figure as shown will help us to memorize the above properties.
S
A
T
C
A
Notations.
Given a triangle ABC, the length of BC, CA, AB are
denoted by a, b, c respectively as shown in the figure. The letter R
usually stands for circumradius（外接圓半徑）.
c
B
R
b
C
a
Area of Triangle.
area of ABC 

1
1
1
ab sin C  bc sin A  ca sin B .
2
2
2
A
Try to prove this formula yourself! A hint is given by the
figure on the right.
h
B
D
C
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Sine Law.
With the notations above, we have
a
b
c


sin A sin B sin C
( 2 R) .
Proof I.
Just note that
follows that
1
1
a
c
ab sin C  area of ABC  bc sin A  a sin C  c sin A 

. It
2
2
sin A sin C
a
b
c


by symmetry.
sin A sin B sin C
Proof II.
Let O be the circumcenter（外心）of ABC and BO meets the
circumcircle（外接圓）at D.
A
D
Note that BCD is a right angle triangle, so
sin A  sin D 
Hence,
O
B
BC
a

.
BD 2 R
C
a
b
c
 2 R and follows that

 2 R by symmetry.
sin A
sin B sin C
Applications of Sine Law.
Example 1. Given a triangle ABC with A  40 , B  80 , and CA = 4.33 cm. Find
the length of BC.
Solution.
By Sine Law,
a
b

sin A sin B
a
4.33

sin 40 sin 80
4.33
a
 sin 40
sin 80
a  2.83 cm (correct to 2 d.p.)

A
40
o
4.33 cm
80
B
o
C
This is typical case. In the examination, C is usually given instead of B . In this
case you can find B (by  sum of triangle) first and then apply Sine Law.
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Example 2. Given a triangle ABC with A  40 , BC = 2.83 cm, and CA = 4.33 cm.
Find B correct to nearest degree.
Solution.
BC
CA

sin A sin B
2.83
4.33

sin 40 sin B
4.33
sin B 
 sin 40
2.83
sin B  0.9835
Hence, B  80 or 100 (correct to nearest degree). The situations are shown in the
following figures.
A
40
A
40
o
o
4.33 c m
4.33 cm
B
TIP.
2.83 cm
C
B
2.83 c m
C
In order to get a more accurate answer, it is better to compute (by your calculator)
 4.33

sin 1 
 sin 40  but not sin 1 0.9835 .
 2.83

Cosine Law.
With the usual notations,
cos A 
b2  c2  a 2
2bc
cos B 
c2  a 2  b2
2ca
cos C 
a 2  b2  c2
2ab
Similarly,
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Applications of Cosine Law.
Example 3.
A triangle ABC is shown in the following figure. Find B .
A
4 cm
2 cm
3 cm
B
C
Solution.
22  32  42
2 23
1
cos B  
4
B  104.48 (correct to 2 d.p.)
cos B 
Example 4.
Find the length of CA in the following figure.
A
2.43 cm
60
B
o
3.72 cm
C
Solution.
2.432  3.722  CA2
2  2.43  3.72
2
CA  10.7037
cos 60 
CA  3.27 cm (correct to 2 d.p.)

Example 3 and example 4 are typical applications of Cosine Law. Roughly speaking,
these are the only two cases for which we use Cosine Law. All the questions in HKCEE,
except these 2 types, can be solved by Sine Law.
We see one more example.
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Carto Wong
Example 5.
Find the length of BC in the following figure.
A
2.43 cm
60
3.27 cm
o
B
40
o
C
Comment.
The easiest method is to find A first and then apply Sine Law. But I want
to introduce you a corollary of Cosine Law:
a  b cos C  c cos B .
Solution.
Using the formula above,
BC  3.27  cos 40  2.43  cos 60  3.72 cm (correct to 2 d.p.) .

The formula a  b cos C  c cos B is out-of-syllabus. It can be used in MC only.

In example 5, the given conditions are redundant（過多的）. If the length of AB (or CA)
is not given, the question is still solvable.
Let us end the discussion by
IQ Question.
Simplify
sin x
.
n
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