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Chapter 10 Conics
why is a^2+b^2 = c^2 this time?
Note: c^2 = a^2 - b^2 (Not exactly Pythagorus for an ellipse)
Chapter 9 Review
notes for polar (Word)
pcr 9ad ,
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.)
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)
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.
You may write a formula sheet in your own handwriting for the test.
Include a unit circle if you do not have it memorized.
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.
Homework Answer Keys:
Answers for Radical Review worksheets Radical Review answer keys
Chapters 1 & 2
Absolute Value Transformations (Google Slides)