Week 5 Prompt

What is trigonometry according to you? (do not use a source for this) WHat professions use trigonometry? Give an example of how it is used in their day to day job. (use a source for this. Everyone should not have the same source!)

9-1

In this section we learned how to find the sides and angles of a triangle. This involved the use of a little thing called SOHCAHTOA. This stans for Sin=Opposite/Hypotenuse. Cosine=Adjacent/Hypotenuse, and tangent= opposite/adjacent. If you invert these you can also find out how to use cosecant, secant, and cotangent. I don't know when I will ever use this outside of school but I'm not gonna complain because this may be the easiest thing we have done so far this year. The steps are simple and the formulas are easy to remember because of SOHCAHTOA. So basically you are given a right triangle and a combination of side lenghts or angles. Then depending on what you are given you must find out the rest of the information required to get the answer.

9-1

This week in advanced math we learned about Chapter 9 section 1. We learned how to solve angles and lines for right triangles and iscolceles triangles. We learned how to take information out of the world problem and plug them into the triangle. The line touching the right angle is the adjacent side, the side opposite the right angle is the hypotenuse, the side not touching the right angle is the opposite side. To find the angle you do not have in a right triangle you do 180 minus 90 minus the other angle that they give you, then you find the missing angle. We also learned SOHCAHTOA which helps to remember that sin= opposite/hypotenuse, cos= adjacent/hypotenuse, tan=opposite/adjacent. With iscoceles triangles we learned that you split them into two right triangles, then you solve for one of the triangles, then you multiply it by 2.

Ex. How to find the other angle.

If you have a right triangle, one angle is 90 degrees, and lets say they give you another angle that is 77.

You have to do 180 minus 90 minus 77 to find the missing degree.

180-90-77= 13

Therefore the missing angle is 13 degrees.

Section 9-1

For these problems to be worked in 9-1, you have to know SOHCAHTOA. This stands for:
Sin= opposite/hypotenuse Cos= adjacent/hypotenuse Tan= opposite/adjacent

To solve a problem, you simply draw your right triangle and add in the degrees that they give you. Once you are done, you look to see what they are asking you to solve, and you look for the appropriate formula. Then you are done. If you can remember SOHCAHTOA, you can remember all the formulas because secant, cosecant, and cotangent are all reciprocals of these.

Opposite is the degrees that is across from the one that is already given to you. Adjacent is the degrees that is attached to the one already given, and hypotenuse is always the degrees across from the right angle.

Solve the triangle ABC.
Angle B- 34degrees Angle b- 26 degrees Angle A- 90 degrees
Those are the angles given, so you need to find angles a, C, and c.

9-1

This week in Advanced Math, we took a test on chapter 8 and moved on to chapter nine. In this chapter, we learned how to use SOHCAHTOA to find different sides of a triangle.
SOHCAHTOA stands for Sin equals Opposite over Hypotenuse, Cos equals Adjacent over Hypotenuse, Tan equals Opposite over Adjacent. Besides those three, Csc, Sec, and Cot are also used. Csc equals hypotenuse over opposite Sec equals hypotenuse over adjacent and Cot equals adjacent over opposite.
But, there are some restrictions. These can only be used with a right triangle. The hypotenuse is the longest side which means it is opposite of the right angle. Never confuse the hypotenuse with the adjacent side.


Example-

If you look out of a third story window 20 feet in the air to the top of a skyscraper 400 feet away and the angle of elevation is 35 degrees you and the top of the skyscraper, how tall is the skyscraper?
In this case, you would take tan 35 and set it equal to y over 400.
Then solve for Y.
Y = 400 tan 35
Y then equals 280.083.
After that you must add 20 because of the 20 feet you were already up in the air.
Your total would then be 300.083

9-1

In Advanced math we took a chapter test, and now we are currently working on completing chapter nine. Chapter nine, and Mrs. Robinson teach us how to find the unknown side and angle of a right triangle. I think this is one of the easiest thing that we have learned so far, because it is almost just a repeat of what we learned in Geometry. They have a word that will help you to do well in this chapter it is known as SOHCAHTOA .

S= sin= Opposite/ Hypotnuse
C= cosine= Adjacent/ Hypotnuse
T= tangent= Opposite/ Adjacent

HERE ARE ALL OF THE FORMULAS TO REMEMBER:

sin= opposite/hypotenuse sec= hypotenuse/adjacent
cos= adjacent/hypotenuse csc= hypotenuse/opposite
tan= opposite/adjacent cot= adjacent/opposite

EXAMPLE:

For triangle ABC

A= 90(degrees) B= 16 a= 34

First, since we already have A and B we need to find C. In order to do that all we are going to do is take 180 (which is the measure of a right triangle) minus 90 your right angle minus 16 the measure of angle B. Which will leave you with 74 and that is angle C.

Next, you have to find b and c. In order to do that you are going to have to take sin, cos, or tan of the angle B using your formula. For this problem you will use sin.

sin= opposite/ hypotnuse
sin 16 degrees= b/34
34sin16= 9.37
Now, you are going to need to find C. You are going to have to use cos= adjacent/ hypotnuse.
cos 16 degrees= c/34
34cos16= 32.68

And that is how you do section 9-1.

Chapter 9

Chapter 9 in Advance Math is a continuation of Geometry. We discovered how to fine the unknown side of a 90 degree triangle. In order to be succesful in this section, You have to remeber SOHCAHTOA.

SOHCATOA -
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
csc=hypotenuse/opposite
sec=hypotenuse/adjacent
cot=adjacent/opposite

Example 1: For a triangle ABC, C=90 degrees, A=30 degrees, a= 24
1.Since angles A and C have already been defined, we have to solve for B. Subtract 30 and 90 degrees from 180 to get your answer. Angle C=60 degrees. All angles have been defined.
2. We were given angle A, so to solve for b and c use the functions above. To solve for b, use sin,which equals opp/hyp- sin30 degrees = b/24
3. Multiply both sides by b and continue solving - b=12
4.Now that we have solved for a and b, we must solve for c by using cos-adj/hyp.
cos30degrees=c/24,
5.Multiply c by both sides and get 12 square root of 3=20.784

Chapter 9 in Advanced Math brings us back to Geometry. We learned how to fine the unknown side of the right triangle. The key to this section is to remember SOHCAHTOA.

SOHCAHTOA
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
csc=hypotenuse/opposite
sec=hypotenuse/adjacent
cot-adjacent/opposite

EXAMPLE 1: For triangle ABC, C=90 degrees, A=26 degrees, and a=38
1). Since we know angles A and C, let’s solve for B. Subtract 90 degrees and 26 degrees from 180 degrees to get your answer. Angle C = 64 degrees. Now we have all of our angles.
2). We are given angle A, so to solve for b and c use the functions above. To solve for c, use sin which is opp/hyp - sin 26 degrees = 38/c
3). Multiply both sides by c and continue solving - c = 86.684 degrees
4). Now that we have a and c, we need to solve for b. We will use tan which is opp/adj - tan 26 degrees = 38/b.
5). Multiply both sides by b and continue solving - b = 77.911 degrees

This week we finished chapter 8 and started learning chapter 9. Chapter 9 takes us back to what we learned in geometry.

SOHCAHTOA
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
csc=hypotenuse/opposite
sec=hypotenuse/adjacent
cot-adjacent/opposite

For most problems you will either use sin, cos, or tan.

Example:
for triangle ABC
A=90 degrees B=24 degrees a=36

Since we already have angles A and B we need to find C.
1) first subtract 90 degrees and 24 degrees from 180 degrees.
your answer is 66 degrees.
2) in order to find angles b and c you need to use angle B to solve. To solve for b you can use sin-opp/hyp. Sin 24 degrees=b/36.
3) 36sin24 degrees=14.643.
4) to find c you need to use cos=adj/hyp.
cos24 degrees=c/36
5) 36cos24 degrees=32.888

Since we have everything worked out your answers are:
A=90 degrees
B=24 degrees
C=66 degrees
a=36
b=14.643
c=32.888

At first I was completely lost until I sat down and worked a few problems. I somewhat remembered SOHCAHTOA from geometry and its very easy when working these problems.

Section 9-1

This week in Advanced Math we started on Chapter 9. In section 9-1, we learned how to find the unknown sides to a right triangle. We already learned this in geometry but now we are taking a few extra steps to solve these problems. The key to this section is to remember SOHCAHTOA. It stands for:

Sine = Opposite/Hypotenuse
Cosine = Adjacent/Hypotenuse
Tangent = Opposite/Adjacent

The reciprocals for these formulas are:

Cosecant = Hypotenuse/Opposite
Secant = Hypotenuse/Adjacent
Cotangent = Adjacent/Opposite

*Hypotenuse is always going to be opposite of the right angle.

Let’s try an example problem:

EXAMPLE 1:

Solve triangle ABC
Angle A = 90° Angle B = 36° a = 28

1) Since we already know angles A & B we must find C. You will subtract angles 90° and 36°from 180° to get an answer of 54°.
2) To get sides b and c you will need to use angle B. To solve for side b,
you will use sine's opposite/hypotenuse. Sin 36° = b/28
3) Multiply to get 28 Sin 36° = 16. 458
4) To get side c you will use cosine’s opposite/adjacent. Cos 36° = c/28
5) Multiply to get 28 Cos 36° = 22.652

Angle A = 90°
Angle B = 36°
Angle C = 54°
Side a = 28
Side b = 16.458
Side c = 22.652

Chapter 9

This week we completed chapter eight and began chapter nine. The beginning of chapter one is trigonometry; solving right triangles. Section 9-1 goes back to chapter seven ;using trigonometry functions, as well as Geometry. A few additional steps are included. One of the most helpful tips to remember when using right triangles is SOHCAHTOA.

Sine: opposite/hyp
Cosine:adjacent/hyp
Tangent: opposite/adjacent

You must also remember that almost triangles' angles add up to equal 180 degrees. There are also special triangles such as the 45/45 and 60/30 that should be remembered.

When looking for the sides of a triangle use either pythagorean theorem ; a^2+b^2=c^2 , or trigonometric functions.

EXAMPLE: in triangle DEF you are given measure of angie D which is 90 degrees, measure of angle E which is 12 degrees, and the length of side E which is 9.
In order to solve this triangle you must find the measure of angle F, and the lengths of sides d and f .

To find the missing angle we must remember that all triangles add to equal 180 degrees. We see that the sum of our given angles is 102. Right angles equal 90 degrees plus the 12 degrees of angle E equal 102. Subtract from 18o , E is equal to 78. Next we must find the sides. In order to find side f we use the trig function tan, opposite/adjacent, of the given 12 degrees.

Tan 12=9/f . Cross multiply and divide to get 42.3. Therefore the length of side f is 42.3.
You now have two options to find the last side. You can 1. Use pythagorean theorem, or 2. Use the trig functions again. It's best to use the trig function again just in case your first answer was wrong.

To find side d you take sine 12 =9/d. Again you cross multiply and divide and you get 43.3. That is the length of side d.

Angles : 90, 12, 78.
Sides: 9, 42.3, 43.3
Now you have solved triangle DEF.

Chapter 9-Section 1

In Chapter 9, section 1 we learn how to solve right triangles. This lesson is a simple repeat of what we had learned in geometry. To help solve these problems and make memorizing the formulas easier you must remember SOHCAHTOA.

SOHCAHTOA is:
Sin (S)= opposite (O)/hypotenuse (H)
Cos (C)= adjacent (A)/hypotenuse (H)
Tan (T)= opposite (O)/adjacent (A)
Once you can remember SOHCAHTOA, you can remember all the formulas. Secant is the reciprocal of sine, cosecant is the reciprocal of cosine, and cotangent is the reciprocal of tangent.

When looking at a triangle, opposite is the degrees across from the angle already given to you. The degrees already attached to the given is the adjacent angle, and hypotenuse is ALWAYS the degrees across from the right angle.

To solve a problem, draw your triangle and plug in the points and angles given to you. Some triangles may be named. Once you have drawn the triangle, find what they are asking you to solve (the missing angles or points), and you look for the appropriate formula. Once you have a degree to each angle you are done. Make sure to label all angles and points on the triangle, if you want credit.

Solve the triangle QRS.

Angle R- 34degrees Angle r- 26 degrees Angle Q- 90 degrees Those are the angles given, so you need to find angles q, S, and s.

In order to find S, simply subtract 180-90-34 to compose an answer.
Angle S= 56 degrees

In order to find q, use the formula for sin which is opposite/ hypotenuse
Angle q= 46.496 degrees

In order to find s, use the formula for tan, which is opposite/ adjacent
Angle s= 38.547 degrees

9-1

This week we started Chapter 9. It is about using trigonometry to find unknown sides or angles of a right triangle. In Chapter 7, we defined the trigonometric functions in terms of coordinates of points on a circle. In this chapter, our emphasis shifts from circles to triangles. To solve right triangles, you must remember SOHCAHTOA which is:

sin theta = opp/hyp
cos theta = adj/hyp
tan theta = opp/adj

As we learned, cosecant, secant, and cotangent are all reciprocals:
csc theta = hyp/opp
sec theta = hyp/adj
cot theta = adj/opp

To solve right triangles, you also need to remember that hypotenuse is opposite of the right angle.
Here’s an example:
EXAMPLE 1: For triangle ABC, C=90 degrees, A=26 degrees, and a=38
A). Since we know angles A and C, let’s solve for B. Subtract 90 degrees and 26 degrees from 180 degrees to get your answer. Angle C = 64 degrees. Now we have all of our angles.
B). We are given angle A, so to solve for b and c use the functions above. To solve for c, use sin which is opp/hyp - sin 26 degrees = 38/c
C). Multiply both sides by c and continue solving - c = 86.684 degrees
D). Now that we have a and c, we need to solve for b. We will use tan which is opp/adj - tan 26 degrees = 38/b.
E). Multiply both sides by b and continue solving - b = 77.911 degrees

That’s it! You’ve solved the triangle. This section is simple if you know your formulas, draw your triangle, and insert your answers in it.

Solving Right Triangles 9-1

This chapter so far, is the easiest one we have had yet. It is pretty much geometry that we already learned, but with a few extra steps. For these problems to be worked, you have to remember SOHCAHTOA. This stands for:
Sin= opposite/hypotenuse
Cos= adjacent/hypotenuse
Tan= opposite/adjacent

If you can remember SOHCAHTOA, you can remember all the formulas because secant, cosecant, and cotangent are all reciprocals of these.
Opposite is the degrees that is across from the one that is already given to you. Adjacent is the degrees that is attached to the one already given, and hypotenuse is always the degrees across from the right angle.

To solve a problem, you simply draw your right triangle and add in the degrees that they give you. Once you are done, you look to see what they are asking you to solve, and you look for the appropriate formula. Then you are done.

Solve the triangle ABC.
Angle B- 34degrees Angle b- 26 degrees Angle A- 90 degrees
Those are the angles given, so you need to find angles a, C, and c.

To find C, you just simply subtract 180-90-34 to come up with your answer.
Angle C= 56 degrees

To find a, you will use sin which is opposite/ hypotenuse
Angle a= 46.496 degrees.

To find c, you will use tan, which is opposite/ adjacent
Angle c= 38.547 degrees.

Well, i am sure i am not the only one having problems with advanced math all together. For me and a few other people this is the easiest yet. Right now we are in Chapter 9 working with right triangles. This brings us back to the things we learned in geometry. Right triangles are all about 90°, Angles, opposites, adjacents, and a hypotenuse. Going along with right triangles is:

SOHCAHTOA; which stands for
Sin Opposite/Hypotenuse
Cosine Adjacent/Hypotenuse
Tangent Opposite/Adjacent

or

csc=hyp/opp
sec=hyp/adj
cot=adj/opp

These formulas are only used with right triangles.
*A hypotenuse is opposite of the right angle, always.
*The smaller letters are always opposite from the capital.

EXAMPLE:

Solve the triangleABC
-AngleC=32° AngleA=90° Anglea=26

On the triangle you need to find angles B,b,and c.
Your man point is angle C and across from that is B. In order to find B you subtract 180-90-# which is 32. To find b, which is adjacent to C, you will use cosine because you already know the hypotenuse (a). To find c, which is opposite to C, you will use sine.

B 180-90-32=68°
b cos32=b/26 (use simple algebra to find b) 26cos32° =22.0493
c sin32=c/26 26sin32°= 13.7779

This lesson is as simple as that. Just know the 3 formulas and you are set.

9-1 Solving Right Triangles

In advance math this week we learned about right triangles. In 9-1 we learned how to solve right triangles and how to identify all the sides of a triangle. The term we learned in 9-1 dealing with right triangle and how to solve. It is called SOHCAHTOA.

SOHCAHTOA stands for:
-Sin theta= opp/hyp
-Csc theta= hyp/opp
-Cos theta= adj/hyp
-Sec theta= hyp/adj
-Tan theta= opp/adj
-Cot theta= adj/opp



All those function which is known as SOHCAHTOA is only used with right triangles. Another thing is the hypotenuse is opposite of the right angle.


In chapter 7, we defined the trigonometric functions in terms of coordinates of points on a circle. In this chapter, our emphasis shifts from circles to triangles. When sides and angles of a triangle are known, you will see that trigonometry relationships can be used to find the unknown parts. This is called solving a triangle. For example, if you know the lengths of the sides of a triangle, then you can find the measures of its angles. This lesson is basically explaining how trigonometry can be applied to right triangles.



Examples of solving a right triangle:

For triangle ABC
Solve the triangle
- they give you
Angle C=90degrees
Angle A=28degrees
A=40
- you are looking for
Angle B
C
B
- finding C you do Sin 28degrees=40/c.
- multiply c on both side you get c sin 28degrees= 40.
- you divide sin 28degrees over.
- you get c=40/sin 28degrees.
- the answer you get is 85.202 which is C.
- Finding B you do Tan 28degrees=40/b.
- multiply b over you get b tan 28degrees=40.
- you divide tan 28degrees over.
- you get b=40/tan 28degrees.
- the answer you get is 75.229 which is B.
- Finding angle B
you do 180-90-28.
- you get 62 degrees.
- you solve the whole triangle and you are done.



Basically if you know all your steps and formula and SOHCAHTOA . Lesson 9-1 is really easy and not hard but just takes time and be careful and know what is your hypotenuse because you can messed up your whole problem if you do not know which one is the hypotenuse but other than that 9-1 is easy and you would not have any trouble at all if you know SOHCAHTOA.

Radians are really important to use in trig because they they are much more simple to work with, rather than working with degrees. When you have radians you are only dealing with pie which can be worked in and out with any type of equation. Where as if you use degrees you are going to have to graph it figure out where the equation is located on the graph and then convert to radians.

When working with theta you ALWAYS want your answer to be in radians. When you are given theta in degrees it is very simple to convert into radians, you and going to multiply by pie over 180 degrees. This will cancel out the degrees leaving you with a faction most of the time and occasionally a whole with pie in the numerator part of the fraction or when dealing with a whole number pie will be right next to the number as if the number were being multiplied by pie.

Examples:

1. (theta)=150(degrees)
150x (pie)/ 180
=5(pie)/6

2. (theta)=270(degrees)
270x(pie)/ 180
=3(pie)/2

3. (theta)=360(degrees)
360x(pie)/ 180
=2(pie)

Week 4 Prompt

Why is it important to use radians in trig? Explain how to convert to radians. Give an example problem.

Pythagorean Identities

One of the things we learned in the week was using pythagorean identities to help simplify trigonometric equations. Some identities are

1 – cos^2θ = sin^2θ
1 – sin^2θ = cos^2θ
sin^2θ – 1 = - cos^2θ
cos^2θ – 1 = - sin^2θ
csc2 x – 1 = cot^2x
sec2 x – 1 = tan^2x
sec2 x – 1 tan^2x = 1
1 – csc^2x = -cot^2x
1 – sec^2x = -tan^2x
tan^2x – sec^2x = -1
csc^2x – cot^2x = 1
cot^2 – csc^2x = -1

The steps to solving these problems are:
1) Check the problem for Pythagorean identities : sin^2θ + cos^2θ = 1
1 + tan^2θ = sec2 θ
1 + cot^2θ = csc^2θ
2) Change everything to sin, cos, or tan.
3) Use algebra to simplify the problem. (FOIL, etc)
4) Plug in anymore identities.
5) Solve the problem using algebra.


sin x (csc x - sin x)
sin x(1/sin x - sin x)
1 - sin squared x
cos squared x

8-3

This week in advanced math we covered most of chapter 8. In 8-1 we learned to find cosine and to plug it into the right quadrant. We also learned the angle of inclination in 8-1. In 8-2 we learned to find the amplitude and the period, along with how to find the 5 points and graph with a period change. To do this we had to divide the points by B then graph them accordingly. In 8-3 we learned an expansion of 8-2.

8-3

Amplitude= max-min/2

y= Asin(Bx+/- C) +/- D

y= Acos(Bx +/- C) +/- D

Steps to graph

1) Get 5 points on the graph and divide by B.

2) Add or subtract C from those 5 points. Do the opposite of what C is in the formula.

3)Move all y values up D or down D.( If D is + go up and vise versa)

Ex. 3Sin(2x+2)- 4

1)o 0/2=0 0-2= -2

2)pie/2 pie/2/2 = pie/4 pie/4-2= -7(pie)/4

3)pie pie/2= pie/2 pie/2 -2 = -3pie/2

4) 3(pie)/2 3(pie)/2/2=3(pie)/4 3pie/4-2= -5(pie)/4

5)2(pie) 2(pie)/2=pie pie-2= -1(pie)

Next you would graph these points on the graph from the orgin since it Sin. Then you would subtract 4 from the y of every one of those points and graph that.

Jeff:chapter 8-2

Last week in Advance math, we learned some of chapter eight. The section that i understood the most from this year so far was section 2 of chapter 8. eight-two seemed to be the easiest concept that we have learned all year. In section eight two, we learn how to find the amplitude and the period.In order to graph the things with a period change you must first:1.find all five points.2.divide by b 3.sketch the graph,which seems to be the easiest part of the process.

the amplitude is the absolute value of the distance from zero to the highest point and half the vertical distance. the period is how long it takes to repeat itself.

Ex:1 y=4 sin1/2x
1.amp:4
2.period=2pi/(1/2)=2pi times 2 = 4 pi
3.find your five points(0,pi/2,pi,3pi/2,2pi)
4.divide by b(1/2)
5.take the reciprocal which gives u 2 and multiply
6.0x2=0
7.pi/2 x 2= pi
8.pi x 2= 2pi
9.3pi/2 x 2= 3pi
10.2pi x 2= 4pi
11.sketch the graph with 2nd set starting at the origin(0,0).

8-5

The easiest section this week was by far section 8-5. It was really just a carryover of 8-1 with some add on with trig identities from 8-4. If this isn’t possible you change into tan or cot by either putting sin over cos or by putting cos over sin. If this step is not required it is just as simple as getting the trig function by itself and taking an inverse. You may have to divide by another number if there is one but if not you find the quadrants where the answers lie and perform the operations to figure them out.

Ex.
7cosA=4
Divide by 7

cosA=4/7
divide by cos or in better words take the inverse

A=cos^-1(4/7)
You get 55.15.

Since cos is positive in the I and IV Quadrant you take the original value, make it negative and add 360.
 Your two answers ar 55. 15 and 304.85

Section 8-5

This week we learned and reviewed sections of Chapter 8. In 8.5 we learned how to use identities to get to be the same. Trig function= if you can’t try to make it tan or cot by sin/cos or cos/sin. When you are working in 8.5 you will have to follow the inverse rules. There are things you can’t do in 8.5 and if u do them you will get the problem wrong. The things you can’t do are: Divide by a trig function when solving to cancel. Cancel from the inside of a trig function.

Ex: 2sin^2theta-1=0
1 trig function and has 1 term.
You move the 1 over.
You get 2sin^2theta=1.
Divide the 2 before sin by each side.
You get sin^2theta=1/2.
Square both sides.
Sin theta= + and – square root ½.
Do sin inverse square root ½.
You get 45.
Find Q2, Q3, Q4.
Q2 you do 180-45=135. Q3 you do 180+45=225. Q4 you do 360-45= 315.
Your

answer is: theta= 45 degrees

, 135 degrees, 225 degrees, 315 degrees.

8-4

We are working on section 8-4 this week in advanced math. I feel that it is one of the easiest things that we have learned all year if you can memorize the formulas. All you are doing is replacing the trig functions with Pythagorean identities. Below I will show you how to do an example step by step. You won’t need to many examples before you get the hang of this section.


GUIDLINES:
1.pythagorean identities
2.move everything to sin and cos or tan if possible
3.algebra to simplify
4.identities again
5.algebra again


First to start off I’ll give you a few easy Pythagorean identities to plug into the trig functions.
csc = 1/sin
cot =1/tan
sec =1cos
tanx=sinx/cosx
cotx=cosx/sinx

EXAMPLE 1.(secx)(tanx)secx= 1/cosx and tanx= sinx/cosx(1/cosx)(sinx/cosx)=sinx/cos^2x=tan^2x is going to be your final answer

The section I understood the most this week is section 8 – 5. It is about using identities to get the same trig function in a equation. If that is not possible, then you can change everything to tangent and cotangent by sin / cos or cos / sin. Follow all inverse rules when working these problems. The things you can not do when working these problems is dividing by a trig function when you try to cancel things. You can not cancel the inside of the trig functions. If you have sin(9x)/ 9, you can not cancel the 9’s. you also can not divide and cancel cos^2x=cosx.
Example 1:
1. Sec^2 theta = 16
2. Square root(sec^2 theta) = square root(16)
3. Sec theta = 4
4. Theta = sec^-1 (4)
5. Sec^-1 is located in the first and second quadrant.
6. Theta = 75.5 degrees, 104.5 degrees.
This section was overall pretty easy once you understand it.

8-1

In section One chapter Eight, we learned how to find the angle of inclination.
For a line, m= tan alpha
If A doesn't equal C for a conic, use tan 2 alpha = B/A-C
If A does equal C for a conic, use alpha = pi/4
Ax^2 + By + Cy^2 + Dx + Ey + F = 0
alpha = angle of inclination
To the nearest degree, find the inclination of line
2x + 5y = 15
line = m = tan alpha
-2/5 = tan alpha
alpha = tan ^ -1 (2/5)
alpha = 21.801
Then one must take the alpha and move it to section II and IV
To go from I to II subtract 180
To go from I to IV and make negative then add 360
So in the end, the answer is
Alpha = 158.199 and 338.199
Find the direction of angle alpha,
X ^2 – 2xy + 3y^2 = 1
First find out if A = C
Since 1 does not 3 you must use the formula tan 2 alpha = B/A-C
So, tan 2 alpha = -2/1-3
Then, tan 2 alpha = 1
2 alpha = tan ^-1 (1)
2 alpha = 45, 225
So, alpha = 22.5 and 112.5

This week in Adv. math we learned different sections in chapter eight. In section 8-4 you are pretty much just replacing all the identities with trig functions. All you have to do is memorize and follow your chart. This is definitely one of the easiest sections if you memorize all your functions.

STEPS:
1. Pythagorean identities
2. Move everything to sin and cosine OR tangent IF POSSIBLE
3. Algebra to simplify
4. identities again
5. Algebra again

Ex.
sin theta (csc theta - sin theta)
sin theta(1/sin theta - sin theta)
1 - sin squared theta
cos squared theta

8-4

Section 8-4 is all about substation. The key to this section is memorizing identities and remembering some algebra. If you aren’t good at memorizing you can derive the identities and just remember 3 of them. If you can memorize them these are the ones you should know:

1 – cos^2θ = sin^2θ
1 – sin^2θ = cos^2θ
sin^2θ – 1 = - cos^2θ
cos^2θ – 1 = - sin^2θ

csc2 x – 1 = cot^2x
sec2 x – 1 = tan^2x
sec2 x – 1 tan^2x = 1
1 – csc^2x = -cot^2x
1 – sec^2x = -tan^2x
tan^2x – sec^2x = -1
csc^2x – cot^2x = 1
cot^2 – csc^2x = -1

The following guidelines will help you solve some of these equations:
1) Check the problem for Pythagorean identities : sin^2θ + cos^2θ = 1
1 + tan^2θ = sec2 θ
1 + cot^2θ = csc^2θ
2) Change everything to sin, cos, or tan.
3) Use algebra to simplify the problem. (FOIL, etc)
4) Plug in anymore identities.
5) Solve the problem using algebra.
Example Problems:

Example 1:
1-cos^2x cos^2x = tan^2x
sin^2x/cos^2x = tan^2x

Example 2:
sec^2x – 1(csc^2x – 1) = 1
tan^2x(cot^2x)
sin^2x/cos^2x(cos^2x/sin^2x) = 1
(Cancel each other out)

This week in advanced math we are working on section 8-4. I think that it is probably one of the easiest things that we have learned all year. All you are doing is replacing the trig functions with Pythagorean identities. I will show you how to do this in the following steps followed by an example. You will only need one example I promise! If you need more than one example then too bad you aren't getting another one.

Now remember this may be easy for me because I have the best advanced math teacher in the entire world and I am a genius but for you it may be a little confusing, but your still only getting one example.

STEPS:
1. Pythagorean identities
2. Move everything to sin and cosine OR tangent IF POSSIBLE
3. Algebra to simplify
4. identities again
5. Algebra again

EXAMPLE #1 and the ONLY one:
1.(secx)(tanx)
secx= 1/cosx and tanx= sinx/cosx
(1/cosx)(sinx/cosx)
=sinx/cos^2x
=tan^2x is going to be your final answer.

8-5

In Advanced Math this week, we learned and reviewed sections of Chapter 8. In 8.5 we learned how to use identities to get to be the same. Trig function= if you can’t try to make it tan or cot by sin/cos or cos/sin.

When you are working in 8.5 you will have to follow the inverse rules.

There are things you can’t do in 8.5 and if u do them you will get the problem wrong.
The things you can’t do are:
-Divide by a trig function when solving to cancel.
-Cancel from the inside of a trig function. (Examples sin (2x)/2=you can’t cancel the 2 and the other one is sin^2x=sinx=you can’t divide and cancel.)

Example: 2sin^2theta-1=0
- 1 trig function and has 1 term.
- You move the 1 over.
- You get 2sin^2theta=1.
- Divide the 2 before sin by each side.
- You get sin^2theta=1/2.
- Square both sides.
- Sin theta= + and – square root ½.
- Do sin inverse square root ½.
- You get 45.
- Find Q2, Q3, Q4.
- Q2 you do 180-45=135. Q3 you do 180+45=225. Q4 you do 360-45= 315.
- Your answer is: theta= 45 degrees, 135 degrees, 225 degrees, 315 degrees.

By looking at 8.5 if you know the rules and everything else. It is really easy but just takes time and effort.

Chapter 8 Section 1

This week we learned chapter 8. This chapter is neither hard nor complicated , but rather tedious. There are a lot of rules you must follow and it is a lot to remember. The first thing we learned was Angles of Inclination, and solving simple trigonometric equations. We later learned identies, and the curves of sine and cosine.
In section 8-1 we learned to find angles of inclination.
Lines must be in y=mx+b form, or slope intercept form. Conics however, they can be in point slope form.


Ex. The inclination of a line is 140°, what is slope.
To find slope”m” we use the formula m=tanα .
m=tan(140) α= angle of inclination*
m=-.8391  final answer .


Ex. The slope of a line is 3/5, find its angle of inclination.
It is given that m=3/5. We are looking for angle of inclination, α .
m=tanα
3/5=tanα
α=tan-1(3/5) * to solve for alpha you must take the inverse.
α= 31°

8-1

In Advanced Math this week, we learned and reviewed sections of Chapter 8. In 8-1 part II, we learned how to find angle of inclinations and direction angles.

To find the angle of inclination, keep in mind that alpha = angle of inclination.
Also, m=tan alpha for a line.
tan 2alpha = B/A-C – if A is not equaled to C
alpha = pi/4 – if A = C

EXAMPLE 1: To the nearest degree, find the inclination of the line 3x+5y=15
A). First thing you need to do is solve for y. So, you will then get y=3/5x-3.
B). Next, you know that m=tan alpha so look at the problem and the number before x will be your tan alpha. In this problem it is 3/5.
C). Now find the inverse.. alpha = tan^-1(3/5) *(Remember, you have to find the inverse of a positive number).
D). Draw your coordinate plane. Your quadrant 1 angle is 40 degrees. Tangent is positive in quadrant 3, so add 180 degrees to 40 degrees to get 220 degrees.

EXAMPLE 2: Find the direction angle alpha, x^2-2xy+y^2=1.
A). According to the notes above, check your A and C first. They are both 1, so your answer will simply be pi/4.

EXAMPLE 3: Find the direction angle alpha, -6x^2-2xy+4y^2=1.
A). According to the notes above, check you’re a and C first. In this problem they are not equaled so you will continue to work the problem out with the function tan 2alpha = B/A-C.
B). Plug your numbers into the function.. tan 2alpha = -2/-6+4 à tan 2alpha = 1
C). tan 2alpha = 1. Now find the inverse.
D). Draw your coordinate plane and place 45 degrees in quadrant one. Since tangent is positive in quadrant 3, add 180 degrees to 45 degrees to get 225 degrees.
E). Now that we’ve found our two angles, divide them by 2. *(Remember, the 2 didn’t disappear so you must divide your angles by 2 to get alpha =).
F). Your final answer is alpha = 22.5 degrees, 112. 5 degrees

This section is pretty much following functions. If you know your functions then all you have to do is plug in the numbers and solve. Also, it is just repetition of 8-1 part I.

Well in Advanced Math this week I am really struggling. This is one of the hardest thing, to me, that we have learned all year. What i am talking about is 8-4. The only thing you are really doing is replacing trig functions with identities. There are guidelines to follow to do so:

1.Pythagorean identities (squared)
2.Move everything to sin and cosine or tangent if possible
3.Algebra to simplify
4.Identities
5.Algebra

You may have to repeat some of the steps in order to get the answer.

The identies are things such as sin²θ + cos²θ = 1
tan²θ + 1 = sec²θ
1 + cot ²θ = csc²θ

Ex: (sin²x-1)(cot²x+1)
*Make sure to turn everything into sine and cosine
=cos²x(cot²x+1) = (-cos²x)(csc²x) = cos²x(1/sin²x)=-cosx/sin²x= -cot²x = tan²x-cot²x

Most likely by looking at the example it makes no sense, but the key to this lesson is to know your identities and rules. The problems vary in many different ways also.

This week in advanced math we learned more of chapter 8. Section 8-4 is all about pythagorean identities. Its difficult until you can learn all of your identities and then working it becomes so much easier.

5 steps:
1) look for any pythagorean identities
2) move everything to sin and cos or tan if possible
3) use algebra to simplify
4) check again for any pythagorean identities
5) algebra

You wont necessarily use all of these steps. They are pretty much guidelines to follow when working these types of problems.

Example:
secxcotx
secx is equal to (1/cosx) and cotx is equal to (cosx/sinx)
the 2 cosx's cancel out and you are left with (1/sinx)
1/sinx is equal to cscx
your final answer is cscx

Doing these types of problems takes a lot of practice and will become easier when you learn all of your identities. All problems you work are different but require the same 5 steps.

8-4

This week we learned more of chapter 8. In section 8-4, we talked about pythagorean identities. It looked pretty easy, but once you have to start doing it on your own, it gets more difficult and takes a lot of practice. Before you can start working any problems, you have to learn these pythagorean identities:

1- cos^2 theta = sin^2 theta
csc^2x-1=cot^2x
1-sin^2theta=cos^2theta
sec^2x-1= tan ^2x
sin^2 theta-1=-cos^2theta
sec^2x-tan^2x= 1
cos^2theta-1- -sin^2theta
1- csc^2x=-cot^2x
csc^2x-cot^2x=1
1- sec^2x= -tan^2x
cot^2-csc^2x=-1
tan^2x-sec^2x=-1
tanx= sinx/cosx
cotx=cosx/sinx

When doing these types of problems, all you are basically doing is replacing and simplifying. There isn't much math involved.

There are 5 steps you follow to get the answer to your problem:
1. see if there are any pythagorean identities
2. if not, change everything to either sin, cos, or tan.
3. use algebra and simplify
4. see if there are any pythagorean identities
5. algebra

Some of the steps repeat and sometimes you won't use them all. It all depends on the problem.

ex 1: tanx - cosx
a. the pythagorean identity that goes with tanx is sinx/cosx.
sinx/cosx - cosx
b. the next step you will use is algebra. you multiply the two fractions and cosx cancels out leaving you with sinx.

sinx is your answer.

The problems vary a lot and there are many different ways to do them, but if you practice them and learn the identities, it makes it much easier. This section becomes one of the easiest if you do these types of problems enough.

Periodic Function

A periodic function is a function that repeats its values in regular periods and intervals. Most significant examples are the trig functions, which repeat over intervals of length 2pi. Periodic functions are used throughout science to describe oscillations,waves,and etc. that exhibit periodicity. Function f is periodic f(x+p)=fx. For all values of x, the constant P is called the period. It has to be positive.P will repeat over intervals of length P,which are referred to as periods.

Ex: the sine function is periodic with 2pi, since sin(x+2pi)=sin(x)
f(x)=sinx has period 2pi,therefore sin(5x)will have period 2pi/5.
Periodic functions can be used for wave splicing.The study of periodic functions can be lead to many real life solutions of real life problems. These problems include planetary motion,sound waves,electric current generation,earthquake waves,and tide movements.

Week 3 Prompt

What is a periodic function? What examples of activities in the real world have periodic behavior?

8-2 Graphing with Period changes

This week in advanced math we learned sections 8-1 through 8-3. In 8-1 we learned to find theta, then to find whether the angle is positive or negative. Once we find those we find out which quadrants the angles are in. In 8-2 we learned to graph with a period change. We also learned to find the amplitude and the period. We then expanded on 8-1, learning that m= tan a for a line Tan 2a=B/A-C if A is not equal to C for a conic. If A=C for a conic section x=pie/4. Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0. And last but not least that a= angle of inclination. In 8-3 we learned to graph the five points. We also learned that amplitude for a graph = max-min/2.

y= Asin(Bx +/- C)+/-D

y= Acos(Bx+/- C) +/- D

8-2

-y= ASinBx or y= ACosBx

-|A| = Amplitude

-2(pie)/B = Period

1) To graph with a period change.

a) Take all 5 points

b) Divide by B

c) Sketch

Ex. y= 2Sin4x

Amplitude=2 Period=2(pie)/4

1)o= 0/4= 0

2)pie/2= pie/2 *1/4= pie/8

3)pie= pie/4

4)3(pie)/2= 3(pie)/2 * 1/4= 3(pie)/8

5)2(pie)= 2(pie)*/4= 1/2(pie)

Chapter 8

I personally understand and think that chapter 8 was much easier than charpter 7 so far. In section 8-1, we are trying to solve for theta or x. When you are solving for theta, you need to get the trig function alone and take the inverse of it.
- Most inverses have two answers, so if you do not end up with two, you know you are doing something wrong. Find where the answer is on the coordinate plane and figure out whether the equation is negative or positive. You need to know these steps in order to be able to solve the problem.
- To get to the quadrant 1 angle, you just take the inverse of the positive or negative.
- If you need the quadrant 2 angle, you make the number negative and add 180 degrees.
- For quadrant 3, you simply add 180 degrees.
- And quadrant 4, make the number negative and add 360 degrees.

For example: solve for theta. 4cos theta + 7 = 6
a. subract 7 from both sides. 4cos theta = -1
b. divide 4 from both sides. cos theta = -1/4
c. now get the inverse of cos. theta = cos ^-1(1/4)
d. cos^-1= 75.522 <------ this angle goes into the first quadrant, so you now have one angle.
e. Since cos deals with x and the first angle is positive, you need to figure out where the next angle will be. In this example, it will be in quadrant 4.
f. Now you make 75.522 negative and add 360 degrees.
g. the answer you end up with is 284.478

It seams that 8-1 is not that much different than chapter 7 aside from a couple of new formulas.

8-3

In advanced math this week we learned the begining of chapter eight. We learned chapters 8-1 through 8-3, which all tied into each other. It was a lot of information to retain but considering that I have the BEST advanced math teacher in the world... I managed to pull through it! In order to work with these sections you are going to have to use sine and cosine, and find the amplitude and period. You will learn about the both of these in the next example.

First let me tell you your formulas:
y = A-sin (Bx+ or – C) +/– D
y = A-cos (Bx + or – C) +/– D
Amplitude (the distance from 0 to the hightest point) = max-min/ 2
Period (how long it takes a function to repeat itself)

STEPS:
1. Follow the steps for period and change the five points.
2. Add or subtract C from the five points.
3. You will move any y values up or down.

EXAMPLE:
y= 4sin (4x-1/2) +4
Amp: 4
Period: 2(pie)/ 4= (pie)/ 2
1.0: 0/4= 0+ 1/2= 1/2
2. (pie)/2: (pie)/2 (1/4)= (pie)/8+ 1/2= (pie+5)/8
3. (pie): (pie)/4= (pie)/4+1/2= (pie+1)/2
4. 3(pie)/2: 3(pie)/2(1/4)= 3(pie)/2+ 1/2= 2(pie)
5. 2(pie): 2(pie)/4= (pie)/2+ 1/2= (pie+1)/2

We learned chapter 8 last week and at some parts i was confused but for the most part i got it. We learned how to solve for theta and for x.

To solve for theta you have to get the trig function by itself and then take an inverse. An inverse has two answers with some exceptions, then you find where the angle is based on the trig function and if the number is positive or negative.

Steps:
Take the inverse of the positive number to find the quadrant one angle. To get to the quadrant two angle you have to make negative then add 180 degrees then you add 180 again to get to quadrant three. To get to quadrant four you have to make negative and add 360.

Ex. Tan theta=1.2
Plug that into calculater and you 50.2
Then you add 180 and you get 230.2

8-1 was pretty understand and just involved remembering your formula and knowing how to plug it into your calculator

Last week we started chapter 8. I had more trouble with 8-1 than 8-2. 8-2 was pretty much finding the amplitude and period with a possible shift change. But learning 8-3 helped with the previous sections.

In case of a period change you must:
1) take all five points
2) divide by whatever b is
3) sketch your graph

Amplitude is the distance from 0 to the highest point and 1/2 the vertical distance.
Period is how long it takes for the function to repeat itself.

Example:
y=2sin3x
since it is sin your graph will start at the origin.
first you must find the amplitude
the amplitude is 2.
to find the period you divide 2pi by b which is 3.
your period is 2pi/3.

Next you must take all the points on the graph and divide them by your period.
0 0/3=3
pi/2 pi/2/3=pi/2x1/3 =pi/6 (when dividing you must multiply by the reciprocal.
pi pi/3
3pi/2 3pi/2/3=3pi/2x1/3 =3pi/6=pi/2
2pi 2pi/3

Once you have found all your points you can sketch your graph.
Since your amplitude is 2 you must make sure your graph repeats itself twice.

8-3

This week in Advanced Math we learned sections 8-1 through 8-3. All 3 sections tie into one another. They all involve sine and cosine. You also must know what amplitude and period is. Amplitude is the distance from O to the highest point. ½ is the vertical distance. Period is how long it takes a function to repeat itself. Here is a look at section 8-3.

Formulas:
y = Asin (Bx+ or – C) + or – D
y = Acos (Bx + or – C) + or – D
Amplitude = max-min/2

Here are the steps to graphs:
1. First you must follow the steps for period. (Change the 5 points)
2. Next you will add or subtract C from the 5 points.
3. Lastly you will any y values up D or down P.

Example Problem:

y = 2sin (2x – ¼) + 2

Amplitude = 2 Period = 2π/2 = π

1. 0: 0/2 = 0 + ¼ = ¼
2. π/2: π/2(½) = (π/4) + ¼ = (π + 1) / 4
3. π: π/2 = π/2 + ¼ = (2π + 1) / 4
4. 3π/2: 3π/2(1/2) = 3π/4 + ¼ = (3π + 1) / 4
5. 2π: 2π/2 = π + ¼ = (4π + 1) / 4

Last week in advanced math we learned some of chapter eight. The thing that I understood most about chapter eight is section eight two. I understood chapter eight more than I did chapter seven.
In section eight two, we learned how to find the amplitude and the period.

To graph things with a period change you must:
find all 5 points
divide by B
Sketch the graph.

The amplitude is the distance from zero to the highest point and half the vertical distance.

The period is how long it takes to repeat itself.

Example 1: y=2 sin1/2x
1. amplitude= 2= 2
2. period=2pi/ (1/2)= 2pi x 2= 4pi
3. find your five points
1. 0
2. pi/2
3. pi
4. 3pi/2
5. 2pi
4. divide by B
1. 0/ (1/2)= 0
2. pi/2/ (1/2)= pi
3. pi/ (1/2)= 2pi
4. 3pi/2/ (1/2)= 3pi
5. 2pi/ (1/2)= 4pi
5. sketch the graph with the second set of point starting at the origin.

Find Amplitude, and Period.

Theres not much I understand in Advanced Math, but on thing that I absolutely know is how to find Amplitude, and Period.

When finding Amplitude, period and the five points, the problem will look like so :
y = # cos # x OR y = # sin # x
To find Amplitude, just get the absolute value of the number before cos or sin.
y = -5 cos 8 x
the absolute value of -5 if 5. Amplitude is 5.

y = -234 sin 876 x
they absolute value of -234 is 234. Amplitude is 234.

To find the period, divide 2 pi by the number right before the x.
y = -24 sin 54 x
2pi/54 = pi/27
period is pi/27

y = -98 cos 90 x

2pi/90 = pi/45

period is pi/45

so YURP :D

In chapter 8 we are learning more to do with graphs. In 8-2, we were taught to find the amplitude, period, and period change which have to do with shifting of graphs.

Amplitude is the distance from 0 to the highest point going vertically on a graph.
Period is how long it take the function (waves) to repeat themselves.

Formulas:
Y=A sin Bx or A cos Bx
A (Absolute value of "A") is the amplitude
2pi/B is the period

In order to graph the period change you must figure out the 5 points.
1. Take the 5 points: 0, pi/2, pi, 3pi/2, and 2pi.
2. Divide each one of these points by B.
*Do not divide by period, remember to take the original B
3.Once you have the 5 points, sketch the graph.

Ex: Find the amplitude, period, and period change of y=-8 sin 4x

First, find amplitude and period.
A=8 P=2pi/4=1/2

Next, to find the 5 points to graph, take orinal 5 points and divide by 4.
1. 0
2. pi
3. 2pi
4. 3pi
5. 4pi

8-1

This past week, we learned how to solve for the angle theta. To me, this is the easiest of all the concepts we have learned thus far. Solving for angle theta is pretty simple to solve.

To solve for the angle theta you have to take the inverse of the trig function. The angle theta can be used by using the trigonometric identities. An inverse has two possible answers.

-To get the quadrant one angle, you take the inverse of the the positive or negative angle.
-To get the angle in quadrant two, make the number negative, and add 180 degrees.
-For quadrant three, add 180 degrees to the angle
-For quadrant four, make the number negative, and add 360 degrees.

Example:

3 cosine(theta)= 1

1. Subtract 3 from one, which will equal -1/3
2. Then do the cosine inverse(1/3)=
70.529
3. We are looking for the positive cosine, which is x.
4. 70.529 degrees is in the first quadrant, so it is positive.
5. Since it is in the first and fourth quadrant, for fourth quadrant, we will do 360-70.529, and it will equate to 289.471 degrees

Answers: 70.529(degrees), 289.471(degrees)

8-1

So far, chapter 8 has been easier for me to understand than chapter 7 was. It is mostly simple math that you just need to put into formulas.

In section 8-1, we are trying to solve for theta or x.
-when you are solving for theta, you need to get the trig function alone and take the inverse of it.
- Most inverses have two answers, so if you do not end up with two, you know you are doing something wrong. Find where the answer is on the coordinate plane and figure out whether the equation is negative or positive.

** You need to know these steps in order to be able to solve the problem.
- To get to the quadrant 1 angle, you just take the inverse of the positive or negative.
- If you need the quadrant 2 angle, you make the number negative and add 180 degrees.
- For quadrant 3, you simply add 180 degrees.
- And quadrant 4, make the number negative and add 360 degrees.

EX. 1: solve for theta. 4cos theta + 7 = 6
a. subract 7 from both sides. 4cos theta = -1
b. divide 4 from both sides. cos theta = -1/4
c. now get the inverse of cos. theta = cos ^-1(1/4)
d. cos^-1= 75.522 <------ this angle goes into the first quadrant, so you now have one angle.
e. Since cos deals with x and the first angle is positive, you need to figure out where the next angle will be. In this example, it will be in quadrant 4.
f. Now you make 75.522 negative and add 360 degrees.
g. the answer you end up with is 284.478

You now have your two angle and your final answer is theta = 75.522 degrees, 284.478 degrees

These problems seem sort of confusing at first, but once you practice them, it gets much easier.

Chapter 8

-Simple Trigonometric Functions
-Sine and Cosine Curves

This week we learned the beginning of chapter 8. In chapter 8 we learned inverses of trigonometric functions. In section 8-2 we learned properties of the curves of sine and cosine (amplitudes, periods), and how to graph them.
Section 8.1
Finding the inverse.

Ex.
3Cosx=1
*Make sure that your calculator is in DEGREES, not radians. You WILL get the wrong answer.

Like any equation, you must get the variable by itself in order to solve. So first divide “cosx’ by 3.
- 3cosx/3 = 1/3
Next, you have to solve for x. X will equal the inverse of cosine.
X=Cos^-1(1/3)
Finally, you calculate (enter into your calculator) the inverse of cosine of (1/3).
Your answer should be 70.528 degrees. Round to the nearest tenth of a degree if not stated.
X=70.5°

Section 8-2
Amplitudes and Periods
The Amplitude of a periodic function is the point in which the function reaches its maximum height.
The period of a function is how long it takes a function takes to repeat itself.
For the functions: Y= A sine Bx
Y= A cos Bx
Amplitude equals the absolute value of A.
Ex. Y=2 sine 3x
The amplitude = absolute value of l 2 l which is 2, therefore Amplitude= 2.
In order to find the period, you must use the formula 2π/B.
The period equals 2 π/3

Chapter 8 Section 2

In section two chapter eight, we learned how to find the amplitude, period, and period change.

The amplitude is the distance from zero to the highest point half the vertical distance.

The period is how long it takes a function to repeat itself.

Y= A sin Bx or A cos Bx

The absolute value of A is the amplitude

And 2 pi/B is the period

To graph a period change:
First, take all five points.
The five points are 0, pi/2, pi, 3pi/2, and 2pi.
Then divide those five points by B.
Finally sketch the graph.

For example,find the amplitude, period, and five points of y= -4 sin 2x

Take the absolute value of A which in this case is -4
I-4I = 4
So, Amplitude is 4

Then, to find the period divide 2pi by B which in this case is 2.
2pi/2 = pi.
So, the period is pi.

In order to find the five points, you must divide the given points by B.
0/2 is 0
Pi/2/2 or pi/2 times ½ is pi/4
Pi/2 is pi/2
3pi/2/2 or 3pi/2 times ½ is 3pi/4
And 2pi/2 is pi

8-1 Simple Trigonometric Eqns.

This past week in Advanced Math, we learned sections 8-1 through 8-3. We learned part I of 8-1 on Friday and part II on Wednesday. 8-1 is about solving simple trigonometric equations. We were taught how to solve simple trigonometric equations and also how to apply them. To solve an equation involving a trigonometric function, we first transform the equation so that the function is alone on one side of the equals sign.

To solve for theta, you get the trig function by itself then take an inverse. Remember, an inverse has two answers with some exceptions! Here are the steps to follow:
1. Take the inverse of the positive number to find the quadrant one angle.
2. For quadrant two, make it negative and add 180 degrees.
3. For quadrant three, just add 180 degrees.
4. And for quadrant four, make the number negative and add 360 degrees.

With that in mind, let’s try an example problem..
EXAMPLE 1: 3 sin theta = 2
A). First, transform the equation so that the function is alone on one side. Dividing 3 from the
left and right, you will then get sin theta = 2/3.
B). Now, take the inverse of 2/3 to find quadrant one’s angle – theta = sin ^-1(2/3)
C). Plugging that into your calculator, you will get 41.8 degrees. Now, draw your coordinate
plane and place 41.8 in quadrant one. We know that sine is positive in quadrants I and II, so solve for those angles.
D). For quadrant II, make 41.8 negative and add 180 degrees. You will get 138.2 degrees. That is your quadrant II angle.

EXAMPLE 2: 2 cos theta = 1
A). Like in example one, divide 2 from the left and right to get cos theta = ½.
B). Now, take the inverse of ½ to find quadrant one’s angle – theta = cos ^-1(1/2)
C). Plugging that into your calculator, you will get 60 degrees. Now, draw your coordinate plane and place 60 degrees in quadrant one. We know that cosine is positive in quadrants I and IV. Since we have quadrant one’s angle, solve for quadrant four’s by making 60 negative and adding 360 degrees. Your answer will be 300 degrees.


These are two examples from section 8-1, part I which we learned on Friday. Part II of 8-1 deals with inclinations and slopes but still requires solving trigonometric equations.

Chapter 8-Section 3

In this lesson we use our prior knowledge from lesson 1 and 2 to carry out the process. Sin and Cos are both involved in the process. We also use the amplitude and period. Amplitude- distance from 0 to highest point, 1/2 the vertical distance.

• The absolute value of A=amplitude
• To find the period use 2π/B

y=Asin(Bx±C)±D or y=Acos(B±C)±D (plug into this formula)

To graph:
• Follow period change with 5 points.
1.0
2.π/2
3.π
4.3π/2
5.2π

• Then add or subract C from those 5 points. (if the eqaution is + C you subtract and vice versa)

• Move all y values up D or down P (if + then up)

When looking at a graph and making an equation, we use a new formula to find amplitude.

-max-min/2= amplitude

y=4sin(πx-1/2)+1
amplitude=4 and period= 2

1.0= 0/π=0 + ½= ½
2.π/2= (π/2) / (1/ π)=1/2+1/2=1
3.3 π/2=(3 π/2)/ (1/ π)=(3/2)+(1/2)=3/2
4.π= π/ π=1 + ½=2
5.2 π=2 π / π=2 +1/2=5/2

Real World Applications of Sine and Cosine wave applications

Sine and Cosine can be used in everyday life in various ways. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Surveying, navagation, and astronomy all rely on sin and cos for the position of objects and other calculations.Music is composed of waves of different frequencies and amplitudes and these can be described using sin/cos. in fact anything involving sound waves will rely on sine and cosine. Ballistic trajectories rely on sin/cos. in physics, sin.cos is used often. Satellites and space flights relies on calculations and conversions to polar coordinates.Gps and cellphones rely on triangulation and formulas involving sin/cos. Signal transmissions involve waves described with sin/cos. Thermal analysis is used to model how things get hot e.g oven,spacecraft. this equation is usually solved using sums of sin and cosine. Also evaporation and [recipitation averages incorporate sin and cosine. Many other means of science/math in the real world incorparate sin/cos in some shape or form.

yahooanswers.com Real life applications of sine and cosine applications/functions

Week 2 Blog Prompt

What types of things can the graphs of sine and cosine be used to model in the real world? Give an example. **Be sure to cite if you use a source!**

adv math

For word problems they have 3 formulas that can be used to solve the equations
s=r x theta
k=1/2 r^2 x theta
k= 1/2 rs

these signs also have alternate meanings:
theta= apparent size
s=diameter of an object
r=distance between an object

ex:1

a sector of a circle has a radius of 8 cm and a central angle of 60 degrees. find its approximate arc length and area.
1st. list the formula
r=8cm theta=60 degrees
s=? k=? s=r x theta
2nd: plug in the formula
s=(8)(60)
s=140 degrees
3rd.plug in the formula to find k
k=1/2 x 8 x 140
k=560 cm

The thing that i understood most about chapter seven is the word problems. it has three formulas that can be used.

1. s=r x theta
2. k= 1/2 r^2 x theta
3. k= 1/2rs

some of the word problem hints is:
theta = apparent size
s= diameter of an object
r= distance between the objects

example 1:

A sector of a circle has a radius of 7cm and central angle of 50 degrees. find its approxiomate arc length and area.
r= 7cm theta=50 degrees
s=? k=?

then plug into a formula:
s=(7)(50)
s= 350 degrees

then plug that into a formula to find k.
k=1/2(7)(350)
k= 1225cm

This past week we basically reviewed everything from the week before. One thing we reviewed was finding an inverse of an angle. Both sin-1 and arcsin mean the same thing. They are both used to find inverse. They are two ways to find inverses: by plugging it into your calculator or working it yourself by drawing a triangle picture. When finding an inverse by using your calculator it is important that you have your calculator set to degrees.

example:
1) sin-1 0.9
plug into your calculator sin-1 0.9 and your answer should be 64.2

On Friday we started 8-1. To solve for theta you have to get the trig function by itself then take the inverse. First you find where the angle is based on the trig function and whether it is positive or negative. By having to leave class early Friday I missed some of the examples on how to do these types of problems.

Find a Reference Angle

7- 4 Finding a Reference Angle

Follow these steps to find a reference angle:
1. Locate which quadrant the angle is in.
2. Decide whether it is positive or negative.
3. Subtract 180 until the absolute value of theta is between 0 and 90 degrees.
4. If your answer is a trig chart angle plug it in. If it’s not you leave it alone or you plug it into your calculator.

Trig Chart Angles:
0 degrees = 0
30 degrees = pi/6
45 degrees = pi/4
60 degrees = pi/3
90 degrees = pi/2

Here are some examples you can try:

Find the reference angles:

Example 1:

sin(690°) = -sin(30°) = -π/6

690° is in the IV quadrant. They are asking for sin, and in the IV quadrant sin is negative. So we subtract 180° continuously from 690°, until we arrive at 30°. We then plug 30° into –sin and get-π/6, our answer.

Example 2:

cos(400°)=cos(40°)

400° is in quadrant I, which for cos is positive. We continuously subtract 180° from 400°, until we get an angle between 0° and 90°, which is 40°. Our final answer is cos(40°).

Area of a Sector

A circle sector is the part of a circle enclosed by two radii and an arc. When finding the area of a sector of a circle, you are actually finding a fractional part of the circle. The percentage is equated by the ratio of the central angle to the entire central angle, which is 360 degrees. To find the area of a sector in degrees, you would do the following steps:

1. First, identify the degree of the sector

2.Then, identify the radius of the circle

3. Multiply the radius (squared) times theta times pie.

EXAMPLE:

Find the area of a sector with a central angle of 60 degrees and a radius of 10. Express answer to the nearest tenth.

A = (theta)(pie)(radius)[squared]

(60)(pie)(10)[squared]

52.35987756

A = 52.4

Finding the area of a sector is fairly easy and useful. When we understand the area of a sector we can see the importance of the relationship of pie, theta, and the radius of a circle. Understanding circle sectors can lead to far more knowledge when learning about circles.