Friday, June 28, 2013

TRIG SUM AND DIFFERENCE IDENTITIES

THE RHYTHM OF TRIG SUM & DIFFERENCE IDENTITIES

               Sin (A + B) = Sin A Cos B + Cos A Sin B
               Sin (A - B) = Sin A Cos B - Cos A Sin B
               Cos (A + B) = Cos A Cos B - Sin A Sin B
               Cos (A - B) = Cos A Cos B + Sin A Sin B
               Tan (A + B) =   Tan A + Tan B 
                                      1 - Tan A Tan B
               Tan (A - B) =   Tan A - Tan B 
                                      1 + Tan A Tan B

It’s no surprise to find a teenager listening to music while studying.  There are two ways to recognize a familiar tune: the notes and the rhythm.  Try it. 

Play the notes and try to identify the song:


Did you recognize Mary Had a Little Lamb?  Even without the quarter notes, half notes, and measures, you would still know the tune.

Try this one with just the pacing:



 The "merrily, merrily, merrily, merrily" probably gave away that this one is Row-Row-Row Your Boat.  If you recognized it, try using rhythm to remember the Trig Sum and Difference Identities.

First, establish the pattern of syllables.  Then clap or tap the syllables for each word.

     Sin (A +/- B) = Sin A Cos B +/- Cos A Sin B

Read this as    SINE  COSINE      COSINE       SIN
Tap it out as    CLAP  CLAP-CLAP CLAP-CLAP CLAP

Once you know the rhythm you can write out sin and cos, fill in the A’s, B’s, and plus or minus symbols. 

     Cos (A +/- B) = Cos A Cos B -/+ Sin A Sin B

    Read as          COSINE        COSINE         SINE     SINE
    Tap it out as     CLAP-CLAP   CLAP-CLAP    CLAP    CLAP

    and fill in the angles and minus or plus.

Notice with these two identities, Sine sum or difference start with ‘sine’ and Cosine sum or difference start with ‘cosine’.  In addition, the Sine formulas have the same sign (+ or -) as the computation of the angles, while Cosines are opposite.

     Tan (A +/- B) =    Tan A +/- Tan B
                               1 -/+ Tan A Tan B


    Read as           TAN      TAN       ONE  TANTAN
    Tap it out as      CLAP   CLAP     CLAP-CLAP-CLAP

When filling in operations, notice that Tangent is Sine over Cosine, so the numerator maintains the operation on the angles but the denominator uses the opposites.

Once you know these identities, you won't have to learn the Double Angle Identities because you could just substitute (A + A) for (A + B). Each Sum and Difference equation can do double duty.  And who wouldn't like the requirement to learn half as much?!?

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PRACTICE

These exercises obviously take liberties with musical notations, but tap out the rhythms and use an unexpectedl part of your brain to remember the Sum and Difference Identities.




TANGENT:
Tan Tan OneTanTan





SINE:
Sine Cosine Cosine Sine




COSINE: Cosine Cosine Sine Sine




Thursday, June 27, 2013

PARTIAL FRACTIONS, the basics

WHY DO WE STUDY PARTIAL FRACTIONS IN ALGEBRA AND PRECALCULUS?

In the early days of Algebra study, we covered “collect like terms” so that the process would be a simple step once we got to FOIL (also called “double distributive” in some schools).  The same thinking encourages us to learn other basic algebraic steps before using them in more complicated mathematics.  In Calculus, we may need to decompose rational expressions before applying the rules of integration.  By doing the initial algebraic work of decomposition now, higher level concepts will be easier to learn and understand.

There are several situations which may be presented in Partial Fraction exercises.  We’ll start with the simple and work toward the more complex.

Simple rational expression:

                2x + 3      
          (x^2 - 7x +10)        Notice that the degree of the numerator is smaller than the degree of the denominator.  This is an important restriction that must be met.  (see below: “What if the Numerator is a higher degree than the Denominator?”)

    STEP 1 - factor the denominator

              2x + 3  
            (x-2)(x-5)

    STEP 2 - using the factors from the denominator, list separate fractions to form an equation. (We don’t know the numerators yet, so let’s just keep them blank until Step 3).

              2x + 3         ?         ?  
            (x-2)(x-5)= (x-2) + (x-5)

    STEP 3 - Substitute letters for the unknown numerators.

              2x + 3         A         B  
            (x-2)(x-5) = (x-2) + (x-5)

    STEP 4 - Expand the right side to have a common denominator.

              2x + 3       A(x-5) + B(x-2)
            (x-2)(x-5) =      (x-2)(x-5)

    STEP 5 - Since denominators on both sides of the equation are the same, set the numerators equal and distribute A and B.
   
              2x + 3    =  Ax - A5 + Bx - 2B

    STEP 6 - Collect like variable terms.

              2x + 3   = (A + B)x - 5A - 2B

    STEP 7 - Create a system of equations by setting coefficients of the variables equal.

            2  = A + B
            3  = -5A -2B 
The reason I prefer this ‘systems’ method to some others is that I can use the matrix function on my calculator to get answers with only a few key strokes and no manual arithmetic.

    STEP 8 - Solve the system.

            A = -7/3
            B = 13/3

    STEP 9 - Seems like a lot of steps so far, but this is the last one.  Substitute A and B values into the equation from Step 3.

              2x + 3         -7/3      13/3 
            (x-2)(x-5) =  (x-2) +  (x-5)


WHAT ABOUT MORE COMPLEX DENOMINATORS?             
  
More factors?  Include more fractions in Step 3....
   
                 2x + 3                           
            (x-2)(x-5)(x-1) = (x-2) + (x-5) + (x-1)

    .... and continue with Steps 4 - 9  (SOLUTION:  A = -7/3, B= 13/12, C= 5/4)

Repeated binomial factors?  Consider the highest degree of the repeated binomial and EVERY SMALLER DEGREE to set up the equation in Step 3....

                2x + 3                    B           C    
            (x-2)(x-5)^2   = (x-2) + (x-5) + (x-5)^2

    ....and continue with Steps 4 - 9.  (SOLUTION:  A= 7/9, B= -7/9, C=13/3)

A quadratic factor?  If the denominator is quadratic, make the numerator of the fraction a binomial...

          2x + 3             A        Bx + C   
    (x-2)(x^2 - 5)   = (x-2) + (x^2 - 5)

    ....and continue with Steps 4 - 9.  (SOLUTION: A=-7, B=7, C=16)

 What if the Numerator is a higher degree than the Denominator? 
AHA!!  Another example of how simple algebra concepts are eventually used again as steps within more complex algorithms!!  To satisfy the requirement that Partial Fractions can only be constructed if the Numerator is a lower degree than the Denominator, we need to break up the rational expression even more by LONG DIVISION.  The Partial Fraction is constructed from the REMAINDER, expressed as a fraction with the divisor as the Denominator.

Wednesday, June 26, 2013

VECTORS IN A NUTSHELL

       

WHY SHOULD I LEARN ABOUT VECTORS?

After so many years of working with the same curriculum, I sometimes need to be convinced of the benefit of learning something new.  I’m relatively sure many students feel the same way.  So when it came time to expand the curriculum to include vectors, I was resistant.  However, a little research pointed out that many of my interests could benefit from understanding vectors.

Of course, vectors are found in science and engineering, but those disciplines have never prompted me to learn vectors in the past.  A math class problem made me realize that the football quarterback is thinking in vectors when calculating where to throw the ball and how much force to use to get it to the intended receiver.  In my youth, I thought of being a weather forecaster, and now I discover that vectors are needed to draw wind maps.  Fats Domino, famous pool player, was expert in calculating vectors to “drop the 5 ball in the corner pocket” by first hitting it with the 10 ball.  In helping a student study for the police exam, I found that investigators use vectors to describe the details of an accident.

Our trigonometric experience with triangles is useful in determining the outcome of opposing forces described as vectors.  For example, you want to swim across the Rock River.  There's a spillway about 100 yards downstream from where you are.  Considering the rate at which the water is flowing, can you swim to the other side before falling over the spillway and possibly incurring bodily harm? We never thought about this when we, as teens, were swimming across the river, but maybe we should have.

Wow, maybe vectors ARE worth learning to prepare for a wide variety of experiences and careers.  So let’s get started...

WHAT IS A VECTOR?

In its most general description, a vector is a line which demonstrates 2 characteristics:
                LENGTH (called "magnitude")
                and
                DIRECTION, indicated by an arrow head at its terminal end

Here's a vector drawn on the coordinate plane:

It starts a S(1,2) and ends at T(6,5).

To find the Cartesian coordinates that would name this vector if it started at (0,0), we would subtract:
                   T - S
and, in this example,  get  (5,3).  From these coordinates, we can calculate the MAGNITUDE by using the Pythagorean Theorem:

                  magnitude, ||ST|| = √ (x^2 + y^2) = √ (5^2 + 3^2) = √( 34)

           Finding ||ST|| from the original components is just like the distance formula:
                                       ||ST|| = √[(6-1)^2 + (5-2)^2]

The coordinates also give us all the information necessary to determine the angle, or direction, of the vector:

                  Tan θ = y/x     or the longer way Tan θ = (5-2)/(6-1)
                                      Doesn’t that fraction look just like SLOPE?  


When the magnitude and direction are given as polar coordinates, (r, θ)....
....the x and y coordinates can be calculated using these equations:
              x = r(cos θ)
              y = r(sin θ)    (Remember from Trig....Cosine is the x axis, Sine is the y axis) 

COMPUTATIONS WITH VECTORS

ADDITION

Let Vector A = (3,8)
      Vector B = (2,4)

To add Vector A and Vector B for a RESULTANT Vector C:

    Add the x values.........3 + 2 = 5
    Add the y values.........8 + 4 = 12

The RESULTANT Vector C = (5,12)
with magnitude of 13.



SUBTRACTION

To subtract Vector B from Vector A for a RESULTANT Vector D:
  
     Subtract the x values.....3 - 2 = 1
     Subtract the y values.....8 - 4 = 4...........Vector D = (1,4) with magnitude of √17

SCALAR MULTIPLICATION

To increase a Vector by a given scale, multiply both x and y by the scale.

     5(Vector A) = 5(3,8) = (15,40)

DOT PRODUCT versus CROSS PRODUCT

DOT PRODUCT (sometimes called the SCALAR PRODUCT or INNER PRODUCT) is found by multiplying the x values together and the y values together and adding the two products.  The answer is a single number.

     Vector A (3,8) • Vector B (2,4) = (3)(2) + (8)(4) = 6 + 32 = 38

If the sum equals 0, the two lines are PERPENDICULAR.  WHY?!?
    Well, let's look at what we know from Algebra I.  If the slopes of two lines are opposite inverses, then the lines are perpendicular.  So, multiplying (as in the Dot Product) the numerator of two perpendicular slopes would be the additive inverse (opposite) of the product of the denominators.  Adding the two would result in a sum of zero.....ergo perpendicular lines. 

CROSS PRODUCT (sometimes called the OUTER PRODUCT)

     (Vector A) X (Vector B) = ||A||•||B||

Although Cross Product will not appear until later in the program, it is important to recognize the differences now, when we're talking about the Dot Product.  Dot Product can be completed on two vectors in a single plane; the Cross Product is only applicable in 3 dimensions.  It provides a vector perpendicular to two other vectors in a plane -- a vector perpendicular to that plane.  Cross Product involves matrices, so brush up on Cramer's Rule and determinants before we get to it later in the curriculum.

OTHER UPCOMING ISSUES WITH VECTORS 

Once we're comfortable with these basic vector concepts, we can explore more complex issues like unit vectors, resolving vectors into their components, vectors in 3 dimensions, and representing scalar values of unit vectors using i, j, and k.