Math 1113 Review

Table 14: Linear and Angular Velocity

Formula
Symbol	v: Linear velocity , w: angular velocity, r: radius of circle
Notice	w must be in radian; v and r must have the same distance unit

Table 15: Units relations

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

Table 19: Graph of Sec and Csc

	Asec(Bx + C) + D	Acsc(Bx + C) + D
Domain	{x|Bx + C ≠ k( π/2), k is any odd number}	{x|Bx + 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 6-Chapter 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 n-gon 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.