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PROJECT OPTION sent via Google Classroom May 7th, due May 29th.  Click for details.

Classes Mondays and Thursdays 8-9

Dial-in: (US) +1 402-921-2262 PIN: 262 116 556#

Classes online Tue/Thurs at 10-11AM

Wed/Fri at 11-12PM.

To join the video meeting, click this link: TamiscalPrecalculuSIS

dial +1 484-854-3263 and enter this PIN: 113 598 551#

Chapter 14

Notes 14.1 Limits

Notes 14.1-14.2 Algebraic Tricks

Notes 14.3 One Sided Limits and Continuity

challenging slope problems                Save

Chapter 5--Logarithms

inverse functions and one to one

exponential doubling

Natural Number e for exponential growth

Notes inverse functions and one-to-one

Review Ch 5 (assignment #29) review Ch5 KEY Chapter 12--Sequences and Series

O

Chapter 4 Polynomial and Rational Functions

p238 17 (and extra)

p232 #49,65

Rational Root Examples

Remainder and Rational Root Theorems

p 218 problems

graphing rational functions ws

Chapter 10 Conics

Hyperbola hyperbola notes

why is a^2+b^2 = c^2 this time?  Ellipse Animation Note:  c^2 = a^2 - b^2 (Not exactly Pythagorus for an ellipse)  Parabola animation intro to parabolas circle and complete the square    Chapter 9 Review

Ch 9 Review Solution KEY

Vectors

vector applications KEY

vector applications solutions

vector applications worksheet

vector notes

vector example problems (Word documents)

Polar Graphs

polar notes

notes for polar (Word)

symmetry tests

polar grids         The sin, cos and tan ratios only work on right triangles.  For obtuse and acute triangles, we have:

The Law of Sines (when given AAS or ASA)

The Law of Cosines (when given SAS or SSS.)

Law of Sines and Cosines KEY

The Side-Side-Angle case could result in two possible triangles which is why SSA is not a congruence condtion.

If the larger side is opposite the given angle (SsA or HL), there is one triangle possible.

If the smaller side is opposite the given angle (sSA), there could be two triangles.

If Sin(angle) > 1, then there is no triangle possible because the third side is not long enough to make a triangle.

Because inverse cosine returns an angle between 0 and 180 degrees, Law of Cosine always gives the right answer.

However, inverse sine only goes up to 90 degrees, so that sin(Arcsin(100)) returns 80 degrees, not 100.

When using the Law of Sines, there may be two possible solutions.  Hence, The Ambiguous Case.

The chapter 7 test for TuTh and WF classes will be small group quests (2-3 persons.)  Also, I will postpone the calculator section and make it takehome.  You may have a unit circle with any formulas you think you'll need in written in your own handwriting.

Chapter 7 Review #2 (7.1-7.2, 7.7-7.8)

Chapter 7 Review2 (Jen's key)

Ch 7 Review2 p1 (2019 key)

Ch 7 Review2 pg2 (Curt's key)

Chapter 7 Review2 p3 calculator

7.1 Inverse Trig Functions

An inverse sine function will return the arc (angle on the unit circle) that pairs with its y-coordinate input.

For this reason, we shall call the inverse of sine an ArcSine.  But two values correspond to y = 1/2 (30 degrees and 150 degrees) because sin(30)= 1/2 and sin(150) = 1/2.  Which angle shall the inverse return? Define Sin(x) with a capital 'S' as a sine function with a restricted domain from -pi/2 to pi/2.

So the arcsine returns the angle on the right half of the unit circle, from -90 to +90 degrees.

Then -pi/2 < ArcSin(x) < pi/2 and its domain is -1 < x < 1.  This covers every possible y-coordinate, or sine value from -1 to 1, on the unit circle.

Which angles do we have to run through to cover every possible cosine value from -1 to 1?

An inverse cosine has to return the arc on the unit circle that pairs with its x-coordinate. Define Cos(x) with a capital 'C' as a cosine function with a restricted domain from 0 to pi (180 degrees.)

So the arccosine returns the angle on the top half of the unit circle, from 0 to +180 degrees.

Then 0 < ArcCos(x) < pi and its domain is -1 < x < 1.  This covers all possible cosine values.

In conclusion, each inverse trig function returns the arc value closest to zero on the unit circle.

This also holds true for ArcTan(x).  Think about which angles you have to run through to cover every possible tangent value (slope of terminal side of angle) from -infinity to infinity.  -90 to +90 degrees, not including the quadrantal angles, is the required range. Therefore,

-pi/2 < ArcTan(x) < pi/2 is the range and the domain of the inverse tangent is all real numbers.

Review 7.3-7.5 KEY

p 482 answer -- double and half angles

You may write a formula sheet in your own handwriting for the test.

Include a unit circle if you do not have it memorized.

formulas to copy ...

Ch 7 Review KEY

Trig Identities KEY

TEAM PC: trig application problems KEY

Notes in class:

QUIZ POSTPONED TO NEXT WEEK.

We'll start Chapter 7 instead Thursday and Friday and flip flop lessons to stay on pace.

Again, no quiz this week.  Next Tues/Wed will be a group quiz on Ch. 6 Part 2 (graphing trig.)

We can also review a bit more.  See email sent for more details.

Graphing tan and cot KEY

Graphing sine and cosine KEY

Graphing sec and csc KEY

Degrees-Minutes-Seconds KEY

Trig Review Ch6 Part 1 KEY

Chapters 1 & 2

Math Modeling 2.6 Examples

pg 115 #8 (rectangle inside semi-circle), p115 #12 (wire into triangle and circle)