Precal Blog

Hey everyone merry Christmas and sorry that this post is late!

I am writing about what we did in class the Monday before break.  During class we were reading and working on the problems in section 4-6.  These problems had to do with inverse trigonometric relation graphs.  An example problem would be y=inverse tangent x.  All of the problems we did followed this basic structure, but they did become more complicated, such as y=cos(inverse tangent x).

In order to solve the tougher problems it is easiest to deal with what is in the parentheses first.  For example if you are given a problem like tan(inverse cos 4/5) then you find that cos of an angle = 4/5 and then you look at what the tangent of 4/5 is.  To find the tangent you can draw a triangle and use the Pythagorean Theorem.  Once you have all sides of your triangle then you can use the definition of tangent to find what it is equal to.  Since we know that cos is adjacent over hypotenuse, we know that 4 is the adjacent leg of the triangle and 5 is the hypotenuse.  When you use the Pythagorean Theorem, the third opposite side comes out to be 3.  To solve for tangent  you look at what two values would be opposite and adjacent to the angle.  When this is done the answer comes out to be 3/4.

This system of seperating the problem and drawing a triangle is really useful and helps me understnad the problems.  So if you worked on the claims, which were 9, 10, 19, 20, and 25, then you can test this method!

I hope you all have a happy holiday and aren’t forgetting to work on your quarter and semester journal!

Sorry guys, I know this is a late post.

Anyways last wednesday in class we had a sub. We continued the two sheets that Mr. Bieniek left for us on tuesday. The first worksheet went back to parametric equations. Remember how you switched your calculator into parametric mode and then you had to deal with a t variable in addition to an x and y variable? well the worksheet gave an example scenario where x measured the horizontal distance from the airport to an airplane, y measured the height of the aircraft, and the third variable t stood for the time it took the aircraft was in the air. We had to use the function x(t) to measure the distance of the plane and y(t) to measure the height of the plane to describe the position of the aircraft.

Then the sheet told us that an asteroid was a simple example of a curve represented by a parametric equation. It gave us functions to plug in for x(t) and y(t). so we had to put our calculators in parametric mode, and plugged in those functions to find the shape of an asteroid. The sheet told us to plug in a=4, and then record values for x(t) and y(t) on the table provided. After we filled in the tables we had to accuratley sketch the graph of an asteroid on the back.

Don’t worry about remembering any of this though. It was just kind of an exploration thing.

On Friday we didn’t have school because of a snow day!!!!!!!!!!!! It was good timing seeing we then had a three day weekend. Although we didn’t have Pre-Calc on Friday, we still have some homework to do for Monday. We have to read section 4-6 and complete a yellow sheet. This section deals with inverses of the six Trig. Functions. It looks at the inverses graphically and there is also a nice table that show the domain, and the range (verbally and numerically) of the functions.

We also have to start working on our quarter journals and semester portfolios. The quarter journal is done the same way we did it for the first quarter. Mr. B talked to us about the semester portfolios and they are a lot more in depth and will not be something that you can finish in one night. So even though its winter break, everyone should get a start on the portfolio.

The scribe for Monday will be Nikolai.

Coverit Live test goes live at 9:30 am.

Mr. Bieniek was here today!

We continued working with the packet that we got from the sub on Wednesday.  The graphs that we created were graphs of, for example, `y=arcsin(x)`, `y=arccos(x),` etc.  If we were to look at these graphs as functions, we would need to restrict the range.  These restricted ranges can be divided in two groups: one of which has a range based on `[-pi/2, pi/2]`; the other, based on `[0, pi]`.  Within each “group,” there are some subtle differences in the range.

The first group consists of inverse sine, inverse cosecant, and inverse tangent.  The ranges for each function are as follows: inverse sine, `[-pi/2, pi/2]`; inverse cosecant, `[-pi/2, pi/2], y<>0`; and inverse tangent, `(-pi/2, pi/2)`.

The second group consists of inverse cosine, inverse secant, and inverse cotangent.  The ranges for each function are as follows: inverse cosine, `[0, pi]`; inverse secant, `[0, pi], y<>pi/2`; and inverse cotangent, `(0, pi)`.

If told to find values for these functions, you would look in quadrants I and IV for the first group and quadrants I and II for group two.  NEVER use values from the third quadrant when dealing with inverse trigonometric functions.  For example, if asked to find the value of `(sin^-1)x=1/2`, you could give either 30 or 150 as answers for arcsine, but you would get a minimal on a test if you used 150 as an answer for the function of `(sin^-1)x`.  This is because 30 is in quadrant I, 150 is in quadrant II, and answers for sine, which is in the first group, must be in either quadrant I or IV.

Tomorrow’s scribe is Tyler.

Today was a very exciting day in pre-calc.

We had a sub-again-so we got more time to work on our homework from yesterday.

This consisted of parametric equations using trig functions and inverse trigonometric functions and equations.  Working  in small groups, we found out that to finding the inverse relation of a sinusoid is the same as finding the inverse function of any other graph. You can do one of two things: reflect the graph over the line y=x, or you can switch the x and y variables around.  Either of these would have the same result.

We were also told to continue working on identity and proofs, and I think that our class is getting much more comfortable with those.

Hey everyone, i know that this is like basically a week late but i didn’t want to just not do it completely!! Okay so last friday (dec. 12) — (MY BIRTHDAY!!) we presented claims.  The claims went really well.  The first few were pretty easy as usual but then things started to get a little tricky.  We came across the conclusion that many times, there are more than one way to solve the problems and we even presented to different ways to do them in class.  As Rosie said, it helps if you break the problem down into its “simplest form” and work from that because that can help you see things in a different way.  Again so sorry about this guys and i promise that this wont ever happen again!!

Today in class (12/16) we received two worksheets, the first being titled Parametric Equations. The sheet defines a parametric equation as one in which x and y are both dependent variables, and depend on the parameter t. The example is given of an equation of a flying airplane, in which t represents time, x(t) represents distance, and y(t) represents height. Each coordinate therefore consists of its own function, and allows a Cartesian graph to show the plane’s flight path with respect to time.

The worksheet then tells us of one type of curve which can be represented by parametric equations, called an asteroid. The x-coordinate of an asteroid is given by the function x(t)=a(cos^3)t, and y-coordinate is represented by the function y(t)=a(sin^3)t. There is then a table given, in which a equals 4, and x(t) and y(t) are to be found for every t value between 0 and 360 (at increments of 30). After graphing these points, an asteroid is formed.

The second worksheet is then entitled Inverse Trigonometric Functions. The first page asks to accurately plot the six circular functions (sine, cosine, tangent, cotangent, secant, and cosecant).  The second page asks then asks to plot the inverse relations of the six circular functions, by reversing the x and y coordinates. Once this is done, you can see that these new inverse relations are not functions because they do not pass the vertical line test (drawing a line vertically down the graph, and because the line passes through more than one point, the graph has more than one output for a single input and is therefore not a function). The third page calls these inverse relations arcsine, acrcosine, arctangent, arccotangent, arcsecant, and arccosecant. It also states that our calculators are programmed to show inverse functions rather than these inverse relations when graphing sin^-1, cos^-1, and tan^-1 (this does not apply to the other circular functions, however). Inverse functions are portions of the inverse relation graphs that hold the qualifications of being a function by passing the vertical line test.

The homework was to complete both worksheets and continue doing identities and proofs from section 4-3.

Today in Precal we started with a question: Is 2sin40 = 2sin80?  We determined that this was false because we know the values of sin30 and sin60 and 2sin30 is not equal to sin60.  The sin of 80 would be equal to sin2(40), or sin2x.

Next, we looked at the area of a triangle.  We named an included angle theta and had x and y on either side.  We had previously known that A=`1/2`bh.  We drew in “h” for the height and determined that the sin`theta`=`h/x`; therefore, h=x`sin theta`.  So the area of a triangle is equal to `1/2 x y sin theta` when we know two sides and the included angle.  From now on we are supposed to accept that this is true and not worry about deriving it every time any more.

Next, we looked at 2 identical isosceles triangles with two sides “x” and an included angle of 80.  On the second triangle we drew in “z” for the height and had two 40 degree right triangles.  We had previously determined that the area of the triangle would be equal to `1/2 x x sin 80`.  We now wanted to solve for sin80.  Since the triangles were the same, the second triangle should have the same area as the first.  We came up with the area of the second triangle by adding the areas of the two separate triangles.  A=`1/2 x z sin40 + 1/2 x z sin 40`.  We set these values equal to eachother, and eventuall came up with sin 80 = 2`z/x`sin40.  `z/x` is equal to the cosine of 40 in the second triangle, so we came up with sin80=2sin40cos40.  From this activity, we came up with a general property, called the Double Angle Property, which states sin2x=2sinxcosx.  This works for any triangle where the angle is divided into two equal angles when the height is drawn in, such as an angle of 40 degrees with two separate right angles with a 20 degree angle.

Next, we had another true false question to answer.  Is sin(`theta`+20) equal to sin `theta` + sin 20?  In the previous problem, we had just determined that sin(20+20) is not equal to sin20+sin20, so this question is false.

How do we figure the areas when the two angles are not equal?  We made two triangles with a 75 degree included angle inside of sides “x” and “y.”  The second triangle was broken up with a height of “z” with two right 30 and 45 degree triangles.  The area of the triangle on the left, as we already determined, is equal to `1/2`xysin75.  The area of the triangle on the right would be equal to `1/2`xzsin30+`1/2`yzsin45.  We set these equations equal to eachother to solve for sin 75, eventually coming up with sin 75=`z/y`sin30+`z/x`sin45.  `z/y`=cos45 and `z/x`=cos30, so the derived answer was sin75=cos45sin30+cos30sin45.  The general rule for this, called the composite argument property, is sinA+B=cosAsinB+cosBsinA.

Finally, our homework for tonight was to redo both of these problems, the first with an angle of 60 or one that can be separated evenly, and the second one with any angle that isn’t equal.  Do these problems without looking at the examples we used today!  I hope this gave a good explanation but let me know if anything needs clarification!

Today in class we focused on the question: Is sin(2x)=2sin(x). We found out that the answer in no and we proceeded to find out what each term was equal to.  We used the area of a triangle to derive these formulas, one with a height added in and the other without. What we found out was that the area of a triangle is equal to: A=1/2x^2sin(80) and a=xzsin(40) with x being a side and z being a height. What we wer actually doing is proving that sin(2x) is not equal to 2sin(x) and that sin(2x) is actually equal to 2sin(x)cos(x). Another formula that we proved wrong was that sin(a+b)=sin(a)+sin(b). We disproved this formula in the same way as the first. What we found that it was equal to is sin(a+b)=sin(a)cos(b)+sin(b)cos(a).