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)