Unit 5-Review Package CRAM-1

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Unit 5:

Trigonometric Transformations

Vocabulary List:

Relation: a pattern between two variables that may be represented as ordered pairs (x,y), a table of variables, a graph, or an equation

Function: a type of relation where each value of the independent variable corresponds to exactly one value of the dependant variable

Inverse Function: has all the same points except the x&y values have been reversed

Invariant Point: points that do not change due to transformation

Vertical Line Test: When you run a line vertically over graph to determine if it is a function

Domain: a set of values of the independent variables

Range: the set of values of the dependent variables

Parent Function: four different functions commonly used in math

Parent Functions:

Quadratic Function: f(x) =x2

Radical Function: f(x) = √x

Absolute Value Function: f(x) =lxl

Reciprocal Function: f(x) =1/x

Asymptote: a line that a curve approaches but never touches

Period: the horizontal length of one cycle of a graph

Cycle: one complete repetition of pattern

Horizontal Axis: an imaginary horizontal line that divides the function halfway between its maximum and minimum

Amplitude: half the distance between the maximum and minimum of a periodic function

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Trigonometric Transformations

(Sine/Cosine)

Sine Function:

o Amplitude (a) is 1 o Horizontal axis is y=0 o Period (T) is 360°

o D: {XER}

o R:{yER/ -‐1<y<1}

Cosine Function

o Amplitude is 1

o Horizontal Axis (HA) is y=0

o Period (T) is 360°

f(x) = sin x y = sin x

Key Points:

(0,0) (90,1) (180,0) (270,-‐1) (360,0)

f(x) = cos x Key Points:

(0,1) (90,0) (180,-‐1)

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o R:{yER/ -‐1<y<1}

Trigonometric Equations:

y = a sin [ k (x-‐d)] +c  Sine Function

y = a cos [ k (x-‐d)] +c  Cosine Function

y = a tan [ k (x-‐d)] +c  Tangent Function *DO NOT NEED TO KNOW*

a  Vertical stretch or compression in y “a” is positive: no reflection

 “a” is negative: reflection over x axis  |a| is the amplitude

 |a| = (max – min) /2

d  Horizontal Translation  “d” is the phase shift

c  Vertical Translation

 “c” fives the Horizontal Axis

k  Horizontal stretch (HA) or compression in x  “k” is positive: no reflection

 “k” is negative: reflection over y axis  “k” gives us the period

 T = 360° / |k|

Equations:

Period = 360° / |k|

Amplitude =(max – min) /2

X intervals = T/4

Max = HA + |a|

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HA (Horizontal Axis) = (max + min) /2 Example Questions:

1) What is the amplitude, phase shift, horizontal axis and period for: y= 2 sin [ 3 (x-‐45° )] + 5

a= 2

d = 45°  Phase shift of 45° c = 5

Period/T = 360° / |k| = 360° / |3| = 120

Horizontal Axis = c HA = 5

2) Find the max and min of y= 3 sin [2 (x-‐25°)] + 4

• c = 4 • HA = 4 • a = 3

Max = HA + |a| Min = HA -‐ |a| = 4 + |3| = 4 -‐ |3| = 7 = 1

3) Find the period of y = sin4x T = 360° / |k|

= 360° / |4| = 90°

Steps to sketch a cycle of y = sin2x 1) Find the k value

K= 2

2) Find the Horizontal Compression= ½

*Remember y = a cos [ k (x-‐d)] +c  Cosine Function

*Sometimes we have to factor first:

For Example: y = -‐2 sin (2x -‐ 10°) + 5

= -‐2 sin [2 (x -‐5°)] + 5

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3) Use the key points for the sine function to find the new key points y= sinx  y = sin 4x

(0,0)  (0,0) (90,1)  ( 45, 1) (180,0)  ( 90, 0) (270,-‐1)  ( 135, -‐1) (360,0)  ( 180, 0 )

4) Graph the key points

Finding the Equation:

Write the equation for sine function with a period of 90° and amplitude of 3 |k| = 360° / T

= 360° / 90° = 4

d = 0 c=0 a = 3 y = 3 sin (4x)

Steps to solving word problems:

1. Sketch a diagram, define variables

2. Determine

a,k,d,c,T, max, min, HA

as given.

3. Calculate missing information from the question.

4. Draw a graph (Cosine or Sine?)

5. Ensure everything is labeled

6. Therefore statement

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Sample word problems:

The tides at Harbour Deep cause the water level to rise to 6m above sea level and drop to 6m below below sea level. One cycle of the tides is completed every 12 hours.

1)If the water level is at sea level at midnight and the tide is coming in (the water is rising), Sketch a graph to show how the depth of the water changes over the 12 hours of one cycle. Assume that at low tide the depth of the water is 2m.

T= 12 hours

Max: 14m

Min: 2m

a= 6

x-‐intervals = T/4

= 12/4

= 3 hrs

k= 30

Create equation based on information above: g(x)= 6sin(30t) + 8

Trigonometric Identities

2 Basic Identities:

-‐Pythagorean Identitiy (PI) -‐Quotient Identitiy (QI)

3 Reciprocal Identities: -‐Cosecant Identities -‐Secant Identities

6m

y

x

14 -‐

8 -‐ 2 -‐

3 6 9 12 15

•

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Pythagorean Identity

: sin

2_ʘ_{+ cos}2_ʘ_{= 1}

Sinʘ = y/r Cosʘ = x/r

LS= Sin2_ʘ_{+ Cos}2_ʘ _{RS= 1}

= (y/r)2 + (x/r)2

=y2/r2 + x2/r2

= y2 + x2 r2 = _1_ 12 = 1 1 = 1

LS = RS

Quotient Identity

: Tan

ʘ = _Sinʘ_ Sinʘ = CosʘTanʘ Cos= _Sinʘ_ cosʘ Tanʘ

Reciprocal Identity:

cscʘ= _1_ secʘ= _1_ cotʘ= _1_ sinʘ cosʘ tanʘ

• P (x,y)

r

y

x

r2_{= x}2_{+ y}2

12_{= x}2_{+ y}2

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Example:

Prove that cosʘtanʘ = 1 sinʘ

LS = cosʘtanʘ RS = 1 sinʘ

= (cosʘ/1) ( sinʘ/cosʘ) sinʘ

= _sinʘ_ sinʘ = 1

LS = RS

Strategies for solving Trig Identities:

-‐Always change everything to sines & cosines first -‐Use the proven identities

-‐Get a common denominator -‐Expand

-‐Factor

-‐Force the denominator on the left to be the same as the one on the right

Substitute: tan = sin /cos

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Across

3. There are 3 ______ identities. 6. “D” Represents the ________ shift. 8. A relation between trig ratios.

9. Half the distance between the max/min 10. Use the vertical line test on these. 11. The same value as the amplitude. Down

1. y =a sin[k(x-d)]+C is the formula for a _____function. 2. calculated with (max + min)/2.

4. "k" gives us the ________.

5. Shifted to the right becomes a Sin function. 7. One repetition of a pattern