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Wednesday, December 9, 2015

Unit Circle: Patterns

In this post, I will show the patterns of values found in the unit circle. I will first derive the values of two triangles, show a pattern of the values, and show a pattern of the signs.

When it comes to trigonometry, knowing the unit circle is key. I think these three things will be very helpful.

Part 1: Derive the values of two triangles

The triangles that are part of unit circle are 45-45-90 triangle and 30-60-90 triangle. These two triangles are what make up all the values on the unit circle except the points on the axis's (i.e. 0, 90, 180, 270, 360). Let's derive the values from these two triangles.

45-45-90 Triangle

Since it is an isosceles, I write with the legs equaling to x and the hypotenuse equal to 1 (see figure 1).
Figure 1. 45-45-90 Triangle.




30-60-90 Triangle

With the 30-60-90 triangle, I can take two of these triangles and for an equilateral triangle (see figure 2).
Figure 2. Equilateral triangle with two 30-60-90 triangles



Now I can find the remaining side of the 30-60-90 triangle.




Part 2: Values that appear in the unit circle.





Let's use equivalent values to show a pattern.
On a calculator, the values look like this:

Part 3: Sign rule of the unit circle

The unit circle is formed by a right triangle (see figure 3). The angle of focus is formed by the horizontal side and the hypotenuse of the right triangle.

Figure 3. Right triangle in Unit Circle

The trigonometric functions can be formed using the coordinates.

This means that the signs of the ratios are based the location of the point on the unit circle. In other words, it depends which quadrant the point is in. Now consider the signs in each quadrant. In quadrant I, the points (x,y) are (+,+). This means all trigonometric functions are positive in quadrant I. In quadrant II, the points are (-,+). So, sine function is positive because the y’s are positive, but the other functions are negative. In quadrant III, the points are (-,-), so the tangent function would be positive, but the sine and cosine function would result in a negative. In the quadrant IV, the points are (+,-). Cosine function would be positive, but the others would be negative.

An easy way to remember the signs rules is with the acronym, A Smart Trig Class. It is an order of the quadrants.
Figure 4. A Smart Trig Class with Quadrants

It signifies which values are positive. In the first quadrant, the values of ALL trigonometric functions are positive. In the second quadrant, the values of the SINE function are positive and the others are negative. In the third quadrant, the values of the TANGENT function are positive and the others are negative. In the fourth quadrant, the values of the COSINE function are positive and the others are negative.

Hopefully, these patterns are helpful for the unit circle. I think the patterns are useful, because it makes the memory work easier. And if you forget, you can always use the patterns to make your own unit circle.  

Determining Odd or Even Functions

Let's talk functions. In this post, I show how to determine if a function is odd or even.

The function is odd f(-x)=-f(x)

The function is even f(-x)=f(x)

Is It an Even or Odd Function?

If you have questions, please e-mail me at jdmathguy@gmail.com.

Factoring Trinomials: Factor Grid Method

Let's talk about factoring trinomials. This method is exhaustive.

Procedure
Ax2+Bx+C

Step 1: List out factors of A and C. Factors of A form the rows and factors of C form the columns.

Step 2:  Setup the grid. Multiply the column by the row and put result in the corresponding square. It forms major squares and minor squares.

Step 3:  Look at the diagonals of each major square.
  •  If C is positive, then find the sum resulting in B. (Outer + Inner)
  •   If C is negative, then find the difference resulting in B. (Outer-inner or inner-outer)

Step 4: Setup the factors by using the grid.
  • First, label the diagonal of the major square with inner and outer.
  •  Second, label the row term (FI – first inner) and column term (LI- last inner) that line up with the “inner” term of the diagonal.
  • Third, label the row term (FI – first outer) and column term (LI- last outer) that line up with the “outer” term of the diagonal.
  • Fourth, using (FO_LI)(FI_LO) setup the factors.

Step 5: Choose the signs.

Mechanics
Binomial times a binomial. It involves a method called FOILing.
(A+B)(C+D)
AC+AD+BC+DB
First+Outer+Inner+Last

On the grid:
First/Last
B
D
A

AD=Outer
C
BC=Inner


Therefore, one can set up their factors this way.
First/Last
B - LI
D - LO
A –FO

AD=Outer
C – FI
BC=Inner

(By simply seeing how they align. C is part of both the first term and inner term, A is part of both the outer term and the first term, etc.)
(FO+LI)(FI+LO)

The other diagonal produces another possible:
(C+B)(A+D)
AC+CD+BA+DB
First+Outer+Inner+Last

On the grid
First/Last
B
D
A
BA=Inner

C

CD=Outer

First/Last
B - LI
D - LO
A –FI
BA=Inner

C –FO

CD=Outer

(By simply seeing how they align. A is part of both the first term and inner term, C is part of both the outer term and the first term, etc.)
(FO+LI)(FI+LO)
Notice on the grids. One grid can contain two possible combinations of the outer and inner while the first and the last remain the same.

Examples
1. x2+5x+6

Step 1: List out factors of A and C
A=1
1x*1x
C=6
3*2, 6*1

Step 2: Setup the grid
First/Last
2
3
6
1
1x
2x
3x
6x
1x
1x
2x
3x
6x
1x

Step 3: Look at the diagonals of each major square.
First/Last
2
3
6
1
1x
2x
3x
6x
1x
1x
2x
3x
6x
1x

Step 4:
First/Last
2 -LO
3 - LI
6
1
1x - FO
2x - outer
3x
6x
1x
1x - FI
2x
3x -inner
6x
1x
Using (FO+LI)(FI+LO), you get (x_3)(x_2)

Step 5:
(x+3)(x+2)

2. 3x2-2x-16

Step 1:
A=3
3x*1x
C=16
1*16, 2*8, 4*4

Step 2: Setup the grid
First/Last
1
16
2
8
4
4
3x
3x
48x
6x
24x
12x
12x
1x
1x
16x
2x
8x
4x
4x

Step 3: Look at the diagonals of each major square.
First/Last
1
16
2
8
4
4
3x
3x
48x
6x
24x
12x
12x
1x
1x
16x
2x
8x
4x
4x

Step 4:
First/Last
1
16
2 - LI
8 - LO
4
4
3x – FI
3x
48x
6x – inner
24x
12x
12x
1x - FO
1x
16x
2x
8x – outer
4x
4x
Using (FO+LI)(FI+LO), you get (x_2)(3x_8)

Step 5:
(x+2)(3x-8)

3. 8x2+114x+81

Step 1:
A=8
8x*1x, 4x*2x
C=81
1*81, 3*27, 9*9

Step 2: Setup the grid
First/last
1
81
3
27
9
9
8x
8x
648x
24x
216x
72x
72x
1x
1x
81x
3x
27x
9x
9x
4x
4x
324x
12x
108x
36x
36x
2x
2x
162x
6x
54x
18x
18x

Step 3: Look at the diagonals of each major square.
First/last
1
81
3
27
9
9
8x
8x
648x
24x
216x
72x
72x
1x
1x
81x
3x
27x
9x
9x
4x
4x
324x
12x
108x
36x
36x
2x
2x
162x
6x
54x
18x
18x

Step 4:
First/last
1
81
3 LI
27 LO
9
9
8x
8x
648x
24x
216x
72x
72x
1x
1x
81x
3x
27x
9x
9x
4x FO
4x
324x
12x
108x- outer
36x
36x
2x FI
2x
162x
6x - inner
54x
18x
18x
(FO_LI)(FI_LO)=(4x_3)(2x_27)

Step 5:
(4x+3)(2x+27)

5. 4x2-81 (special type – Difference of Squares)

Step 1:
A=4
4x*1x, 2x*2x
C=81
1*81, 3*27, 9*9

Step 2: Setup the grid
First/last
1
81
3
27
9
9
4x
4x
324x
12x
108x
36x
36x
1x
1x
81x
3x
27x
9x
9x
2x
2x
324x
6x
54x
18x
18x
2x
2x
162x
6x
54x
18x
18x

Step 3: Look at the diagonals of each major square.
First/last
1
81
3
27
9
9
4x
4x
324x
12x
108x
36x
36x
1x
1x
81x
3x
27x
9x
9x
2x
2x
324x
6x
54x
18x
18x
2x
2x
162x
6x
54x
18x
18x
Notice: With a difference of squares, it is the major square that has the same term in each box. This happens because the middle term is 0x.

Step 4:
First/last
1
81
3
27
9 - LI
9 -LO
4x
4x
324x
12x
108x
36x
36x
1x
1x
81x
3x
27x
9x
9x
2x - FI
2x
324x
6x
54x
18x –inner
18x
2x - FO
2x
162x
6x
54x
18x
18x -outer
(FO_LI)(FI_LO)=(2x_9)(2x_9)

Step 5:

(2x-9)(2x+9)