Review: 4.5-Graphs of sine and cosine functions

Properties of sine and cosine functions
1. The domain is the set of real numbers.
2. The range is the set of y values such that −1≤ y ≤1
3. The maximum value is 1 and the minimum value is –1.
5. Each function cycles through all the values of the range 2π over an x-interval of 2π
6. The cycle repeats itself indefinitely in both directions of the x-axis.

The standard equations for sine and cosine are:

y = a sin b (x - c)
y = a cos b (x – c)

• |a| is the amplitude of the sine or cosine graph. 
• The amplitude describes the height of the graph.
• “b” affects the period of the sine or cosine graph. 
• “c” indicates the phase shift of the sine graph or of the cosine graph. 
• The x-coordinate of the key point is c.

The standard graphs of sine and cosine are: 

Math joke of the day: 

Review: 9.4-Mathematical Induction

Mathematical induction is a way of proving an equation.

These are the steps of mathematical induction:
1)Prove the statement is true at the starting point (n = 1).
2) Assume the statement is true for n.
3) Prove the statement is true for n + 1.
4) If the LHS and RHS equate, the statement is true for “all n element of natural numbers,” which can be denoted as ∨ n € |N.

The Principle of Mathematical Induction

• Let P(n) be a statement involving the positive integer n.

o If P(1) is true, and

o 2) the truth of P(k) implies the truth of P(k + 1), for every positive k, then P(n) must be true for all positive integers n.

• To apply the Principle of Mathematical Induction, you need to be able to determine the statement P (k + 1) for a given statement P(k). It is important to recognize that both parts of the Principle are necessary components to validate your conclusion.

Example: 

• Prove: 1 + 3 + 5 + 7… + (2n – 1) = n^2
• Step 1: Prove the statement is true for n = 1.

o LHS: (2n – 1) = (2(1) – 1) = 1.

o RHS: n^2 = 1^2 = 1.

o Because the LHS = RHS, the statement is true for n = 1.

• Step 2: Assume the statement is true for n.

o 1 + 3 + 5 + 7… (2n – 1) = n^2 ‡ true.

• Step 3: Prove the statement is true for n + 1.

o 1 + 3 + 5 + 7… 2 (n + 1) – 1 = (n + 1)^ 2.

o Here, you are replacing each n for n + 1 in the equation.

o 1 + 3 + 5 + 7… 2n + 1 = (n + 1)^2.

o LHS: n^2 + 2n + 1 = (n + 1)^2.

o Here, 1 + 3 + 5 + 7 can be replaced by n^2 because in Step 2 you stated that the

o (n + 1) (n + 1) = (n + 1)^2

o (n + 1)^2 = (n + 1)^2.

• Step 4: State your conclusion.

o Because the LHS = RHS, the statement is true for n + 1. Therefore, the statement equation is equal to n^2. is true for ∨ n € |N “all n element of natural numbers” by mathematical induction.

Math joke of the day:

Review: 9.5-Binomial Theorem

Binomial: a polynomial that has two terms; (x+y)n

Expansion Formulas: 
(x+y)0=1
(x+y)1=x+y
(x+y)2=x2+2xy+y2
(x+y)3=x3+3x2y+3xy2+y3
(x+y)4=x4+4x3y+6x2y

Alternate Method: Pascal's Triangle

• Alternately, you can use Pascal's triangle to expand any given polynomial
• Using pascal’s triangle using the number of the exponent you can use the row that corresponds to that number to find the answer and put declining powers on the first term and increasing powers on the second term. 
• This is easier to explain using an example: 

Math Joke of the Day: 

A mathematical children's book

Review: 5.2-Verifying Trigonometric Identities

Sabrina W. did her presentations on verifying trig equations by using trig identities. This is a hard lesson and it really helps to practice, practice, practice. There are only so many tips that Miss V and Sabrina can give us, the rest is critical thinking.

Rules for verifying trigonometric identities:
1. Work with only one side the equation
2. Factor an expression, add fractions, square binomials, or create a monomial denominator
3. Look for opportunities to use the fundamental identities
4. Convert all terms to sines and cosines if the preceding guidelines do not help

To verify trig equations, you really need to know your trig identities, especially your Reciprocal identities, Pythagorean identities, Quotient identities, Co-Function identities, and Parity identities. Because of the trig quizzes, you should have all these identities memorized, but if you don't or you just need some review, all the trig identities can be found in the back of or textbook and were also posted on edmodo.

Math joke of the day:

Review: 8.1-Matrices and Systems of Equations

Teddy did his presentation on matrices and systems of equations. Here is a quick run-down on this section.

We can find the solutions of matrices by getting them into row-echelon form using the elementary row operations and/or back substitution.

Elementary row options
1. Interchange two equations
2  Multiply an equation by a nonzero equation
3. Add a multiple of an equation to another equation.


Row-echelon form: the necessary form we need for augmented matrices and system of equations.
1. Rows consisting of zeroes belong at the bottom
2. First nonzero has a 1
3. For each row the leading 1 in the higher row is to the left of the lower one.

Here are some examples of matrices in row-echelon form:

How to solve system of equations through Gaussian elimination with back substitution.

1. Get the matrix in row-echelon form using elementary row operations
2. Use back substitution to solve for each variable.

Gauss-Jordan elimination

1. Obtain the reduced row-echelon form using elementary row operations.
2. Variables are equal to the coefficients on the right.

Math joke of the day:

Review: 6.4-Vectors and Dot Products

One of the first review presentations was Darron's on vectors and dot products. Here is an overview this topic. 

Dot Product of Two Vectors
-The dot product of u = <a,b> and v = <c,d> is u • v =  ac + bd 

-The properties of the dot products of two vectors are as follows: Let u, v, and w be vectors in a plane or in space and let c be a scalar.

1. u • v = v • u 2. 0 • v = 0 3. u • (v + w) = u • v + u • w 4. v • v = ll v ll ²  5. c(u • v) = cu • v = u • c


The Angle Between Two Vectors 

If ø is the angle between two nonzero vectors u and v, then cos ø = u • v / ll u ll ll v ll

Vectors u and v  are orthogonal if u • v = 0 

Vector Components
Let u and v be nonzero vectors such that u = w + twhere w and t are orthogonal and w is parallel to v. The vectors w and t are called vector components of u. 

The vector w is the projection of u onto v and is denoted byw = proj v uThe vector t is given by t = u - w
Let u and v be nonzero vectors. The projection of u onto v isproj v u = (u • v/ ll v ll²)v
Work

W = ll proj PQ F ll ll PQ ll                      Projection Form W = (cos ø) ll F ll ll PQ ll = F • PQ              Dot Product Form

Math Joke of the day:
Q: Why didn't the quarter roll down the hill with the nickel?
A: Because it had more cents.

Interesting Math Stuff #14: Knot Theory

A mathematical knot is a knot that cannot be undone because the ends are joined together. In math lingo, a mathematical knot is a "embedding of a circle in 3-D Euclidean space." Two knots are equivalent if they can be transformed from one form to another via a deformation of R^3 upon itself. This basically means that if the knot could be transformed to the second know without cutting the string of passing through the string itself, then the knots are equivalent. 

Here is an picture of one of the simplest types of mathematical knots, called a trefoil knot. 

This a more complex mathematical knot. 

Math joke of the day:

12.3: Tangent Lines and introduction to derivatives

A tangent line is a line at point P that has the best approximating of the graph at that point. Here are some examples of graphs with points of tangency.

For this lesson it will also be important to remember slope. 

Slope is defined as change in x/change in y. 

The equation used to find the slope of line tangent to a graph at any given point is: 

To solve, you will need to substitute the given equations into the formula. 

Here are some examples of some problems: 

Another word for the slope of the line that is tangent to the graph is derivative. The derivative is denoted as f'(x) and read as f prime x. 

Here are some examples of finding the derivative: 

One shortcut to finding the derivative is the polynomial derivatives shortcut, but it will only work for equations in polynomial form. This can be summed up as "bring down the power and decrease by 1."

Here are some examples of finding the derivative by the polynomial shortcut: 

Math joke of the day: