REFERENCE LIBRARY
有教僑華頁一第張六第日八十月九年子壬歷
1973
朝日橋
24 OCT 1972
WAH KIU YAT PO
CITY HALL
二期星
日四廿十月十年二七九一股公年一十六國民華中
英文中學會考試題預習專欄
cos D
sin
堅道英文書院主編.
min
tan
tan 0
C
夏季橋業
i.e. 1+ cot-S
數學科
- cosec C
Example (1).
Given sec C
2m,
MATHEMATICS
find the remaining trigonomt- ical ratios in term of n.
3008
Lesson 1.
approaches U, LUFK approcher
and length of PQ approaches
that of QR while PR approaches 0
solution:
This course provides the can-
didates of H.K.E.E.
and it begins.
By Pythagoras theorem,
=(m2+1)2 -(2m)2
The angles
and 60°
zin QPR = 95
COS QER PB
sin 90°
Lesson 1
This course provides the cana- idates of H.K.5.E. 1973 a geneal revision on mathematics (only syll abus A) and it begins with trionon. etry
Consider a right angled tringl ABC where z ABC 90° and let 2 ABO
Then we have the following
ratios:
sin
now,cos -
adjacent side
ypoten
opposite side
hypotenuse
AC
consider the equilateral angl POR of side"a" and ew FM Perpendicular to QR
From geometry we claim that Mi
is the mid-point of QR and PM bisects 2QPR
/ QPM = 4RPM = 30°
By Pythagoras theorem,
All the above result are summarized by the following table
Angla
45° 6° 90°
O
72 112
Caufen o
NEVA //3.
12 112
133
Exapple
tan QPR = 3
tan 90 OR
adjacent side
hypotenuse
sin
opposite side.
hypotenuse
ta2=3/4 82
opposite side
adjacent side
cosec
PM
sin
cot
adjacent side
posite side
tan
opposite side
side
Now consider
PM Bin 60
hypotenus e.
cos 60°
sec
tan
adjacent side
BC
tan 60°
COSECO
ypotenuse opposite gide
Example (2)
Prove that
sin 10
tan + sect tan - secr+
COS JO°.
PM
√3/22
sin COB C
Solution L.H.S.
·Beca
Ian 30° ON 570 B
(3) The angle
Consider the igosceles
PQR with PRO-90 and PR
POR QPR 45
Find the value of
3tan 300 +4/3 cos
2000 45°
1/3 sin 60°
Solution: tan 300 v
cos 30 - 3/2
sec 45°-= √2
sin 60° = √3/2
the above expression.
=3(1/J3)2 + (√3)2 *(√2) - 1/30/2
=3x1/3+4/3x3/4-x2-x
RQ
-1 -2 = 3/4
It should be noted that:
(1)The above trigonometrical ratios
of an angle are numerical quah tities
(ii)The sine and cosine of an
angle can never be greater than unity; the secant and cosecnt of an angle can never be less than unity,while the tangent and cotangent are unlimited be- tween any values.
(iii) sin (=
Bing
cos.
sin
COS
sin
COS&
009. sinc
COS
toos.e
(multiplying both the numberato
and denominator by cos ()
sin +1 - Cost:
#Bosk(sin +1
cos ( sing+ COSE
tiplying both the and denominator by cos
tan
coso
cos (sing
-(1-911
numberator
Pythagoras theorem
PR
rem
"Again,,
,let us refer to the abve triangle, by Pythagoras thepre we have
CB +
cos (1+sing)= (1+sin() (1-sin()
cos (sin + cos
1+ sing
ACOB
R.H.S
PQ =
Dividing throughout by
AB
-(1)
Dividing (1) throughout by cos
sin Q Cos
COS
1.e. tan +
Sec
+
Dividing (1) throughout by sin
gonometric ratios of the anles 0,30°45°, 60°, 90°
| The most important trigonometrio
ratios are sine,cosine, tangents since the remaining three can be obtained from them by taking the corresponding reciprocals.
(1) The angle 0°
Consider PQR right angled at R and LPQR where is small.
As tends to 0,P and R tend to connoid i.e. FR tenās to 0,and length of PQ approaches that of QR.
PR sin
$11 U
009 45°--
tan 45
4) The angle 90°
Suppose sing
2+2mu +2n2
find the value of tan
2/ Jiven cos + Bin
COS
x
find that
soo
COS
cos
sin D
Suppose sin Q- msing and cos
ncos X,find the value of tan in terms of m and n
rove the following identities
Consider the triangle PQR riht angled at R, and PQR – S where
Ole small
(b) ta
Asino ces
Cesard - Coldsna
Cosse α + cető