Table 14: Linear and Angular Velocity
Table 15: Units relations
1 mile = 5280 feet 
1 feet = 12 inches 
1 hour = 60 mins 
1 min = 60 sec 
Table 16: Radian Formula . s: length of arc, r: radius, P:
perimeter of sector, A: area of sector,
θ: angle in radian
5 Test 5 Chapter 8.3, 8.4, 8.5, 8.7, 9.1, 9.2
5.1 8.3 Graph of Sine and Cosine
Table 17: Graph of Sine and Cosine

Asin(Bx + C) + D 
Acos(Bx + C) + D 
Domain 

Range 
Period 
Amplitude 
Median 
Phase of Shift
(Starting Point) 


Table 18: Graph of Tan and Cot

Atan(Bx + C) + D 
Acot(Bx + C) + D 
Domain 
{xBx + C ≠ k(π /2), k is any odd number } 
{xBx + C ≠ kπ , k is any even number} 
Range 
(−∞,∞) 
(−∞,∞) 
Period 
π/B 
π/B 
Table 19: Graph of Sec and Csc

Asec(Bx + C) + D 
Acsc(Bx + C) + D 
Domain 
{xBx + C ≠ k( π/2), k is any odd number} 
{xBx + C ≠ kπ , k is any even number} 
Range 


Period 
2π /B 
2π /B 
Table 20: Chart of Special Angle
Table 21: Maximum Points
Table 22: Minimum Points
Table 23: Useful formula
6 Test 6Chapter 9.4 9.5 10.1
Contents: Inverse Trig Function , Law of sine and cosine
1. Domain and Range (table (??)):
Table 24: Inverse Function
Function 
arcsin(x) 
arccos(x) 
arctan(x) 
Domain 

Range 
Quadrant

1st and
4th 
1st and
2nd 
1st and 4th 
Table 25: Useful Properties
Properties 
Examples 



trig(a) = x doesn’t necessary implies
a = arctrig(x) 
but 
trig(arctrig(x)) = x 
sin(arcsin(x)) = x 
arctrig(trig(x)) = x, iff x ∈
the range of arctrig 
arccos(cos(x)) = x, iff x ∈ [0, π] 
if
the range of arctrig,
arctrig(trig(x)) = α , where α ∈ the range
such that trig(x) = trig(α ). 
because
and 
Example 6.1 Find the exact value of
.
Analysis: let
then by the definition of the range of arccos, α ∈ [0, π]. And
by taking cos at both side, we have
We know, if trig(α ) = trig( β ), then ref( α ) = ref( β).
Hence we have
since is in the 3rd quadrant, and cos in 3rd
quadrant is negative , it implies
cos( α ) < 0. If combined with the fact that α can either be in the
1st or 2nd quadrant, we deduce
that α must be in the 2nd quad. In summary α is an angle in 2nd
quadrant with reference angle
. Hence it must be
Example 6.2 Find the exact value of
.
Solution : Let
=>
where
or, α in 1st or 4th quadrant.
1. Reference angle
Moreover, is in the
2nd quadrant, where tan value is negative. Hence
2. α is in the 4th quad where tan value is also negative.
3. So
Example 6.3 Find the exact value of
sin(2arctan(−3/4))
Solution:
Let arctan(−3/4) = α, then
where or, α is
either in the 1st or 4th quadrants.
We can narrow the range of α, by observing that tan( α ) < 0, which
implies α can not be in the
first quadrant but in the 4th quadrant.
Since,
sin(2arctan(−3/4)) = sin(2 ) = 2sin(α )cos( α),
we need to find out sin( α ) and cos(α ), of which we know α in
4th quadrant, and tan(α ) = −3/4.
By using the assistant right triangle in the 4th quadrant, we have
sin( α ) = −3/5,
cos(α ) = 4/5,
sin(2α ) = 2(−3/5)(4/5).
6.1 Law of Sine and Cosine
Table 26: Law of Sine and Cosine
Name 
Rule 
Usage 
Law of Sine 

two angles + one side 
Law of Cosine 

two sides + one anlge 
Law of Cosine 

three sides 
Table 27: Useful Knowledge
Sum of the 3 angles of a triangle is π or
180deg. 
Sum of the n angles in a regular ngon is (n − 2)π
.
That means each angle is 
Largest angle opposites to largest side, smallest
angle opposites to smallest side. If A > B > C, then a > b > c. 