Precalculus: ตรีโกณมิติ & จำนวนเชิงซ้อน

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Senior High 3

**ผิดพลาดประการใด ขออภัยด้วยนะคะ

2025.11.04 公開

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ページ1:

Precalculus
sahe sizes
Trigonometry
terminal side
coterminal
kez
elevation
positive
angle
Initial side
Σ <8<*
negative
2
0<<1/1
depression
angle
obtuse
acute
radian
0.5
180° = 2 rad
T<< 3x
<< 2x
α+ß
z
complementary
sin (2h+0) sine
x+p
= π
supplementary
cos (2n+ 0) = cose
Dsin
R, Rsin = [1,1]
tan (+0)
- tane
Dcos
R, Roos-1,1]
odd function
sin (-)-sine
nez, 0GR
even function
reference angle
acute part of coterminal
compare to horizontal
Hypotenuse
Opposite
adjacent
sin (90-0) cose
tan (90-0) cote
sec (90-0) = cosec
1 + tan 20
1+ cot² 6
= coseco
cos (-) cos
tan (-)-tane
cosec (-) cosec
sec (-6) Sec
cot (-e) = -cote
Sin 20+ cos² = 1
=
Sec
Graph
y=
ydasin (bx-c) symmetric with respect to the origin
d+acos (bx-c) symmetric with respect to the y-axis
amplitude lal
half of distance between
minimum and maximum
ax m
a><
→ flip m
010→
1. consider only 1 cycle
bx-c 20
bx-C = 2
2. divide period into 4 intervals
Maximum, minimum, intercept +2
3. consider vertical translation
b>0
xb>₁ mm
period 2
|bl
bas m
b<a → sin (bx) = -sin(-bx)
cos (bx) = cos (bx)
101-
[b]
sin
↑d
Chorizontal shift
phase shift
C>O
COS
C=O
←
-101-
-d vertical translation
do upward
deo downward
banx
sin X COSX 70
COS X
cot x = cos x
Sin X
sinx po
1. consider 1 cycle
x-(2x-1); n€ Z
divide into each interval
2
tan(-x) = - tanx, cot(-x) = cotx
nez
ว
\+
devide into each intervals
3. vertical asymptotes
tan
a
bx-c = -
bx-c =
bx-c = 0
cot
I
bx-c = π
4. devide into 4 intervals
-amplitude, intercept, + amplitude
5. consider vertical translation

ページ2:

COS X
Sec x =
X
1
x = (2n-1), nez
COSX 70
CSC X
1
z
sin x #0
,
'
sin X
1. Consider 1 cycle
x = ht, nez
2. sec generate graph cos
period
bx-c=0
bx-c=2
Danped geometric graph
a
csc generate graph sin
3. vertical asymptotes
intercept of sin-cos graph
4. consider vertical translation
y=xcosx
4R
-1 sin X 1
f(x)
= Xsinx between lines y=x
y
xsinx
-|x| xsin x|x|
1. sketch y = f(x), y = − f(x)
y=-x
2
2 sketch
y sin (bx)
y=0, x = x
y= cos(bx)
y= f(x), x=+hy f(x), xxnx
Inverse geometric graph
One to one function
cos [0,] [1,1]
pass
horizontal & vertical line test
Consider only 1 cycle
arccos [1,1][0,1]
=cus > x = arccosy
Sin
A
←+
-#
minimum
arcsin [11]]
y = sin x
,
x = arcsiny
arcsin(sinx) xe,
arccos (cosx) x [0]
arctan (tanx) xe ()
Application
-
declare variables
bearing
a.
ton (-)-
→ R
a
tan (arccosx)
arcban R→
(14)
y=tonx, x=arctany
1. let arccos x = a
Cosa X
06 [0,1]
2. sketch triangle to find the missing side
* result in radian
acute angle, degree
measure from y-axis (N,S)
N 30°E
arcsin (x)+ arcsin(x) = π
arcsin (-x)=-1
- arcsin(x)
arcsin (x) + arccos(x) =

ページ3:

COS (A+B)
z
Cos Acos sin Asin B
Cos (A-B) COSA cos B + sinAsin B
Sin (A+B) sinAcos B+cos Asin
z
sin(A-B) sin Acos B-cos Asin B
z
tan A+tan B
tan (A+B)
fan (A-B)
tan A-tan
Sin (2A)
Cos(2A)
2sin AcosA
Cos²A-sin²A
2tan A
tan (2A) = 1-tan² A
sin A - 1-cos 2A
2
sin()
Z
+1-COSA
2
Cos()=1+COSA
tan()=sinA
1+COSA
1-tan Atan B
1+fanAtan B
sin A+ sinB 2sin(A+B) Cos/A-B)
sin A-sin B =
COSA +CosB
sin (A+B) cos (A-5)
2
2 cos A+B sin
2
(18) sin (A-B)
2
=2cos A+B cos A-B
2
| Cos (A-B)
2
sin
2
COSA-COSB = -2sin(A+B) sin (A-B`
Complex number
Z = a +
a+bi
✓
Re(z) Im(Z)
Complex plane
imaginary
axis
real axis
2
Cos²A
z
1+COS2A
2
tan² A = 1-COS2A
N
Law of sine
1+cos2A
sin Asin B
N
(cos (A-B) - COS(A+B))
COSA COSB = (cos(A-B) + cos(A+B))
Sin AcosB = 1 (sin (A+B) + sin(A-B))
Z
cos Asing (sin (A+B)-Sin (A-15)
a
Sin A sinB
b C
sinC
Heron's formula
S =
a+b+c
Law of Cosine
Ca2+62-2abcosC
a²= b²+c²-2bccos A
b2 = a²+c² 2accosB
Area-s(s-as-b)(s-c)
Z+2, (a+c)+(b+d) i
i; n = 1 mod 4
i1; n = 2 mod 4
-1; h = 3 mod 4.
1 ; n = 0 mod 4
Re(Z) = Re(Z₂)
Z₁ =Z2
Im(z) = Im(22)
ZZ =
= |2/2
ZW
2,2₂ = (ac-bd)+(ad+bc)i
z
-2 = (-a)+(-b)i
a
b
α²+b²
a²+ b²
Z+W=W+Z
ZW-WZ
Z+(W+V) = (z+W) +V Z(WV)-(ZW)V
Z+0 = Z
2+(-2)=0
complex conjugate = a-bi
Modulus z=a+b²
arg (2)
A
12-2
22-1
Re(z) = 2+2
Im (z) = z-z
Z - Z
Z₁ = Z1
Z+W = Z+W
ZW - ZW
Z-W = Z-W
Z/W=Z/W
Z z ne Nu{0}
Z
|W|2
ZZ = |z²
2=-2=2
2-21
|2w| = |2||W|
z0
12=0Z=0
Iz-wl distance between Z, W
Z+W≤2+W
M-Z|||M|-|Z
Zi = -1+i
Z=1+i
IM(z)
IM (2)
1
Z₁₂ =
a-rcose
b-
z=a+bi
b-
-λ== a+bi
ZZ,
r₂ (cos(0,+2)+ isin(0,₁+0₂))
(cos (0,0)+ isin (0,− 0₂))
Izl
Izl
i-cos (90) + isin (90)
b=rsine
→ Re(z)
Re(z)
a
===α-bi
===α-bi
rectangular form
polar form
z= r^(cos(-9)+isin (-e)
Zr (cos(-6)+isin(-1)
De Moivre's formula z-r" (cos(ne) + isin (ne))
cose-sine+isinecose = (cose + isin0)² = cos(20) + isin (20)
zi=-1-i
z=u" = (a+bijh
z = a+bi
z=r (cose +isine)
√r (cos(0 + 2xk) + isin (0+ 2xk"
ke {0,1,1}
n
r = √12 = √√√a²+ b²
tang - ba
Im (Re)
arcose
In distinct root
b-rsine
Z(Re)
+2n
h
z
027
coterminal with
n