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Study Guide

1000 Study Guide - Summer 2018, Comprehensive Midterm Notes -


Department
Economics
Course Code
1000
Professor
simo
Study Guide
Midterm

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1000
MIDTERM EXAM
STUDY GUIDE
Fall 2018

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www.nafham.com
if y=logax then x=ay
xlog bx=b
log
aa=1
logaax
( )
=x
alog ax=x
logaMN
( )
=logaM+logaN
logaMp
( )
=PlogaM
log
a
M
N
=log
aMlogaN
aN=1
aN
Sx=xa+bi
( )
[ ]
xabi
( )
[ ]
=x22ax +a2+b2
( )
cos
2x+sin2x=1
sin 2
( )
=2sin
( )
cos
( )
cos 2
( )
=cos2sin2
cos 2
( )
=12sin2
cos 2
( )
=2cos
21
sin 2
= ± 1cos
2
cos 2
= ± 1+cos
2
tan 2
( )
=2tan
1tan2
tan 2
=1cos
sin =sin
1+cos
log
ax=ln x
ln a
sin +sin =2 sin +
2cos
2
sin sin =2 cos +
2sin
2
cos +cos =2 cos +
2cos
2
cos cos =2 sin +
2sin
2
sin +
( )
=sin cos +cos sin sin
( )
=sin cos cos sin
cos +
( )
=cos cos sin sincos
( )
=cos cos +sin sin
ax
2
+bx +c=0
x=b±b
2
4ac
2a
Page 1
Math Reference Trigonometry & Analysis James Lamberg
cos π
2+x
= sin x
cos π
2x
=sin x
cos π ± x
( )
= cos x
cos 3π
2+x
=sin x
cos 3π
2x
= sin x
sin π
2±x
=cos x
sin π + x
( )
= sin x
sin π − x
( )
=sin x
sin 3π
2±x
= cos x
tan π
2+x
= cot x
tan π
2x
=cot x
tan π + x
( )
=tan x
tan π − x
( )
= tan x
tan 3π
2+x
= cot x
tan 3π
2x
=cot x
for every
element of
there exists
such that
therefore
Q since
¬ not
and
or
d derive
integrate
x=a evaluate with x =a
proportional to
p precedes
f follows
congruent to
union
intersection
subset
superset
proper subset
propor superset
C (fancy) compliment
implies
double implication
~ negation or ¬
( )
Q.E.D: Quod Erat Demonstandum
"that which was to be proved"
tan
( )
=tan tan
1+tan tan
tan +
( )
=tan +tan
1tan tan
Mid =x
2
x
1
2,y
2
y
1
2
Dist =x
2
x
1
( )
2
+y
2
y
1
( )
2
A = 1
2r2, sector area
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www.nafham.com
c2=a2+b22ab cos
b2=a2+c22ac cos
a2=b2+c22bc cos
sin
a=sin
b=sin
c
K=1
2bc sin
y=Asin x -
( )
+K, > 0
Amplitude = A=Mm
2
Period = 2π
Phase Shift =
Frequency = 1
period =2π
Critical Points = period
4
Unit Circle
cos, sin
( )
0,0°=1,0
( )
π
6,30
°=3
2,1
2
π
4,45
°=2
2,2
2
π
3,60
°=1
2,3
2
π
2,90
°=0,1
( )
2π
3,120°= 1
2,3
2
3π
4,135°= − 2
2,2
2
5π
6,150°= − 3
2,1
2
π,180
°= −1,0
( )
7π
6,210
°= − 3
2,1
2
5π
4,225
°= − 2
2,2
2
4π
3,240
°= 1
2,3
2
3π
4,270
°=0, 1
( )
5π
3,300
°=1
2,3
2
11π
6,330
°=3
2,1
2
Page 2
\Math Reference Trigonometry & Analysis James Lamberg
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