Precal Blog

Today we talked about transformations and identities and how to solve them using pythagoean, reciprocal, and quotient properties.

Reciprocal: 

sec x=1/cos x

csc x=1/sin x

cot x=1/tan x

Pythagorean:

This property is based off the pythagorean therorem which is a^2+b^2=c^2.  On a unit circle u^2+v^2=1 which means that cos^2x+sin^2x=1. 

Sin^2X/Cos^2X=cos^2X/cos^2X=1/cos^2X which give you one of the other ways to write the property.  It gives you tan^2X+1=sec^2X

Sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2x which give you the other way to write the property as 1+cot^2x=csc^2x

Quotient:

tan x = sin x/cos x = sec x/csc x

cot x= cos x/sin x = csc x/sec x

Example Proof:

Prove algebraically that csc?*cos^2?+sin?=csc?

Proof:                                                                                      (begin by writing proof:)

csc?*cos^2?+sin?                                                          (start on more complicated side)

=csc?(cos^2?+sin?/csc?)                                         (factor out csc?)

=csc?(cos^2?=sin?*sin?)                                         (reciprocal prop)

=csc?(cos^2?+sin^2?)                

=csc?                                                                                   (pythagorean prop)

csc^2?+sin?=csc?, QED

The next scribe is emma

Today in class we talked about transformations and identities.   We use Pythagorean, reciprocal, and quotient properties to solve them.

Pythagorean Property: this is derived from the Pythagorean theorem, a^2 + b^2=c^2.  On a standard unit circle, the hypothesis, 1 (c in Pythagorean theorem), equals u^2  (a^2) plus v^2 (b^2) .  Since we know already that (u,v)=(cos x, sin x) we can say that cos^2 x + sin^2 x = 1. This is how the Pythagorean property is derived.  There are also two other forms of the Pythagorean property.  To get them divide the original by sin^2 x and you get cot^2 x + 1 = csc^2 x. Get the other one by dividing it by cos^2 x: 1 + tan^2 x = sec^2 x.

Reciprocal Property: we have already had experience with this.

sec x=1/cos x

csc x=1/sin x

cot x=1/tan x

Quotient Property: The quotient property is simply a combination of the six trig functions we know.

tan x = sin x/cos x = sec x/csc x

cot x= cos x/sin x = csc x/sec x

Transformations: make one side of the equation equal the other side by using the Pythagorean property, reciprocal property, and/or quotient property.

This is the example we did in class:

To get sin x * sec x * cot x to equal 1, we need to use the reciprocal and quotient properties.  Since we know sec x = 1/cos and cot x= cos x/ sin x, we can cancel terms and come out with our answer.

Identities: prove one side of the equation equals the other by choosing one side to work on, usually the more complicated one.  The other difference from transformations is that you can’t assume one side equals another, that is what you are trying to prove.

Example:

In this identity, we took the more complicated side and factoring out csc x.  Then canceling terms will prove it equals csc x.

The next scribe will be reversearp………………………………jk its really streim

So sorry this is late too, i kinda forgot to do the post…lo siento.

Friday in calss we did more examples of arcCosine and arcSine problems but this time we added a range in which to find values.

The first problem we did was 2cos?+?3        [0   720] as our range

we solved it like before by taking the inverse cosine of (?3/2) and negative inverse cosine of (?3/2).  our answers wer 150 and -150.  tehn we found the period of the equasion(360) and added and subtracted untill we found all values that fit our criteria.  they were 150 510 and 210 570.

Next we solved a problem with arcsine.  5sin ?X=2.   [-2    4]

We also solved this like a normal arcsine equasion.  inverse sin(2/5) / ?  and ?- inverse sin(2/5) /?.  we found our answers to be .130… and .869…

we found our period of the equasion, 2, and added and subtracted to get all the numbers that fit into our range.  they were -1.869… .130… and 2.130…. for the other value they were -1.131… .869… and 2.869…

we did more practice problems untill the class was pros at it.

thanks(:

Sorry this is a few days late but nobody in our class decided to do the post so I took it.

The arcsine and arccosine functions are the inverse functions of sine and cosine, respectively.

We have used these functions previously in figuring out the measure of an angle.  But now we began discussing writing an equation using arcsine and arccosine in terms of x, and then, with a known y-value, solving for x.

As everyone knows, a sinusoidal graph contains infinite points at any certain y-value that is in the range of the function.  However, if we were to solve on our calculators we are only given one answer.  This is because our calculator only uses one period of the sinusoid when getting our answer.  For sine, that period is from -90° to 90° (in degrees) or from -pi/2 to pi/2 (in radians).  For cosine, that period is from 0° to 180° (in degrees) or from 0 to pi (in radians).  As a result, we have to make adjustments so we are able to get all the possible solutions for a specific y-value.

So as an example of sine, say we have the equation y=-1+2sin3x and we needed to solve for when y=0.  If you would like to see the graph take a look at taylor’s post because I don’t have it.  But as you notice, there are 2 values for when y=o in each period.  So we need to rewrite the equation and solve it.

0=-1+2sin3x

-1/2=sin3x    Now comes the hard part, you must take the inverse sine of the  equation, but you must also take pi-the inverse sine of the equation.  This must be done because you have to also take the supplement of the angle to find the other point at which the y-value is the same.  And because in radian mode 180° equals pi, you take pi minus the inverse sine.  So then you should have two equations which are

sin^-1(-1/2)=3x                and                       pi-(sin^-1(-1/2))=3x

x=(sin^-1(-1/2))/3          and                       x=(pi-(sin^-1(-1/2)))/3

x=pi/18                                and                       x=5pi/18

Those are the two values for one period of the sinusoid.  To get any other values simply add the period, which in this case is pi/3, to each of those values.

Essentially, the same concept is applied to equations using cosine.

The only difference is when trying to find the second point of the period, instead of taking pi minus the inverse cosine, you simply do negative inverse cosine.  Once again, you can look at the graph in taylor’s post to help you out.  If the same equation from above was cosine you would do the same process up until you took the inverse cosine and split it into 2 equations. For cosine you would do

cos^-1(-1/2)=3x          and          -cos^-1(-1/2)=3x

From there you would simply divide by 3 for both equations to get the x values.  Also, the same rule applies to cosine in that to get additional x points for that y-value simply add the period.

The hardest part for arcsine and arccosine is knowing the rules for when you apply those functions.  Other than that it shouldn’t be too bad.  It’s simply building on what we have already learned

The next scribe will be…Joel.

Today in class we learned about arcsine and arccosine. For those of you who are confused at what those words even mean, here are some definitions.

arcsine – the inverse function of sine

arccosine – the inverse function of cosine

Most of us have already used these functions and didn’t even know it. When we are trying to find the measure of an angle but you already know the value that it creates, you take the inverse sine or inverse cosine to find the angle.

Today in class, however, we took this concept and went a little bit further. We had to rewrite an equation in terms of x (we are using radians by the way) and with a known y-value, we had to solve for x.

There are infinite values for x because sine and cosine functions are periodic, meaning they repeat. Our calculators solve this problem by only using one period for sine, x -values from -90° to 90° (or -pi/2 to pi/2 in radians). In the case of cosine, our calculators use 0° to 180° (or 0 to pi for radians) as the period.

So lets get back to what we did today. We had to rewrite an equation. Let’s take this equation for example:

y=-1+2sin3x    and solve for x when y=0

If you look at the graph, there are infinite values for x but the one value that is solved for when we plug in the equation is always in the first period.

Let’s rewrite the equation.

0=1+2sin3x

-1=2sin3x

1/2=sin3x

sin^-1(1/2)=3x      This step is the most important

(sin^-1(1/2))/3=x

Now when we solve this, x should equal about .1745 or pi/18…(the dark filled-in circle on the graph)

Now to find the other point in the same period (open circle), the process gets a little more complicated.

pi – sin^-1(1/2)=3x You must take the supplement of the angle (because pi is equal to 180°, we must subtract from pi to get the supplement) so that means that the new angle is 5pi/6 because pi-pi/6 = 5pi/6.

5pi/6=3x

Then divide by 3

(5pi/6)/3=x

This value is about .87266 or 5pi/18.

When trying to find the second value on a cosine graph, you must take its opposite, not the supplement.

So for instance, imagine it were cosine instead of sine, the x value would end up being pi/9 (dark on the graph) which would mean the opposite value would be -pi/9 (open circle on graph).

The equations would look like this x=(cos^-1(.5))/3 and x=(-cos^-1(.5))/3

The hardest part about arcsine and arccosine is the writing of the equation. Once you understand how to write the equation, everything should make sense. Remember you have most likely been using arcsine and arccosine but now Mr. B is trying to trick us by making us graph them, apply transformations, and use them in radian mode.

The next scribe has to be Jocelyn. (sorry you’re the only one left)

Hi everyone, first of all sorry that I can’t be available for a Cover It Live tonight.

Also, I will not be including any cotangent functions on the exam tomorrow – that should be less for you to worry about.

Read the next post as well…

You have determined the period of your sinusoid. Now you need to write an equation. Since the “B” value tells you how many cycles of your transformed function fit into an untransformed version of your graph you can’t just write the period in the equation.
You have to figure out by what factor has the untransformed function been dilated to get the the transformed one.

So, for example, the period of your cosine function is 38 degrees. The untransformed function has a period of 360 degrees. So, by what factor has 360 been dilated to get 38? Essentially 360x = 38. The answer is 38/360 yes? Maybe simplify to 19/180. Now, 19/180 is the
dilation factor but as always with horizontal transformations, what is actually in the equation in the inverse of the transformation. So our final equation is y = a + c[cos 180/19 (theta - d)].

Please spend time connecting what we did in chapter 1 to this current dicsussion – we are doing the exact same thing: you are determining the factor by which the preimage was dilated in order to get the image. Then you are taking the reciprocal of that and putting it into the equation. If the mystery graph was dilated by 2, then you wrote 1/2 in the equation.

What about radians? No difference. Say your sine function has a period of pi/6. By what factor did we dilate 2pi to get pi/6? Essentially 2pi x = (pi/6) and x = (pi/6)/(2pi). This gives us the dilation factor of (1/12) so in the equation we are writing 12.

Work backwards now: if we see 12 in the equation then the dilation factor is (1/12), so (1/12) of 2pi is (pi/6) which is our period.

All of the above holds for tangent and cotangent except you have to remember that untransformed tangent and cotangent have periods of pi or 180 degrees – not 2pi, 360.

Well… Today in class we worked on Problem Set 3-5,  and learned how to compare radians to degrees. To solve most of the problems, you could have used the plate tools we made in class, mainly if you haven’t memorized the radian values. Here’s some examples of the beginning problems to get you started.

60°

To solve this, all you have to know what sixty degrees is equivalent to in radians. If you don’t know the answer off the top of your head, you can refer to your plate, and see that the answer is ?/3.

45°

For this one, you do the same thing you just did for the previous problem, and the answer is ?/4.

We also did problems involving the six trigonometric functions, but in radian mode.

An example of this would be sin2. In Radian mode it would equal .9093.

Another example is tan3. In Radian mode it would equal -.1425.

So that’s all we did in class today.

Hello Honors Pre-Calculus Hour 3! Sorry for the delay in my post, Emily took my scribe because I was absent and I said I would take hers in return…and I subsequently forgot. But better late than never hopefully Thank you for your patience.
At any rate, on Wednesday, December 2nd we worked on the Problem Sets for chapters 3-4 and 3-5.
The next scribe will be…Jake (assuming he hasn’t gone yet and that this counts as my post )

Today we contintinued our paper plate activity from yesterday. We started the day with our tape already folded into 1/3’s and continued to fold our paper until we had 1/12’s. After distinguishing these markings on the tape we then marked them on our paper plates. Our final product of the paper plate is referred to as a unit circle. The image below is similar to the markings we made on the plate.

Source: http://www.mathsisfun.com/geometry/unit-circle.html (This link also has interactive tools that explain relationship between radians and degrees)

This image also shows the calculations of sine and cosine for each radian measure.

After finishing our unit circles we began making the connections with how radians relate to degrees.

60° = ?/3 radians

Since we are still dealing with the same proportions, sin and cos of both 60° and ?/3 radians will be the same.

sin(60°)= 0.866025….

sin(?/3)= 0.866025…..

This also means that the graphs for sin and cos in degrees will appear the same as the graphs for sin and cos in radians. The only difference will be the horizontal scale.

Homework for tonight: Read and fill out yellow sheets for 3-4 and 3-5

*Exam Next Tuesday

The next scribe will be Emily