10-3 Simplify

This week in advanced math we learned chapter 10-3 and 10-4 along with reviewing chapter 10 and testing on it. In 10-1 you have the formulas sin(alpha + beta)= sin(alpha)cos(beta) + cos(alpha)sin(beta) and cos(alpha + beta)= cos(alpha)cos(beta) - sin(alpha)sin(beta). In 10-2 you have the tangent formula tan(alpha + beta)= tan(alpha) + tan(beta)/1-tan(alpha)tan(beta). Then there's 10-3.

10-3 Formulas

Doouble Formulas

Sin 2(alpha)= 2sin(alpha)cos(alpha)

Cos2(alpha)= cos^2(alpha)-sin^2(alpha) or 1-2sin^2(alpha) or 2cos^2(alpha)-1

tan2(alpha)= 2tan(alpha)/1-tan^2(alpha)

Half Formulas

sin alpha/2=+/- the square root of 1-cos(alpha)/2

cos alpha/2= +/- the square root of 1+ cos(alpha)/2

tan alpha/2 = +/- the square root of 1-cos(alpha)/1+cos(alpha) or sin(alpha)/1+cos(alpha) or

1-cos(alpha)/sin(alpha)

Ex. Simplify 2sin(90 degrees)cos(90 degrees)

1) Sin2(90 degrees)

2)sin 180 degrees

3= 180-180= sin(0)

4)=0

happy boo day!

We did not learn to much new material this week. We had a chapter test this week on chapter ten, and had a sub for two days, the wonderful Ms. FeFe. I missed two days this week, which sort of put me behind, and for this reason i feel i didnt do to well with this chapter. We could ask questions on problems we were not sure how to do. This chapter only had 4 sections in it, and out of them the first section was definately the easiest. There were only two formulas to learn. It was pretty basic math, and calculators were still not allowed. Still had to know the trig chart if you had any intentions of finishing a problem.


The two formulas were for sin and cos:
Sin ( alpha + beta ) = ( sinalpha ) ( cosbeta ) + ( cosalpha ) (sinbeta)

Cos ( alpha + beta ) = ( cosalpha ) ( cosbeta ) – ( sinalpha ) (sinbeta )

Example:Find the exact value of cos 150

Cos ( 90 + 60 ) = cos90 cos60 – sin90 sin60

(0)(1/2) – (1)(square root 3/3)

0 – square root 3/3

Answer: Square root3/3

So this past week we didn't learn anything new. We basically just reviewed chapter 10 and took a chapter test. I tried my best to catch on to this chapter but i felt horrible all week. So here's some examples.

EXAMPLE 1:
sin75 degrees cos 15 degrees+cos 75 degrees sin 15 degrees
first you must replace it with a formula!
sin(alpha+beta)
sin (75+15)
=sin 90 degrees
TRIG CHART!
your answer is 1.

EXAMPLE 2:
cos 5pi/12 cos pi/12-sin 5pi/12 sin pi/12
replace formula!
cos(alpha+beta)
cos(5pi/12+pi/12)
cos 6pi/12
=1/2
TRIG CHART!
your answer is pi/3.

HAPPY HALLOWEEN!?!?!

First off I want to wish everyone a HaPpY hAlLoWeEn!!!!! This week in advanced math we did not learn anything new. We reviewed chapter 10 on Monday and Tuesday, took a chapter test on Wednesday and Thursday and had a sub Thursday and Friday. Our chapter test was on sections 1-4. The easiest section was 10.1 because you just have to know two formulas and you just have to plug it in. You also needed to know you trig chart.


The formulas for section 10-1 are:

Sin (α +/- β) = (sin α) (cos β) +/- (cos α) (sin β)
Cos (α +/- β) = (cos α) (cos β) -/+ (sin α) (sin β)


Find the exact value of sin 15degrees
Alpha=45 degrees
Beta= 30 degrees

Sin (45degrees-30degrees) = (sin 45degrees) (cos 30degrees)-(cos45degrees) (sin 30 degrees)
(Square root of 2 over 2)(Square root of 3 over 2) – (Square root of 2 over 2) (1/2)

Square root of 6 over 4 –square root of 2 over 4= square root of 6- square root of 2 over 4.

Your answer is Sin 15degrees= square root of 6-square root of 2 over 4.
Because it isn’t on trig chart u leave it at that if it was then you can simplify it.

You see how simple this section is.


Remember know your formulas and know your trig chart. You basically need to master your trig chart and know it in order to do some problems because some stuff are plug from trig chart like sin 45 for example is square root of 2 over 2. NO calculators !!!

Happy Halloween!

First off I want to wish everyone a HAPPY HALLOWEEN!!!!! This week in advanced math we did not learn anything new. We reviewed chapter 10 on Monday and Tuesday, took a chapter test on Wednesday and Thursday and had a sub Thursday and Friday. Our chapter test was on sections 1-4. All for sections had to do with memorizing formulas. If you did not use a formula for every problem on the test you did it wrong. Section 1 was the easiest section of the 4 because you just had to memorize 2formulas. Here are the sine and cosine formulas:

Sin (α +/- β) = (sin α) (cos β) +/- (cos α) (sin β)
Cos (α +/- β) = (cos α) (cos β) -/+ (sin α) (sin β)

Let’s try an example:

Find the exact value of sin 75°

1. Sin (45° + 30°) = (sin 45°) (cos 30°) + (cos 45°) (sin 30°)
2. (square root 2/2) (square root3/2) + (square root 2/2) (½)
3. (square root 6/4) + (square root 2/4)
4. square root 6 + square root 2/ 4

HAPPY HALLOWEEN!!!

So the week before last, I was crazy sick. This week people took tests and hung out with Mrs. FeFe. That means that I really didn't learn anything. I stole notes from Lauren Louque, but I really don't understand them that well.

Here's an example of things I have in my notes.

From 10-4
cos2x = cos x
2cos^2x-1 = cosx
2cos^2x-cosx-1 = 0
2cos^2x-2cosx+cosx-1
2cosx(cosx-1) + (cos-1)
(2cosx+1) (cosx-1)
cosx = 1/2

I'm still having trouble with this stuff. But hopefully I can get some help.

Anyway,

HAPPY HALLOWEEN!!!

Chapter 10 Section 1

Just as everybody else has been saying, we haven't really learned much this week, we just reviewed and tested and Chilled with FE FE. But one of the things that I ACTUALLY learned, was Section 10-1. Even though this sections does not permit the use of calculators, it still is very simple and easy to understand.

The formulas in 10-1 are as followed:

cos (alpha +/- beta) = cos alpha cos beta+/- sin alpha sin beta

sin (alpha +/- beta) = sin alpha sin beta +/- cos alpha cos beta

Now lets get to work!

Find the exact value of sin 15*
Alpha = 45 *
Beta = 30 *

Sin (45 * - 30 *) = (sin 45*) (cos 30*) - (cos 45*) (sin 30 *)
(Square root of 2 \ 2)(Square root of 3 \ 2) – (Square root of 2 \ 2) (1 \ 2)

Square root of 6 \ 4 – square root of 2 \ 4= square root of 6 - square root of 2 \ 4.

= Sin 15* = square root of 6-square root of 2 \ 4.

Now how easy was DAT! hehe

Also, here a little tip. Make sure you keep a keen eye for trig chart functions, if you miss one, then you may not be allowed to answer your problem correctly.

Happy Halloween

HAPPY HALLOWEEN EVERYONE :D
This week we didn’t really learn anything new. We reviewed and took our Chapter tests on Chapter 10. In 10-3 we were given double and half angle formulas. I would type all the formulas out, but I honestly don’t feel like it. Everyone should have them in their binders or even better, KNOW THEM by now.

EXAMPLE 1: 2sin30 degrees cos30 degrees
A). According to our formulas, this is the right side of the sin 2 alpha formula.
B). So, replace it with the left – sin 2(30 degrees)
C). Your answer is sin 60 degrees which is in your trig chart.
D). It reduces to square root of 3 / 2.

EXAMPLE 2: cos squared 15 degrees – sin squared 15 degrees
A). According to our formulas, this is the right side of the cos 2 alpha formula.
B). So, replace it with the left – cos 2(15 degrees)
C). Your answer is cos 30 degrees which is in your trig chart.
D). It reduces to squre root of 3 / 2.

Have a great weekenddddd!

Chapter 10

This week we really did not learn much new. We had a chapter test this week on chapter 10, and had a sub for two days. We continued to perfect chapter 10 in the beginning of the week and we could ask questions on problems we were not sure how to do. This chapter only had 4 sections in it, and out of them the first section was easiest. There was only two formulas that we had to learn and it was pretty basic just substituting. The two formulas were for sin and cos:

Sin ( alpha + beta ) = ( sinalpha ) ( cosbeta ) + ( cosalpha ) (sinbeta)

Cos ( alpha + beta ) = ( cosalpha ) ( cosbeta ) – ( sinalpha ) (sinbeta )

Example:

Find the exact value of cos 150
Cos ( 90 + 60 ) = cos90 cos60 – sin90 sin60
(0)(1/2) – (1)(square root 3/3)
0 – square root 3/3
Square root3/3

We are suppose to blog every week so.. here i go. To be honest, we really did not learn anything new this week. On Monday, we continued section ten four. We reviewed homework problems, and also learned how to solve the problems in our calculators. I have a Ti-89 calculator which is sort of different from the ti-84 and COMPLETELTY different from the In-spires. So I had to Google directions for solving ten four problems. Its not hard either, my calculator just has the functions hidden in different areas. So anyways , I am blogging about section ten four. First of all, section ten four is pretty much like chapter eight. Its so much like it that we did not even take notes on it! We just used the same ones from chapter eight, and did some examples. Although the two sections are pretty much the same, the section from chapter ten is a little more difficult because it involves more complicated identities incorporated with trig equations. To be honest again, this section is really hard. And I can not give an example because I do not know how to do any of them ( other than the ones we completed in class). Sorry, I know I’m a horrible blogger :/

Week 2 Blog Prompt

How do you solve a quadratic trig equation? Give 2 examples.

10-2

This week in advanced math we started on Chapter 10 and took some quizzes on Chapter 10. In Chapter 10-1 we learned some of the formulas for cos(alpha +/- beta) and sin(alpha +/- beta). We also learned Chapter 10-2, which involves formulas for tan(alpha + beta) and tan(alpha - beta). And the same for cot. Then we learned 10-3 which involves Double formulas and Half formulas. For ex. double formulas such as Sin2(alpha) =2Sin(alpha)Cos(alpha). And Half formulas such as Sin alpha/2 = +/- the square root of 1-cos(alpha)/2. This is a chapter full of formulas and you will make an A or F based on if you know the formulas or not.

10-2

simplify tan 25 degrees + tan 5 degrees/1 + tan 25 degrees(tan 5 degrees)

=tan(25 degrees + 5 degrees)

=tan 30 degrees

=square root of (3)/3

10-2

This week we are working in chapter 10. Chapter 10 section 2 is about plugging into one side of the formula when given an another one. They have many different formulas that you will need to learn in order to figure out the problems. Also you will need to know how to find the reference angle and also you will once again need to know your trig chart. If you don't know these things you will fail once again.

FORMULAS:
tan(alpha+beta)= tanalpha+tanbeta/1-tanalphatanbeta
tan(alpha- beta)= tanalpha-tanbeta/1+tanalphatanbeta

*cot is the reciprocal
i.e. cot(alpha+beta)=1-cotalphacotbeta/cotalpha+cotbeta
ect...

EXAMPLE:
simplify
tan90-tan60/1+tan90tan60
tan(90-60)
tan(30)
answer:squareroot(3)/3

This week we were introduced to Chapter 10 Section 2, which is a lot like Section 1.

We replaced and simplified formulas, but did not use sine and cosine for this section. Unfortunately, we are still not allowed to use calculators in this section, i know its tragic.

Below are some formulas you may need to remember:
Tan ( alpha + beta ) = tan alpha + tan beta/1- tan alpha tan beta
Tan ( alpha – beta ) = tan alpha – tan beta/ 1+ tan alpha tan beta

Cot is the opposite (reciprocal) of tan. So basically just flip the formula around to get cot.

Example : simplify
tan 100degrees – tan 10 degrees/ 1+ tan 100 degrees tan 10 degrees

This formula matches with the first one I gave, so you replace the right side with the left side and get: tan ( 100 – 10)

Subtracting is the next step: tan90degrees
Since tan 90 is on your trig chart that is the answer you will get undefined as your answer.

MATH

Hey everybody. GO TIGERS!!!!! GO SAINTS!!!!!! GO MATH!!!!!! This week we learned about a million more formulas and we took a quiz. These formulas are mostly used in correlation with sine, cosine, and tangent. Apparently these formulas also have different forms known as double angle and half angle formulas. Both of these types of formulas can be manipulated to help find what you are looking for. Primarily these formulas are used to find values of (Alpha+/-Beta).  Everyone knows what happened this week and I don't have to say it. If anyone has any problems please remember that that is not an option and to please talk to someone. I love all of ya'll. peace.

10-2

Section 10-2

This week in advanced math we started section 10-2. Section 10-2 is similar to section 10-1. The only difference is we use tangent formulas instead of sine and cosine formulas. There are two formulas you need to remember for tangent:

Sum Formula:
Tan (alpha + beta) = tan alpha + tan beta/1 – tan alpha tan beta

Difference Formula:
Tan (alpha – beta) = tan alpha – tan beta/1 + tan alpha tan beta

Here’s an example:

Example:
Simplify this difference formula

Tan 140 degrees - Tan 95 degrees/ 1 + Tan 140 degrees Tan 95 degrees
1. Replace the right side of the formula with the left
2. Tan (140 degrees – 95 degrees)
3. When you subtract them you get Tan 45 degrees. Tan 45 degrees is in your trig chart as 1.

*Cotangent is the reciprocal of tangent. Reciprocal means opposites so you just flip problem and use the same formula.

Section 10-2

This week in section 10-2, we learned about the tangent formulas. In this section, the objective was to develop and apply formulas for tan (alpha +/- beta). You still can not use your calculators for anything in this section either. These are the formulas you need to learn and remember:

Tan ( alpha + beta ) = tan alpha + tan beta/1- tan alpha tan beta
Tan ( alpha – beta ) = tan alpha – tan beta/ 1+ tan alpha tan beta
Cot is the opposite (reciprocal) of tan. So, it would be the same formula, just flipped.

Ex: Suppose tan alpha = -1/2 and tan beta = 1/3. Find tan (alpha + beta).
A). Since the problem asks you to find the sum, use the sum formula.
B). Replace the left side with the right and plug in your alpha and beta.
C). -1/2 + 1/3 / 1 – (-1/2)(1/3)
D). -1/7 is your answer.

10-2

This week we completed the nine week grading period and worked on chapter ten; Trigonometric Addition Formulas. Many things from chapters 7-9 (which we learned in the first nine weeks) are helpful and very useful in chapter ten such as: the Trig Chart, converting radians to degree and vice versa, and of course the identities. Like in section 10-1, no calculators are to be used in neither 10-3 nor 10-4. Section 10-2 is pretty much the same as 10-1, but a different formula is used for tangent.

The formula for tangent is as follows:
*imagine a as alpha, and b as beta ;)
Tan(a+b) = tan a+ tan b/1-(tan a)(tan b)
Tan (a-b) is the reciprocal
Tan (a-b) = tan a-tanb/ 1+(tan a)(tan b)


Example:
Find tan(a+b) when a=30 b=15
Tan(a+b)=tan(30+15)
=Tan(45)
Tan(45) is on the trig chart, yay!
Because tan 45 is one, the answer is one.
=1
And that’s all there is to it:)

This past week we went into 10-2. Chapter 10-2 is similar to 10-1 with alpha and beta. Since 10-1 was sine and cosine, 10-2 is tangent.

FORMULAS:
tan(alpha+beta)=tan alpha+tan beta/1-tan alpha tan beta
tan(alpha-beta)=tan alpha -tan beta/1+tan alpha tan beta

Cotangent is just a reciprocal of these formulas.

EXAMPLE:
simplify.
tan130 degrees-tan40 degrees/1+tan130 tan 40
*first you must find which formula matches and replace it.
tan(alpha-beta)
tan(130-40)
=tan 90 degrees
TRIG CHART!
your answer is undefined.

I had no problems with 10-1 and 10-2. It's basically plugging in and replacing formulas.

10-2

This week we learned 10-2, we were introduced to tangent formulas. In 10-2 is basically just like 10-1 but 10-2 is not dealing with sine and cosine. 10-2 is dealing with tanganent and cotangent. Same method in 10-1 is used in 10-2.
10-2 formulas are:
Tan (alpha+/-beta) =Tanalpha+/-Tanbeta/1-/+ (tanalpha) (tan beta)
Cot (alpha+beta) =1+ (cotalpha) (cotbeta)/ (cotalpha) + (cotbeta)

Examples:
Tan110degrees-Tan50degrees/1+tan110degreestan50degrees
=Tan(110degrees-50degrees)
=tan 60 which is on the trig chart which your answer will be square root of 3.

Tan(x+pi/4)
Tanx+tanpi/4/1-tanxtanpi/4
=Tanx+1/1-tanx=your answer

Basically 10-2 is not hard. If you know your formulas you will do well and you will master 10-2. 10-2 and 10-1 are the same know formulas and plug it in to solve it. If you mastered the formulas you will breeze right through this section. REMEMBER you can’t use a calculator. If you use a calculator you will get a zero! Just know your formulas and you will do fine.

10-2

In chapter 10-2, we pretty much did the same thing as 10-1 except with different formulas. We still replaced and simplified, just without the sin and cos. You still are not allowed to use calculators for anything in this section either. The formulas you need to remember are:

Tan ( alpha + beta ) = tan alpha + tan beta/1- tan alpha tan beta

Tan ( alpha – beta ) = tan alpha – tan beta/ 1+ tan alpha tan beta

Cot is the opposite (reciprocal) of tan. So, it would be the same formula, just flipped.

Example : simplify
tan 100degrees – tan 10 degrees/ 1+ tan 100 degrees tan 10 degrees

This formula matches with the first one I gave, so you replace the right side with the left side and get: tan ( 100 – 10)

Subtracting is the next step: tan90degrees

Since tan 90 is on your trig chart that is the answer you will get undefined as your answer.

10-2

This week we’ve learned several sections of Chapter 10. In section 10-2, we were introduced to tangent formulas. In this section, the objective was to derive and apply formulas for tan (alpha +/- beta).
Sum Formula:
tan (alpha + beta) = tan alpha + tan beta / 1 – tan alpha * tan beta

Difference Formula:
tan (alpha – beta) = tan alpha – tan beta / 1 + tan alpha * tan beta

EXAMPLE 1: Simplify, tan 120 degrees – tan 60 degrees/ 1 + tan 120 degrees tan 60 degrees
A). According to the sum/difference formulas, this is a difference formula.
B). Replace the right side with the left.
C). tan (120 degrees – 60 degrees)
D). tan 60 degrees is in your trig chart. So, your final answer is square root of 3.

EXAMPLE 2: Suppose tan alpha = -1/2 and tan beta = 1/3. Find tan (alpha + beta).
A). Since the problem asks you to find the sum, use the sum formula.
B). Replace the left side with the right and plug in your alpha and beta.
C). -1/2 + 1/3 / 1 – (-1/2)(1/3)
D). -1/7 is your answer.

Week 8 Blog Prompt

Why is it important to know the sum/difference & Double/Half angle identities? Also, if you use a citation it needs to be different for each person. Too many people are copying answers from people and posting them.

10-1

This week was exam week. We did handouts on chapter 7-9 to get us ready for our exam. Friday we learned about section 10-1. In 10-1 you are not allowed to use a calculator if u do you will get a zero! We learned sine and cosine formulas.
Cos formula= Cos(alpha +/- beta)=(cosalpha)(cosbeta)-/+(sinalpha)(sinbeta)
Sin formula= sin (alpha +/- beta)=(sinalpha)(cosbeta)+/-(cosalpha)(sinbeta)

Alpha and beta comes from the trig chart and add or subtract to get the angle you are looking for.

Example:
Find the exact value of sin 15degrees
Alpha=45 degrees
Beta= 30 degrees

Sin (45degrees-30degrees) = (sin 45degrees) (cos 30degrees)-(cos45degrees) (sin 30 degrees)
(Square root of 2 over 2)(Square root of 3 over 2) – (Square root of 2 over 2) (1/2)

Square root of 6 over 4 –square root of 2 over 4= square root of 6- square root of 2 over 4.

Your answer is Sin 15degrees= square root of 6-square root of 2 over 4.
Because it isn’t on trig chart u leave it at that if it was then you can simplify it.

This section is easy and not hard if you know your trig chart and everything you shouldn’t have any problems at all.

Start of Chapter Ten

In the first part of this chapter, 10-1, it involves sine and cosine formulas used to simplify and find exact values of trigonometric functions. No calculators are allowed to be used for this section. Well isn’t this just delightful. Anyway like I was saying, lesson 10-1 is showing us how to find the exact values and how to simplify trig functions. Alpha and beta come straight from your trig chart and +/- to get the angle you are looking for, no worries. All you pretty much need to know for this section is the trig chart and the two formulas below.

Cos(alpha +/- beta) = cos alpha cos beta -/+ sin alpha sin beta
Sin(alpha +/- beta) = sin alpha cos beta +/- cos alpha sin beta

Ex:
Find the exact value of sin 50 degrees.
1. There are two degrees from the trig chart that add up to equal 35, so those are the two numbers you will use. 45and 15
Alpha = 60
Beta = 45
2. Since it is asking for sin, you will use that formula.
Sin(60 + 45) = sin60 cos45 + cos60 sin45
3. You fill in everything on the right of the equal sign with the trig chart.
(square root 3/2)(square root 2/2) + (1/2)(square root 2/2)
4. Square root 6/4 + square root 2/4
5. The Final answer is square root 4 + square root 6/8.

FOOTBALL!!!!!!!!!

Yeah buddy rolling like a math geek I hope I passed all my exams this past week. Yeah. Yeah. Yeah. Yeah. Alright I know that was really stupid but I’m feeling great right now, so I don’t care. Even though I don’t care about the Saints they still woon and even though varsity lost to SCC, JV WON 12-8 BABY, AND I PLAYED BOTH WAYS SO I’M A SAVAGE. Seriously, it was a team effort and we all played really hard. On to Math. We worked on our packets all week, which I think I didn’t finish all of them. Sorry. Anyway, we also learned a new section on Friday that seems pretty easy, but it will require rememorizing the Trig Chart. The exam was not overly hard. I either new something, had half a clue, or didn’t have a clue. You will be able to see that on my test. I knew almost everything that had to do with triangles, but the equations were another story. I’m pretty sure I still did a good guessing job though. Well, it’s late and I’m going to sleep. Goodnight everybody, love ya’ll, except you Vicken. Jk, jk. You’re a cool guy Vicken.

10-1

This week in advanced math we completed a nine weeks. Through this nine weeks we learned chapters 7-9. We learned everything from converting degrees to radians to navigation. From the trig chart and radius circle to the Laws of Sine and Cosine. The thing I grasped best from these chapters were all of chapter 7, all of chapter 8 except for 8-4, and all of chapter nine except for navigation. Now we are in chapter 10. In chapter 10 we are learning how to solve for alpha and beta using the formulas from chapter 9. This is Mrs. Robinson's favorite section in the book.

Ex. Find the value of cos(20degrees)cos(10degrees)= sin(20degrees)sin(10degrees)

cos(20+10)=cos30degrees

cos30degrees= square root of 3/2

This past week was exam week. We did packets of chapter 7-9 to prepare us for our exam. which was really hard! On friday we started learning chapter 10. In chapter 10 you dont use calculators!

formulas:
cos(alpha +/- beta)=cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta)=sin alpha cos beta +/- cos alpha sin beta

In order to work these types of problems you must know your trig chart!

EXAMPLE:
Find the exact value of sin 30 degrees.
alpha would be 90 and beta would be 60.
alpha is usually the bigger number especially when subtracting.

sin(90-60)=sin 90 cos 60-cos 90 sin 60
since you are subtracting in the front you must subtract in the back.
TRIG CHART!
=(1)(1/2)-(0)(square root of 3/2)
=1and 1/2-0
sin 30 degrees=1 and 1/2.

CHAPTER TENNNN

This past week in Advanced Math, we did not get a chance to learn much because it was exam week. We worked on our study guide and reviewed for exams all during the week. On Friday, we started Chapter Ten Section One.

This may sound very scary, but calculators are not allowed.
This is how to find sine and cosine of a number with no calculator.
For cosine
cos(alpha + or - beta) = cos alpha * cos beta - or + sin alpha * sin beta
For sine
sin (alpha + or beta) = sin alpha * cos beta + or - cos * sin beta

In order to survive in this chapter you MUST remember your trig chart. I know, I know, this section just keeps getting scarier.

Here's an example of finding the exact value.

Cos 75
Alpha is 45 and Beta is 30
Then plug it in to the formula.
cos(45 +30)= cos 45 * cos 30 - sin 45 + sin 30
Then simplify
(Square root of 2/ 2) (Square root of 3/ 2) - (square root of 2/ 2) (1/2)
Combine
Square root of 6 over 4 - square root of 2 over 4
So,
Cos 15 = square root 6 - square root of 2 over 4

This section is not very difficult, but it just takes time to memorize and go through all the steps.

This past week was exam week so we really didn't have much time to learn anything new. On Friday however we learned section 10-1.In this chapter we are not allowed to use calculators. The purpose of this section is to find exact values and to simplify, you must know the trig chart in order to learn this section.

You only need to know to Formulas:
Alpha and Beta come straight out of the trig chart

Cos(alpha +/- beta) = cosalpha cosbeta -/+ sinalpha sinbeta

Sin(alpha +/- beta) = sinalpha cosbeta +/- cosalpha sinbeta


Example:
Find the exact value of sine 150
a). Look on your trig chart and see if there are any two degrees that add up to 150. 90 and 60 add up to give you 150.
b). They want sine so plug into your sine formula. Sin( 90 + 60) = sin90 cos60 + cos90 sin60
c). Plug in from your trig chart. (1) (1/2) + (1) (square root 3/2)
d). Your final answer is 2

Chapter 10-Section 2

This week we only learned one new lesson. This was a good thing because it would have been tough to grasp new meterial and study for exams. We started 10-1 this Friday. In this chapter we are not allowed to use calculators. The purpose of this section is to find exact values and to simplify. IF YOUR GOING TO SURVIVE THIS SECTION YOU HAVE TO KNOW YOU TRIG CHART! THERE IS NO WAY AROUND IT!

There are only two formulas to remember:
Cos(alpha +/- beta) = cosalpha cosbeta -/+ sinalpha sinbeta

Sin(alpha +/- beta) = sinalpha cosbeta +/- cosalpha sinbeta

Alpha and Beta come straight out of the trig chart.


Ex:Find the exact value of sin 105 degrees.

1. There are two degrees from the trig chart that add up to equal 105, so those are the two numbers you will use. 60 and 45
Alpha = 60 Beta = 45

2. Since it says find exact value of sin , you will use the sin formula.

Sin(60 + 45) = sin60 cos45 + cos60 sin453

You fill in everything on the right of the equal sign with the trig chart.
(square root 3/2)(square root 2/2) + (1/2)(square root 2/2)4. Square root 6/4 + square root 2/45

The Final answer is square root 6 + square root 2/4

10-1

This past week we took nine weeks exams, so we worked on our study guides for chapters 7, 8, and 9 all week. After our exam on Thursday we started chapter 10 on Friday. In section 10-1 we learned how to find the exact values of trig functions and how to simplify them. This section does not allow us to use our calculators. All you need to know for this section is the trig chart and two formulas.

Formulas:
cosine (alpha -/+ beta) = cosine alpha cosine beta -/+ sine alpha sine beta
sine (alpha +/- beta) = sine alpha cosine beta +/- cosine alpha sine beta

Alpha and Beta come straight out of the trig chart

Example:
Find the exact value of sine 150
1. Look on your trig chart and see if there are any two degrees that add up to 150. 90 and 60 add up to give you 150.
2. They want sine so plug into your sine formula. Sin( 90 + 60) = sin90 cos60 + cos90 sin60
3. Plug in from your trig chart. (1) (1/2) + (1) (square root 3/2)
4. Your final answer is 2

Well, for me this week advanced math was pretty wild trying to learn everything for the exam. The one thing we had learned this week, thankfully, was a little bit easier than normal. In 10-1 we are allowed to use no calculators, in that case you must use only formulas and what you know from your head.

10-1 is all about the trig chart, alpha, beta, and the trig chart. Alpha(α) and Beta(β) come form the trig chart and add or subtract to get the angle you are looking for.

Formulas:
cos(alpha +/- beta)= cosαcosβ -/+ sinαsinβ
sin(alpha +/- beta)= sinαcosβ +/- cosαsinβ

When using formulas, to know if to add or subtract you figure out if you are adding alpha and beta or subtracting them to get angle. Which ever one you choose you go by those top symbols. (For example, adding 30 and 15 to get sin45; you would add again because that is the top symbol.

Example:

Find the exact value of sin15degrees

The two trig chart angles 45 & 35 subtract to get fifteen.
Alpha=45 Beta=30
By plugging that into the sin formula you get..
sin(45-30)-sin45cos30-cos45-sin30
Plug into trig chart which gives you..
=(Squaredroot2/2)(Squaredroot3/2)-(Squaredroot2/2)(1/2)
=Squaredroot6/4-Squaredroot2/4=Squaredroot6-2/4
Answer: sin15Squaredroot6-Squaredroot2/4

There are many way you will have problems in this section, you just need to remeber the formulas. Remeber: you always replace your left with right when needed to find the exact value or the formula given.

10-1

This week was more about reviewing than learning. We had our exam on Thursday, so we only learned something new on Friday. On Friday we started Chapter 10. Section 10-1 is about finding exact values of trig functions and simplifying them. We are NOT allowed to use calculators in this section. It all comes from your trig chart and we should all know that by now since we learned it in Chapter 7.

There are two formulas you have to know in this section:
cos(alpha -/+ beta) = cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) = sin alpha cos beta +/- cos alpha sin beta

*alpha and beta come straight from your trig chart and +/- to get the angle you are looking for, no worries.

EXAMPLE 1: Find the exact value of sin30 degrees.
A). What two numbers +/- to get 30.. 90 & 60! So your alpha will be 90 and your beta will be 60.
B). Since the problem is asking for the exact value of sin, we’re going to use the second formula.
So, plug your alpha and beta into it – sin(90 degrees – 60 degrees) = sin 90 degrees * cos 60 degrees – cos 90 degrees * sin 60 degrees.
C). Now that we have our formula filled in, replace the trig functions with values from the trig chart – (1)(1/2) – (0)(square root of 3/2)
D). Your exact value is ½.

Other than finding the exact value, you will be asked to simply like the ones we did in class on Friday. All you’re doing is replacing what’s on the right with what goes on the left side. This section will come easy to you if you know your formulas and trig chart!

10-1

This was a very tough week for me studying for exams, so I am glad we only learned one new lesson. We started 10-1 on Friday. The worst part about it is that we are not allowed to use calculators. The purpose of this section is to find exact values and to simplify. In order to do this section, you MUST know your trig chart. There are only two formulas to remember:

Cos(alpha +/- beta) = cosalpha cosbeta -/+ sinalpha sinbeta
Sin(alpha +/- beta) = sinalpha cosbeta +/- cosalpha sinbeta

Alpha and Beta come straight out of the trig chart.

Ex:
Find the exact value of sin 105 degrees.

1. There are two degrees from the trig chart that add up to equal 105, so those are the two numbers you will use. 60 and 45
Alpha = 60
Beta = 45

2. Since it is asking for sin, you will use that formula.
Sin(60 + 45) = sin60 cos45 + cos60 sin45

3. You fill in everything on the right of the equal sign with the trig chart.
(square root 3/2)(square root 2/2) + (1/2)(square root 2/2)

4. Square root 6/4 + square root 2/4

5. The Final answer is square root 6 + square root 2/4

10-1

This week we have been taking exams therefore, we have only learned one new lesson which is lesson 10-1. In this lesson we are NOT allowed to use calulators. Lesson 10-1 is showing us how to find the exact values and how to simplify trig functions. So hopefully by now you get that we CAN'T USE CALCULATORS. If not then idk what to tell you.. but I do know your going to fail. Also if you do NOT learn your trig chart you are going to fail.

FORMULAS:
sine(theta+/- beta)=sinetheta*cosbeta+/-costheta*sinebeta
cosine(theta=/-beta)=costheta*cosbeta+/-sinetheta*sinebeta

EXAMPLES:
cos 90 cos 45+ sine 90 sine 45
cos( theta- beta)
cos(90-45)
cos(45)
squareroot(2)/ 2

10-1

10-1 Formulas for Cosine(α±β) and Sine(α±β)
So section 10-1 is about sine and cosine formulas used to simplify and find exact values of trigonometric functions. NO CALCULATORS ARE TO BE USED IN THIS SECTION. They will give you the wrong answer anyway. It is also necessary, better yet MANDATORY that you remember, memorize or what ever it is that that you need to to know the trig chart.
The formuals for sin and cos are as follows :
cos(α-β)= cosα cosβ + sinα sinβ
cos (α+β)= cosα cosβ - sinα sinβ

sine (α-β)= sin α cosβ – cosα sinβ
sine(α+β)= sin α cosβ + cosα sinβ

Notice, when using cosine, if the left side of the formula is using cos(α-β), then the right side or the formula is using addition and vice versa.

Some problems require you to simply simplify.
Ex.
Cos 105 cos 15 + sin 105 sin 15
Cos (α-β)
Cos (105-15)
Cos (90)
0


Other problems ask you to solve them, or “find exact value”.
Sin(105)
α= 45 β=60
sin ( 45+60)
= sin45 cos60 +cos45 sin60
=(sqrt2/2)(1/2)=(sqrt2/2)(sqrt3/2)
=sqrt2/4 + sqrt6/4
Sin 105 = sqrt2 + sqrt6/ 4

:)

Recap

What concept do you feel like you have mastered this nine weeks?

I must admit..having Adv. Math and mRs. Robinson teaching it was a wake up call. I really had to get use to not getting babied by a teacher quickly because very few teachers have put the responsibility on us, This class forced me to study every night although i didnot want to. Ihope towards the beginning of the school year i will be strapped and geared for calculus.. I feel like i have mastered Ch.7 and 8.. graphing amplitudes aND periods.trig chart work.and finding coterminals, reference angles and exact value. The method i enjoy most is converting degrees to radians vice versa

i Sruggle the most with Ch.9.. the triangles seem to confuse me because i do not remember alot from geometry..and i have a tendency to mix up formulas at times.

AHA..studying ALOT more often and capitilizing on the freetime alot better by actually working rather than daydreaming 60% of the time. Ive realized that math is a subject that u MUST practice consistently in order to become better at it( unless ur jimmy neutron). Once i learn how to prepare my self for your class..maybe..hopefully ill do better.

Week 7 Blog Prompt

What concept do you feel like you mastered this nine weeks? What concept did you struggle with the most? Why? What can you change this nine weeks in your study habits, etc to improve your grade?

Chapter 9 cont'd

So this week we pretty much wrapped up chapter nine from last week. We basically spent this week in review for our exams. This is seventh hour so I think (by think i mean pretty sure, almost positive) our exam is Thursday. This means that we have a little extra time to catch the hang of anything you don't have completely down packed by now.
As usual the chapter opens with the more easier subjects. 9-2 consists of finding the area of triangles. This hater is relatively simpler than the other sections because it has a rather simple formula that is quite easy to
remember:
( 1/2bh )for right triangles.
( 1/2ab(sin(angle btwn)) for non right.

Section 9-3 illustrates the law of sines.
This formula can only be used with non right triAngles. * it can only be used when you have the angle opposite leg identified.
( sin (angle) ) / opp leg = ( sin ( angle2) / opp leg 2
To solve cross multiply and divide.

If you have neither paired sides/Angles you have to use Another formula which acts as a last resort. The lA of Cosines is used when you are looking for the area of non right triAngle Nd you do not have a pair.

Opp leg^2= adj leg^ 2 +adj leg^2 - 2(adj leg)(adj leg)cos(angle btwn)

Example:
a.5 in b.2 in ang.C 35

To find area of this nonright triangle use the formula 1/2ab sine(ang)
1/2(5)(2) sin(35)
* enter in calc as in ( in degree mode )

Answer:2.87 inches squared
* area is always squared!

:)

9-3 & 9-4 & 9-5

In lesson 9-3, you learned about Law of Sines. When you dont have a right triangle you are going to want to use the Law of Cosines, because you can't use SOHCAHTOA. You are only going to be able to figure the triangle out correctly is if you know the opposite leg and angle value. We learned that our final result is Law of Cosines. We do Law of cosines when you do not have a pair in a non-right triangle. We also learned navigation in lesson 9-5 which also ties in with the different Laws. In Navigation it depends what quadrant you land in as to which direction you are pointed. If your in quadrant one then your north east. If your in quadrant two then your north west. If your in quadrant three then your south west.

FORMULA:
sin(angle)/ opposite leg=sin(angle2 / opposite leg 2
Ex.
A=50(degrees

) B=55(degrees) a=10
now you are going to draw your triangle
now subtract 50 and 55 from 180 to give you your third angle.
C= 75(degrees)
use the formula
sin75 / 10=sin50/x
xsin75= 10sin50
10sin50/ sin75
= 10.249

This week in advanced math we learned about Law of Cosines and Law of Sines. Law of Cosines was more difficult than Law of sines. I understand both of them.

Law of Cosine’s formula is:
(Opp. Leg)^2 = (adj. Leg) ^2 + (adj. leg) ^2 -2 ( adj. leg) x (adj. leg) x cos( angle between )

Example :
Suppose triangle ABC had 2 sides measuring c = 2cm and b = 4cm. The angle between them had a measure of A = 100 degrees. Find the third side.

First, you draw your triangle.

Second, you check to see if you can use law of sines. When you see that you can not, your last option is Law of cosine. Then, you plug into the formula.

a^2 = (2)^2 + (4)^2 – 2 (2) x (4) x cos(100 degrees)

Then, plug the entire equation into your calculator to get your answer.
a = square root((2)^2 + (4)^2 – 2 (2) x (4) x cos(100 degrees))

a = 4.773 cm

GO SAINTS!!!!!!

Ok I’m starting off this blog by saying GO SAINTS!!!!!!! WOOOOHHHHHOOOOO. How good are they? They are almost as good as Tulane. OMG!!!!!! Anyway, this week we finished up Chapter 9. This chapter wasn’t really hard if you can remember the formulas. Law of Sines is a formula used when there is a pair, which consists of an angle and its opposite side measure, and it is used to either find another angle or another side length. After that we learned about the Law of Cosines. This is a last resort formula used when SOHCAHTOA cannot be used and there is not a pair for Law of Sines to work. This extremely long formula can be manipulated to find either the opposite side from an angle or the angle itself. We also learned how to do Global Navigation. I think this is an extremely useful lesson. Whether you’re just out fishing or captaining a Navy vessel this skill will apply.

9-4

This week in advanced math we learned how to do law of cosines. This law is used when you don’t have a pair in a non-right triangle.

The formula for solving for law of cosine is (opposite leg) ^2= (adjacent leg) ^2+ (adjacent leg) ^2-2(adjacent leg) (adjacent leg) cosine (angle between).


Example: problem dealing with the law of cosine.

Suppose 2 sides of a triangle have lengths 3cm and 7cm and the angle between their measures is 130 degrees. Find the 3rd side.

1. You do x squared=7squared+3squared-2(3) (7)Cos 130 degrees
2. Everything is squared so when you plug it all into your calculator you must square root all of it.
3. You get 9.219
4. 9.219 is your 3rd side.



The entire chapter was about solving all kinds of different triangles. Law of Cosines was one of the easiest sections in chapter 9. You use law of cosines when you cannot use SOHCAHTOA or law of sines. All you have to do to solve the triangle is plug in the formula:

This week in advanced math we learned how to do law of cosigns. It was kinda difficult but not really. Next week we have exams so we will be doing alot of reviewing.

LAW OF COSINES:
the formula: (opposite leg)^2=(adjacent leg)^2+(adjacent leg)^2-2(adjacent leg)(adjacent leg)cosine(angle between)


Lets say two sides of a triangle have lengths of 6cm and 12cm and the angle between is 60 degrees. What is the third side?
1)first you must draw your triangle and label it!
2) since we already know our 2 angles and the angle between, we can plug it into the formula.
x^2=6^2+12^2-2(6)(12)cos60 degrees.
3) everything is squared so when you plug it all into your calculator you must square root all of it.
4) your answer is 10.392

NINE FOURRRRRRR

This week in Advanced Math, our seventh hour honors advanced math class finished learning about and later tested on chapter nine. In section four chapter nine, my class learned about the Law of Cosines. This law is used for non-right triangles which is a triangle without a 90 degree angle. [For a 90 degree angle use SOHCAHTOA.]
The formula is: (opposite leg)^2 = (adjacent leg)^2 + (adjacent leg)^2- 2 (adjacent leg) (adjacent leg) cos (angle between)

So, for this section only one example is necessary. And the example is………

Suppose two sides of a triangle have lengths 4cm and 8cm and the angle between them measures 140 degrees. Find the third side.

A). Draw your triangle and label it.

B). We know that 4cm and 8cm are our adjacent legs and 140 degrees is our angle . So plug these numbers into the formula – x^2=8^2 +4^2 -2(8)(4)cos140 degrees

C). Since everything is squared, you must find the square root. Square root everything and plug
it into your calculator.

D). Your answer is 11.4

This past week we finished up Chapter 9. Chapter 9 has been the easiest so far for me. In 9-3 we did Law of Sines and in 9-4 we did Law of Cosines.

LAW OF COSINES:
the formula: (opposite leg)^2=(adjacent leg)^2+(adjacent leg)^2-2(adjacent leg)(adjacent leg)cosine(angle between)
Lets say two sides of a triangle have lengths of 6cm and 12cm and the angle between is 60 degrees. What is the third side?
A)first you must draw your triangle and label it!
B) since we already know our 2 angles and the angle between, we can plug it into the formula.
x^2=6^2+12^2-2(6)(12)cos60 degrees.
C) everything is squared so when you plug it all into your calculator you must square root all of it.
D) your answer is 10.392

This upcoming week will pretty much be an exam review. I feel pretty confident with all these sections so far.

9-4

Law of Cosines

This week in advanced math we finish up chapter 9. We didn’t really learn anything new because exams are next week. We began to review chapters 7, 8, and 9 for our exam next Thursday. Chapter 9 was probably the easiest of the three. The entire chapter was about solving all kinds of different triangles. Law of Cosines was one of the easiest sections in chapter 9. You use law of cosines when you cannot use SOHCAHTOA or law of sines. All you have to do to solve the triangle is plug in the formula:

(opposite leg)2 = (adjacent leg)2 + (adjacent leg)2 – 2 (adjacent leg) (adjacent leg) cos(angle b/w)

Example 1:

Triangle ABC a = 2 b = 6 and angle C = 80°
1. Draw your triangle and label your sides.
2. Plug into your formula c2 = 22 + 62 – 2 (2)(6) cos 80°
3. Square root everything and plug it into your calculator
4. Your answer will be 5.99

9-3

This week we finished chapter 9, but we did not really learn anything new. It was more of a review. Chapter 9 was one of the easiest for me because there are only 3 formulas that you needed to remember which are Law of Sine, Law of Cosine, and SOHCAHTOA. Law of sine was pretty simple, but not commonly used. You use it when you have a pair of angles.

The formula for law of sine:
Sin(angle) / opposite pair times sin(angle) / opposite pair

Example one: triangle ABC a= 7 c= 3 A= 52 B= 68
Say you are asked to find b. Your first step would be to draw the triangle.

Sin(52 )/ 7 times sin(68) / b

Now, you have to cross multiply, which gives you bsin(52) = 7sin(68).

Divide sin(52) from both sides.
b= 7sin(68)/sin(52)

All you have left to do is plug it into the calculator exactly how you see it.
Answer: b= 8.2

9-4

We finished up and tested on Chapter 9 this week. Exams are next week so we will begin reviewing. One of the easiest sections in Chapter 9 for me was 9-4. It was about Law of Cosines. It is used when you don’t have a pair in a non-right triangle.

The formula is: (opposite leg)^2=(adjacent leg)^2 +(adjacent leg)^2-2(adjacent leg)(adjacent leg)cos(angle b/w)

EXAMPLE 1: Suppose two sides of a triangle have lengths 4cm and 8cm and the angle between them measures 140 degrees. Find the third side.
A). Draw your triangle and label it.
B). We know that 4cm and 8cm are our adjacent legs and 140 degrees is our angle . So plug these numbers into the formula – x^2=8^2 +4^2 -2(8)(4)cos140 degrees
C). Since everything is squared, you must find the square root. Square root everything and plug
it into your calculator.
D). Your answer is 11.4

EXAMPLE 2: In triangle ABC, a=8, b=6, and angle c=50 degrees.
A). Draw your triangle and label it.
B). We know that 8 and 6 are our adjacent legs and 50 degrees is our angle. So plug these numbers into the formula - x^2=8^2 +6^2 -2(8)(6)cos50 degrees
C). Square root everything and plug it into your calculator.
D). Your answer is 6.2

This section is pretty simple because if you don’t have a pair to use Law of Sines then you just use Law of Cosines. Then all you do is plug your numbers in and solve.

Week 6 Prompt

How is SOHCAHTOA in Ch. 9 and The unit circle in Ch. 7 connected? Give examples of the connection.

9-4 Law of Cosines

This week in advanced math we learned that we will not always be able to use SOHCAHTOA. We learned that our final result is Law of Cosines. We do Law of cosines when you do not have a pair in a non-right triangle. We also learned navigation in lesson 9-5 which also ties in with the different Laws. In Navigation it depends what quadrant you land in as to which direction you are pointed. If your in quadrant one then your north east. If your in quadrant two then your north west. If your in quadrant three then your south west. And if your in quadrant four then your in south east. But back to Law of Cosines, here is the formula.

The formula for Law of Cosines is (opp. leg)^2= (adj. leg)^2 + ( adj. leg)^2 -2 (adj. leg) (adj. leg) cos (angle b/w).

Ex. If there is a triangle and the opposite side is 4, the adjacent side is 6 and the angle is 140 degrees.

4^2= 6^2 + 6^2 -2 (6)(6) cos(140)

Then that is how you will set it up.

No more right triangle :(

Now we don't get right triangles as often anymore. We can no longer use the short cut of SOHCAHTOA. Now we have to use either the law of sines or the law of cosines. You can only use the law of sines if you have a matched pair of an angle and a side. If you do not you must use the law of cosines. These two formulas are extremely useful.

The equation for law of sines is: Sin(angle)   Sin(angle2)
                                                  opp leg  =opp leg 2
                                         This equation is used by plugging in and cross multiplying

The equation for Law of Cosines is:
      (opp leg)^2=(adj leg)^2+(adj leg2)^2-2(adhleg)*(adj leg)*cos(angle between)

Now I don't know who sits around and tries to figure out this stuff but they obviously had a good reason to do it. They must have expected this to be used.

9-3

This week in advanced math we are learning lesson 9-3.. Law of Sines. When you dont have a right triangle you are going to want to use the Law of Cosines, because you can't use SOHCAHTOA. You are only going to be able to figure the triangle out correctly is if you know the opposite leg and angle value.
FORMULA:
sin(angle)/ opposite leg=sin(angle2 / opposite leg 2

REMEMBER: always cross multiply to solve.

EXAMPLE:
A=50(degrees) B=55(degrees) a=10
now you are going to draw your triangle
now subtract 50 and 55 from 180 to give you your third angle.
C= 75(degrees)
use the formula
sin75 / 10=sin50/x
xsin75= 10sin50
10sin50/ sin75
= 10.249

This is a really confusing section.

The thing I understood most this week was section 9 – 3, Law of sines. The law of sines can only be used with non –right triangles. The only time you can use law of sines is when you have or are given one side, and one angle that are opposites of each other. This is the thing that you try before you use the law of cosines.
The formula is (sin (angle) / opposite leg) = (sin (angle 2) / opposite leg)
To solve the formula you have to cross multiply.
Example 1:
Triangle ABC, angle B = 120 degrees, b = 10, c = 8.
Sin ( 120 degrees ) / 10 = sin(C) / 8
10sin ( C ) = 8 sin( 120 degrees )
10 sin ( C ) = 6.928
Sin ( C ) = 6.928 / 10
C = Sin^ -1 ( 6.928 / 10 )
C = 43.8522 degrees

9-3 is all about the Law of Sines. The law of sine is only used with NON-RIGHT TRIANGLES. You are only about to use the law of sines when you have an angle and an opposite leg value that you know.

Example: An engineer wants to determine the distances from points A to B to a inacessible point C. From direct measurement the engineer knows the AB=25cm AngleA=110° AngleB=20°. Find AC&BC

a.You draw out your triangle from what you already know. Once draw you see the you do not know legs a or b and AngleC.
b. To find angle c subtract the two angles known by 180°.
180-110-25=50°
c.Plug into formula..
sin50/25=sin110/AC
d.Cross multiply to get BCsin50=25sin10
e.Get BC by itself by dividing 50.
f. Answer for angle BC is 25sin110/sin50 which equals 30.667
-Next solve or AC-
a.ACsin50=25sin20
b.Divide to get AC alone.
c.Answer is 25sin20/sin50 which equals 11.162

9-3

This week we learned 9-3. It was explaining and how to solve with law of sines. You use this whenever you can’t use SOHCAHTOA to solve the triangle.

Law of sines is used with non-right triangles and only use when you have an angle plus opp leg values you know.


The formula for law of sines:
Sin (angle)/opp leg = Sin (angle 2)/opp leg 2

You cross multiply to solve law of sine.



Ex:
AB=25m
Find AC and BC
1. Draw the triangle.
2. Find AC.
3. Sin 20/AC= Sin 50/25.
4. You get 25 sin (20) =sin (50) x.
5. You divide sin 50 on both sides.
6. You get 11.612m which is AC.
7. Find BC
8. Sin 50/25= Sin 110/BC
9. You get BC Sin 50= 25 Sin 110.
10. Divide Sin 50 on both sides
11. You get 25 Sin 110/ Sin 50
12. You get BC= 30.667m.



Basically if you know your formula and everything 9-3 is really easy but it can be really difficult if you do not know your formula. Know your formula and everything and you should be good and comfortable with 9-3.

This week we learned 9-3. Law of Sines. You use law of sines when you cant use SOHCAHTOA to solve the triangle.
you can only use law of sines when you have an angle and opposite leg value that you know.

Formula for Law of sines:
sin(angle) / opposite leg=sin(angle2) / opposite leg 2
you cross multiply to solve.

EXAMPLE:
A=45 degrees B=60 degrees a=14
**first you must draw your triangle!
subtract 45 and 60 from 180 to find your 3rd angle.
C=75 degrees.
set it up into your formula.
sin(75) / 14=sin(45) / x.
cross multiply.
xsin75=14sin45
set it up to divide.
14sin45 / sin 75.
plug it into your calculator and you should get an answer of:
x=10.249

Its really not that hard to work these problems. The hardest part is taking the points and making a triangle to solve from word problems.

9-3 & 9-4

We learned the law of sine and the law of cosine this. I think that the law of cosine is much easier to do because you basically just plug all of your work into the calculator. It is hard to learn what formulas to use with which triangles. There is also SOHCAHTOA. You cannot use the law of cosine on a right triangle. The only time you can use it is when you do not have any pairs on a non - right triangle. Most of these start off with a word problem where you have to draw your own triangle, so it can get a little tricky.

The formula for law of cosine is (oppleg)^2 = (adjleg)^2 + (adjleg^2) – 2(adj leg) (adj leg) cos (angle between)

Ex : Suppose 2 sides of a triangle have lengths of 10cm and 8 cm and the angle between them measures to be 140 degrees. Find the 3rd side. Once you draw your triangle, you plug the numbers into the formula. X^2 = 10^2 + 8^2 – 2 (8)(10)cos140 The first step is to get rid of the squared on the x, so you square root it.The next step is to put it in your calculator. X= 96.281 ends up being your answer.

As i pretty much say in every blog Advanced Math is getting harder and harder. Everything you learn is accumulative and has to be remembered in order to solve other problems. Right now we are still in Chapter 9 working on triangles, which trigonometry is.

9-3 is all about the Law of Sines. The law of sine is only used with NON-RIGHT TRIANGLES. You are only about to use the law of sines when you have an angle and an opposite leg value that you know.

Formula: sin (angle) = sin (angle 2)
______________ _________________
opp leg opp leg 2

TIP: you must always cross multiply to solve.

Example: A civil engineer wants to determine the distances from points A to B to a inacessible point C. From direct measurement the engineer knows the AB=25cm AngleA=110° AngleB=20°. Find AC&BC

1.You draw out your triangle from what you already know. Once draw you see the you do not know legs a or b and AngleC.
2. To find angle c subtract the two angles known by 180°.
180-110-25=50°
3.Plug into formula..
sin50/25=sin110/AC
4.Cross multiply to get BCsin50=25sin10
5.Get BC by itself by dividing 50.
6. Answer for angle BC is 25sin110/sin50 which equals 30.667
-Next solve or AC-
1.ACsin50=25sin20
2.Divide to get AC alone.
3.Answer is 25sin20/sin50 which equals 11.162

Each problem is different but you always follow the same rules. Sometimes you may have to find angles, and don't forget when solving for an angle you do the inverse.

9-3

9-3 Law of Sines

Law of Sines is a method used to solve triangles that cannot be solved using SOHCAHTOA. Right now we are going to learn Law of Sines. Law of Sines is used with non right triangles. You can only use it if you have an angle and an opposite leg value you know. The formula to solve these triangles is: Sin(Angle)/Opposite Leg = Sin(Angle 2)/Opposite Leg 2 *Cross multiply to solve
Let’s try an example problem:

Example 1:
In triangle ABC angle B = 128°, side b = 14 and side c = 6. Determine whether or not angle C exists. If it does exist find all possible measures of angle C.

Sin 128°/14 = Sin C/6
6 Sin 14 = 14 Sin C
Sin C = 6 Sin 128°/14

At this point you will take the inverse = C = Sin-1(6 Sin 128°/14)
C = 72.435

Since you are taking the inverse you want two answers. So make your answer negative and add 180°.
-72.435 + 180 = 107.565

Your final answers are 72.435 & 107.565

9-2

We are in chapter nine in advanced math right now. In section 9-2, we learned how to find the area of a triangle given the lengths of two sides and the measure of the included angle. There are two formulas in this section – one for a right triangle and one for a non-right triangle.

1/2bh – only right triangles
1/2absin(angle b/w) – non right triangles

EXAMPLE 1: Two sides of a triangle have lengths 6 cm and 3 cm. The angle between the sides measures 70 degrees. Find the area of the triangle.
A). First, draw your triangle of course.
B). Label the two sides with 6 cm on one and 3 cm on the other.
C). Since we are not told it is a right triangle, we’re going to use the second formula
(1/2absin(angle b/w).
D). Plug in all of your numbers given. (1/2(6)(3)sin70 degrees)
E). Multiply to get 9sin70 degrees.
F). Your answer is A = 8.46 cm squared

EXAMPLE 2: In triangle ABC b=3, c=5, and a=60 degrees. Find the area.
A). Draw your triangle and label the sides.
B). Plug your numbers into the second formula, since we aren’t told it’s a 90 degree angle.
(1/2(3)(5)sin60 degrees).
C). Your answer is A = 6.5 *We aren’t given any units, so leave it alone.

Law of Sine and Cosine

This week in advanced math we started chapter 9, we learned law of sine and law of cosine. You may consider the law of cosine easier simply because after you have plugged all angles and sides into the formula you just plug that same formula into your calculator to get your answer. You must examine your problem very closely in order to know which law to use, be sure that all requirements of a formula are there in your triangle. It is hard to learn what formulas to use with which triangles. Along with law of sine and law of cosine there is SOHCAHTOA which involves a right triangle.

Law of Sines is used with a non-right triangle and an angle and opposite leg value that you know.
Formula: sin(angle) / opp leg = sin(angle 2) / opp leg 2
**cross multiply to solve

Ex 1: In triangleABC, angleS=126°, s= 12 and t=7 Determine whether angleT exists.
1.First draw and label the triangle.
2.Now we take our formula and plug our values into it.

Sin126°/12 = sinT/7 (multiply by 7 on each side)
7sin126°/12 =sinT
T= sinINVERSE(7sin126°/12)
T=28.159°

Law of Cosines is used when there is no pair in a non-right triangle.
Formula: (oppleg)^2 = (adjleg)^2 + (adjleg^2) – 2(adj leg) (adj leg) cos (angle between)

Ex 1: Suppose 2 sides of a triangle have lengths of 10cm and 8 cm and the angle between them measures to be 140 degrees. Find the 3rd side.

Once you draw your triangle, you plug the numbers into the formula.

X^2 = 10^2 + 8^2 – 2 (8)(10)cos140
Now we must square root the and the right side as well to get x.
Once that is done you can plug it into your calculator, make sure when you plug this into your calculator you put parenthesis.

And TADAAAAA X= 16.928

Chapter 9-4 Laws of Cosine

This week in Advance Math, the class learned the Laws of Sine and Cosine. I find the Law of Cosine to be much simpler and easier to comprehend because the majority of your work is done via calculator. The formulas are somewhat difficult and easy to confuse for one another. One method you can use is SOHCAHTOA(exclusive to right triangles). Law of cosine cannot be used for right triangles. it is only to be used on no right triangles without pairs.

The formula for law of cosine is (opp.leg)^2=(adj.leg)^2 + (adj.leg^2) - 2(adj. leg) (adj.leg) cos.

Ex:1 Two sides of the triangle have lengths of 12 cm and 6 cm and the angle between them measures to be 150 degrees. Find the 3rd side. Once you draw the triangle, plug in the numbers.

x^2=12^2 + 6^2- 2(6)(12)cos150.

step1: eliminate the squared x by taking the square root which leaves you with x.
step:2 plug into calculator

Answer(x):36

9-4 Law of Cosines

This week we learned the law of sine and the law of cosine. The law of cosine is much easier to do because you basically just plug all of your work into the calculator. It is hard to learn what formulas to use with which triangles. There is also SOHCAHTOA.

You cannot use the law of cosine on a right triangle. The only time you can use it is when you do not have any pairs on a non - right triangle.

Most of these start off with a word problem where you have to draw your own triangle, so it can get a little tricky. The formula for law of cosine is (oppleg)^2 = (adjleg)^2 + (adjleg^2) – 2(adj leg) (adj leg) cos (angle between)

Ex 1: Suppose 2 sides of a triangle have lengths of 10cm and 8 cm and the angle between them measures to be 140 degrees. Find the 3rd side.

Once you draw your triangle, you plug the numbers into the formula.

X^2 = 10^2 + 8^2 – 2 (8)(10)cos140

The first step is to get rid of the squared on the x, so you square root it.

The next step is to put it in your calculator.

X= 16.928 and that is your answer.