Monday, October 31, 2011
o Reciprocal Identity:
§ Ex)\[sin\theta =1/csc\theta \]
· (applies to all six)
o Quotient Identity:
§ Ex)\[tan\theta =sin\theta /cos\theta \]
· (applies to all six)
o Even Odd Identities
§ Ex)\[sin(-\theta )=-sin\theta \]
· (applies to all six)
o Pythagorean Identities
§ Three different identities
· 1) \[sin^2\theta +cos^2\theta =1\]
· 2) \[tan^2\theta +1=sec^2\theta \]
· 3) \[cot^2\theta +1=csc^2\theta \]
· Solving Identities:
o Ex) \[(cot\theta +csc\theta)(tan\theta -sin\theta)=sec\theta -cos\theta \rightarrow (1-cot\theta)(sin\theta)+(csc\theta)(tan\theta-1)\rightarrow -(cos\theta /sin\theta )(sin\theta )+(1/sin\theta)(sin\theta /cos\theta )\rightarrow -cos\theta +1/cos\theta \rightarrow sec\theta =cos\theta =sec\theta -cos\theta \]
Thursday, October 27, 2011
Ratios
sine x = (side opposite x)/hypotenuse
cosine x = (side adjacent x)/hypotenuse
tangent x = (side opposite x)/(side adjacent x)
In the figure below, sin A = a/c, cosine A = b/c, and tangent A = a/b.

Reciprocal Ratios
cotangent x = 1/tan x = (adjacent side)/(opposite side)
secant x = 1/cos x = (hypotenuse)/(adjacent side)
cosecant x = 1/sin x = (hypotenuse)/(opposite side)
Cofunctions
sin x = cos (90o - x)
tan x = cot (90o - x)
sec x = csc (90o - x)
cos x = sin (90o - x)
cot x = tan (90o - x)
csc x = sec (90o - x)
Example:
1. Problem: Find the function value of
cot 60o.
Solution: Use the cotangent's cofunction
identity to rewrite the problem.
tan (90o - 60o)
tan 30o
The tangent of 30o is
(SQRT(3))/3
Monday, October 24, 2011
Section 4.1 Radian and Degree Measure
Thursday, October 20, 2011
Section 2.7 Graphing rational functions

Graph f(x)=x/x^2-x-2
Monday, October 17, 2011
2.6 Rational Functions and Asymptotes
Where N(x) and D(x) are polynomials and D(x) is not the zero polynomial.
Domain of a Rational Function: The domain of rational functions of x includes all real numbers except x-values, that will make the denominator zero. 


Horizontal Asymptotes:

Intercepts:
Y-Intercepts:
Things to Consider:
- Asymptotes are lines and therefore should be written in proper notation, such as x=a and y=b
- It is possible for a rational function to have multiple x intercepts and asymptotes
- It is possible for a rational function to lack both vertical and horizontal asymptotes
- It is possible for a rational function to lack both x and y intercepts
Tuesday, October 11, 2011
Complex Numbers Section 2.4
2.5 The Fundamental Theorem of Algebra
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- The first-degree polynomial f(x)= x-2 has exactly one zero: x=2
- Counting multiplicity, the second-degree polynomial function
has exactly two zeros: x=3 and x=3
- In some third degree polynomial functions there will be exactly three zeros, but the graph will only show the real zeros as intercepts.
- To find the zeros, first find possible rational zeros then find the real zeros by graphing the equation, and finally use synthetic division to determine algebraically that the zeros are on the graph.
- find a fourth-degree polynomial function with real coefficients, that has -1, -1, 3i as zeros
- from the Linear Factorization Theorem it can be written as
- or simply, let a=1 to get
- first factor the polynomial into the product of two quadratic polynomials
- By factoring over the reals, you have:
- because complex zeros occur in conjugate pairs, you know that 1-3i is also a zero of f.
- both x-(1+3i) and x-(1-3i) are factors of f(x). Multiplying these two factors produces [x-(1+3i)][x-(1-3i)]=[(x-1)-3i][(x-1)+3i]
- then you would use long division to divide
into f(x) and then get
- therefore, you have
- and conclude that the zeros of f are 1+3i, 1-3i, 3 and -2
Sunday, October 9, 2011
Section 2.3 Real Zeros of Polynomial Functions
One of them is long division. Long division is harder than synthetic division.


