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!)

Feroz - Blog 5

9 - 1: Geometry with a hint of retarded-ness.

Solving right triangles using SOHCAHTOA.

For the sake of taking up space, I'm going to explain what this stands for. Ahem..

SinΘ = opposite/hypotenuse

CosΘ = adjacent/hypotenuse

TanΘ = opposite/adjacent

CscΘ = hypotenuse/opposite

SecΘ = hypotenuse/adjacent

CotΘ = adjacent/opposite

Note: And easy way to remember Csc, Sec, and Cot is that they're the opposite of thier respective trig functions, but you probably already knew that.

Example time.

In triangle WTF, angle W = 90, angle T = 12, and t = 12. Find w and f.

This is gonna be hard for me to explain without a triangle in front of me so don't expect anything great.

To find w, you would use:

Sin 25 = w/18

So w = 18sin25
which is 7.607

Now to find f, use:

Cos 25 = c/18

c = 18cos25
which is = 16.314

There you go. You just solved the triangle WTF.

Blogggggg # Idk I lost count...

This week we talked about triangles and SOHCAHTOA from geometry but we went more in depth with it I guess. To use SOHCAHTOA the triangle has to be a right triangle. SOHCAHTOA is just a crutch to help people remember the functions to use to solve for the triangle. SOH stands for Sine=opposite/hypotenuse. CAH is Cosine=adjacent/hypotenuse and TAN is short for Tangent=opposite/adjacent. Hypotenuse is the leg opposite the right angle. Adjacent is going to be the measurement of the leg stemming from the angle you are using, and Opposite is just the measure opposite of the angle you are using. You just plug in the ones you have and solve for the triangle. So all you do is look at the triangle and choose the function that you need and then you're good to go. Easy right?

Mary's blogg #6

this week we learned about right triangles, the most important ones to remember are sin, cos,and tan. An easy way to remember these is SOHCAHTOA.

sin(theta)=opposite/hypotenuse

cos(theta)=adjacent/hypotenuse

tan(theta)=opposite/adjacent

you can find all the parts of the triangle with just the ones on top, but if they ask for something specific like csc, sec, or cot,you go by this

csc(theta)=hypotenuse/opposite

sec(theta)=hypotenuse/adjacent

cot(theta)=adjacent/opposite

this will only work if you are given a RIGHT triangle, where the hypotenuse will be opposite of the right angle.

When you are given an angle and the right angle, you will go and find the other angle inside the triangle, but you should still always use the one you are given incase you have made a mistake, because if you made a mistake on that, your whole problem will be WRONG!

Example.

A=90 C=65 and a=18

you draw your triangle and use 180-90-65, to get your B angle.

then you use cos to find c

and you use sin, to get b

This week we went over section 9.1 which talked about right triangles. Some easy ways to remember your formulas is by remembering SOHCAHTOA.

sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite

You only use these functions when you are trying to deal with right triangles. The hypotenuse is the opposite of the right angle.

Ex.

For the triangleABC find the value of b to 3 significant digits

opp leg= 40 degrees

adj leg=?

b tan 28 (degrees)= 40/b

b tan28(degrees)=40

b= 40/tan28=

b=75.2 degrees

Dom's blog

This week, we did chapter 9-1. It talked about law of sins and solving for right triangles

SOCAHTOA is one easy way to remember all of these formulas.

sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite


These functions are only to be used when dealing with right triangles. The hypotenuse is opposite of the right angle.

For example: one leg of the right triangle ABC is 4 while the hypotenuse is 5 & angle B is 20 degrees. Find the second leg. First you draw the right triangle. Then you do sin20=opposite/hypotenuse. Which is sin20=x/5. mulitply 5 on both sides to get x=5sin20. which gives you 1.7

Kaitlyn's Blog #6

I will be going over section 9-1

In this section, we learned how to find the angles and sides of a triangle. There are many formulas for this so it would make it easier to remember SOHCAHTOA!

sin(theta)=opposite/hypoteneuse sec(theta)=hypoteneuse/opposite
cos(theta)=adjacent/hypoteneuse csc(theta)=hypoteneuse/adjacent
tan(theta)=opposite/adjacent cot(theta)=adjacent/opposite

Example: A=90 degrees, B=52 degrees, a=5

First you want to find your other angle, angle C. To do this you would subtract 90 and 52 from 180. angle C=38

Then you want to find side c. Use your angle 52, and since c is adjacent and your other side is the hypoteneuse, then you would find this by using cos=adjacent/hypoteneuse.
side c=3.078

Then you need to find side b. Use your angle 52, and since b is opposite and your other side is the hypoteneuse, you would find this by using sin=opposite/hypoteneuse
side b=3.940

Helen's Blog #6

I'll be going over section 9-1.

*Always remember SOHCAHTOA

Sin(theta)=opp/hyp
Cos(theta)=adj/hyp
Tan(theta)=opp/adj
Csc(theta)=hyp/opp
Sec(theta)=hyp/adj
Cot(theta)=adj/opp

You only use these functions when you are trying to deal with right triangles. The hypotenuse is the opposite of the right angle.

Example:

In Triangle ABC, Angle A= 90 degrees, Angle B= 25 degrees, and a= 18, Find b and c.

To find b you would use:
sin 25= b/18

b= 18sin25
= 7.607

To find c you would use:
cos25=c/18

c=18cos25
=16.314

In Triangle DEF, Angle D=90, Angle E= 12 degrees, and e=9. Find d and f.

To find f you would use:

tan12=9/f

f=9/tan12
=42.342

To find d you would use:

sin12=9/d

d=9/sin12
43.288

Nathan's Blog

I will go over the notes from chapter 9, which involved right triangles. I understood section 9-1 really well.

sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite
An easy way to remember all of this is SOCAHTOA.

These functions are only to be used when dealing with right triangles. The hypotenuse is opposite of the right angle.

I will give a word problem example.

The safety instructions for a 20 ft. ladder indicate that the ladder should not be inclined at more than a 70 degree angle with the ground. Suppose the ladder is leaned against a house at this angle. Find the distance from the base of the house to the foot of the ladder and the height reached by the ladder.

Draw a triangle with a 70 degree angle and a 20 ft. hypotenuse. Then you draw a little house next to it. Little a is the distance on the ground and little b is the height.

So you get:
cos70 degrees=a/20
a=20cos70=6.840 ft.

For b:
sin70=b/20
b=20sin70=18.794 ft.

That is basically it for 9-1. It was an easy section.

Malorie's Blog

9-1 Right Triangles

In this section we learned about right angles and SOHCATOA. It was pretty easy because I remembered most of it from Geometry with Coach Joe.

In order to do this, you must know that follow formulas:

Sin Theta=Opp/Hyp Csc Theta=Hyp/Opp
Cos Theta=Adj/Hyp Sec Theta=Hyp/Adj
Tan Theta=Opp/Adj Cot Theta=Adj/Opp

(*Hint: this is only used with right angles and Hyp is opposite the 90 degree angle.)


Example:

In Triangle ABC, Angle A = 90 degrees, Angle B = 30 degrees, and a=15. Find b and c.

In order to figure out c, you must do cos b=c/15
c=12.99

In order to figure out b, you must use the pythagorean theorem.
b=7.5


So, now you know how to figure out the measurements of right triangles.

Tori's blog

This week our class went over Chapter 9 Section 1. For me, this lesson was really easy because it was pretty much a repeat of what I learned last year in Geometry.
We learned how to solve right triangles using sin, cos, tan, sec, csc, and cot. And we also used SOHCAHTOA to help us remember how to solve them.
SOHCAHTOA stands for: sine opposite hypotenuse//cosine adjacent hypotenuse//tangent opposite hypotenuse. Also we had to remember, the hypotenuse was opposite of the right angle, and a triangle equaled 180 degrees. We also had to keep in mind that these guidelines will only work for right triangles.

For example:: 1=3 2=4 <=40 <=90

To solve for the hypotenuse, you can use the Pythagorean triplets, which means that the hypotenuse equals 5. Now, to solve for the opposite angle you can just subtract 40 from 90, which equals 50.

Lawrence's blog

This week, we did chapter 9-1. It talked about solving for right triangles and law of sines.

Right triangle:
1. hypotenuse is opposite of the right angle
2. area= 1/2 base times height
3. SOH CAH TOA
-sin theta =opposite/hypotenuse
-csc theta=hypotenuse/opposite
-cos theta=adjacent/hypotenuse
-sec theta=hypotenuse/adjacent
-tan theta=opposite/adjacent
-cot theta=adjacent/opposite

Ex: In triangle ABC, angle A=28 degrees, angle C=90 degrees, side a=49, what is side b?

tan28degrees=40/b
btan28degrees=40

b=75.229

Ex: In triangle ABC, angle A=50 degrees, angle C= 90 degrees, side c=10, what is side a?

sin 50 degrees= a/10

=10 sin 50 degrees= a

a= 7.660

Law of Sines:

sinA/a = sinB/b = sinC/c

Law of sines is used when you know pairs in non-right triangles

To use the law of sines, you are setting up a proportion

Ex: in triangle ABC, angle A=115 degrees, side a=123, side b=16, what is angle B?

sin115 degrees/123 = sin theta/16

123sin theta=16sin115 degrees

sin theta=16sin115 degrees/123

theta =sin inverse (16sin115 degrees/123)

theta =6.771 degrees

Ex: in triangle ABC, angle A=30 degrees, angle B=15 degrees, side a=4, what is side B?

sin 30 degrees/4 = sin 15 degrees/x

x sin 30 degrees = 4 sin 15 degrees

x= 4 sin 15 degrees/ sin 30 degrees

=4 sin 15 degrees/ 1/2

x= 2.071

Charlie's.

This week we reviewed for our chapter test on Monday, then actually took the test on Tuesday and Wednesday; but on Thursday and Friday we took and reviewed 9-1 notes.
We have the 'word' that's going to help us remember...
( SOHCAHTOA )
... that
sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent
then by knowing that sec is the reciprocal of cos we know that
sec= hypotenuse/adjacent &we know that csc is the opposite of sin
so csc=hypotenuse/opposite.
These formula's only work with right triangles.
We also did SUPER long word problems.

*For example on how to do this:
one leg of the right triangle ABC is 4 while the hypotenuse is 5 & angle B is 20 degrees. Find the second leg. First you draw the right triangle. Then you do sin20=opposite/hypotenuse. Which is sin20=x/5. mulitply 5 on both sides to get x=5sin20. which gives you 1.7

Taylor's 5th (or 6th I have no clue XD) Blog

These are my notes from lesson 8.1.

In this lesson we will be solving for THETA. In order for you to be able to solve for Theta you need to get the trig. function by itself and then you need to take an inverse of it. An inverse will have two answers, but there may be exceptions to this rule if the question asks you to just find one quadrant. Any way, you need to find where an angle is based on the trig function and then you need to tell if it is negative or positive.

These are the steps:

First, you need to take the inverse of the number so that you can find the First Quadrant angle.

Then you simply need to remember this in order to find the other answers:

Quadrant 2, you need to make the first quadrant angle a negative and add 180 degrees to it.

Quadrant 3, you just need to add 180 degrees to the first quadrant angle.

Quadrant 4, you need to make the first quadrant angle a negative and add 360 degrees to it.

Now let's solve some problems!!

(REMEMBER I DO NOT KNOW HOW TO PUT A GRAPH ON HERE SO PLEASE NOTE THIS. WHEN I SAY DRAW A QUARDNENT PLANE DO SO AND CHECK THE ONES THAT I TELL YOU TO OK!!)

Solve for Sin A= 3/8 for A.

Sin A = 3/8 First, write the equation down.

A= Sin^-1(3/8) Then put the equation in inverse form like I just did.

A= 22.024 degrees Then you find the inverse. Ok, now draw you quadrant plane. Now since our Sin is positive we need to put a check in the quadrants where Sin or Y would be positive. Which is the First and Second Quadrant. So since we have the First Quadrant already we just need to find the Second Quadrant.

-22.024 degrees + 180 degrees= 157.976 degrees Finally you just use the formula that I gave you to find the angle of the Second Quadrant and this is your answer.

So your answers should be 157.976 degrees and 22.024 degrees.

Let's try cosine next.

Solve for Cos H = 4/6 for H.

Cos H = 4/6 First, write the equation down.

H= Cos6-1(4/6) Then put the equation in inverse form like I just did.

H= 48.190 degrees Then you find the inverse. Ok, now draw you quadrant plane. Now since our Cos is positive we need to put a check in the quadrants where Cos or X would be positive. Which is the First and Fourth Quadrant. So since we have the First Quadrant already we just need to find the Fourth Quadrant.

-48.190 degrees + 360 degrees= 311.81 degrees Finally you just use the formula that I gave you to find the angle of the Fourth Quadrant and this is your answer.

So your answers should be 48.190 degrees and 311.81 degrees .

That is what I learned and my notes on Lesson 8.1.

Nicala's Blog

in chapter 9 section 1 we are learning about right triangles.



when work on right triangles you use SOHCAHTOA

which means sin equals opposite over hypotenuse and cosine is the opposite of sine

cosine equals adjacent over hypotenuse and secine is the inverse of cosine

tangent equals opposite over adjacent and cotangent is the inverse tangent



Hints: SOHCAHTOA can only be use with right triangles


hypotenuse is opposite the 90 degree angle.

example

when doing a problem you have solve for every side of the triangle

you must draw the triangle so that you dont get confused

and you should use parts of the triangle that you had to start off with.

Week 4 Prompt

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

Mary's blog #5 or whatever.

This week we learned about solving for (theta) by using these steps:

1. Pythagorean Identities

2.move everything to sin & cos or tan if possible

3.Algebra to simplify

4. Identities

5. Algebra

Ex.

sec-sinxtanx

there are no identities that apply to this problem so you can move on to step 2

you use the identities to change sec to (1/cosx) and tan to (sinx/cosx)

your new equation is (1/cosx) - sinx(sinx/cosx)

which simplifies to (1/cosx) - (sin^2x/cosx)

which cancels and leaves you with cosx as your answer

8-5

for this section, you follow the same steps but you use inverses

the steps for inverse are

1. divide by a trig function when solving to cancel

2. cancel from the inside of a trig. func.

ex. 2sin^2Θ-1=0

you subtract one on both sides

you get 2sin^2Θ=1

then you divide by to

you get sin^2Θ=1/2

then you square root each side

then you get sin Θ=square root of 1/2

then you take the inverse of 1/2

you get Θ=sin^-1(square of 1/20

and you put that in your calculator and get 45degrees

then you have to move it to each quadrant because when squareroot a number you get a positive and a negative number so you look for the quadrants where sin is positive and negative which means all four quadrants

so your answers are 45, 135, 225, 315

yay!

Feroz's Epic Blog #4

Blog time. I'll be covering 8-5 today, just like everyone else.

Instead of doing retarded graphs, we get to go back and do those equation things from 8-1. You know, where had to get the trig function by itself and stuff.

However, unlike everyone else, I won't be posting," THINGS YOU CAN'T DO EVER DO OMG." Instead, just make sure you know not to divide by a trig function when you're solving to cancel something out, and not to cancel from the inside the trig function.

Cue the example:

2sin^2Θ-1=0

Add the one, 2sin^2Θ = 1

Divide by two, sin^2Θ = 1/2

Find the square root, sinΘ = √1/2 (neg or pos)

The inverse would equal 45.

Now, unlike 8-1, you must find the degrees of all the quadrants, not just positive or negative.

So, using those rules from 8-1, do w/e you need to get sin to each quad.

Q1 = 45

Make it negative and then add 180, Q2 = 135

Just add 180, Q3 = 225

Make it negative and add 360, Q4 = 315

These are your final answers. Hope this helped you in some way. Good luck on the test this week guys, I'm sure you'll need it. (Especially you, David)

Tori's blog.

When doing 8-5, you have to use identities and inverse to solve for the variable in the problem. And if you cannot make the trig functions the same, try to change them to tan (sin/cos) or cot (cos/sin).

Things you CANNOT do:
- dividing by a trig function when solving to cancel
- cancel from the inside of a trig function
EX. 1: sin (2x)/2 => the 2s cannot cancel
EX. 2: sin 2x=sin x => you cannot divide and cancel


2 Sin ² θ-1 =0
2 Sin ² θ=1
Sin ² θ= ½
Sin θ= +/- √ ½
θ= Inverse Sin (√ ½)
QI: 45 °
QII: 135°
QIII: 225°
QIV: 315°

In section 8-5 these are some of the things that i learned.

In section 8-5, you use identities to get to the same trig function if you can't try to make it tan or cot by sin/cos or cos/sin. You also have to see if the trig function has the same inverse or not.

These are some of the things you cant do!

1.you cant divide by a trig function when solving to cancel
2.you also cant cancel from inside the trig function

Ex. sin( squared)x- sinx= cos(squared)x

sin(squared)x- sinx= 1-sin(squared)x

2sin(squared)x- sinx-1=0

(2sin(squared)x- 2sinx)+ (sinx-1)= 0

2sinx (sinxx-1)+(sinx-1)=0

(2sinx+1) (sinx-1)=0

2sinx+1=0

x=sin(inverse)(1/2)

and that is some of the things that I learned in section 8-5

Malorie's blog

8.4 relationships among the functions

In order to do this, you must know the following:

(Reciprocal Relationships)
csc theta= 1/sin theta
sec theta= 1/cos theta
cot theta= 1/tan theta

(Relationships with Negatives)
sin(-theta)=-sin theta and cos(-theta)= cos theta
csc(-theta)=-csc theta and sec(-theta)= sec theta
tan(-theta)=-tan theta and cot(-theta)=-cot theta

(Pythagorean Relationships)
sin^2 theta+cos^2 theta= 1
1+tan^2 theta= sec^2
1+cot^2 theta^2theta

(Cofunction Relationships)
sin theta= cos (90 degrees-theta) and cos theta=sin (90 degrees-theta)
tan theta= cot(90 degrees-theta) and cot theta=tan(90 degrees- theta)
sec theta=csc(90 degrees-theta) and csc theta= sec(90 degrees-theta)

In order to do this you must also follow these steps:

Step 1) look for Pythagorean identities

Step 2) Move everything to sin and cos or tan if possible

Step 3)Use algebra to simplify

Step 4)Look for identities again

Step 5)Use algebra (combining, dividing, factoring, and foiling fractions)


Example :

1-Sin^2 theta/ tan^2

1- sin^2 theta/ 1/ sin^2 theta/ cos^2 theta = 1- cos^2 theta(sin^2theta)/ sin^2 theta

1-cos^2 theta= sin^2 theta

Final answer= sin^2 theta

Nathan's Blog

In section 8-5, you use identities to get to the same trig function; if you can't try to make it tan or cot by sin/cos or cos/sin.

THINGS THAT B-ROB AND THE BOOK SAID WE CAN'T DO:
1.) Divide by a trig function when solving to cancel.
2.) Cancel from the inside of a trig function; Ex. sin (2x)/2: YOU CANNOT CANCEL THE 2's

Here's an example:

sin^2 x-sinx=cos^2 x In Radians
sin^2-sinx=1-sin^2 x
2sin^2 x-sinx-1=0
(2sin^2 x-2sinx)+(sinx-1)=0
2sinx(sinx-1)+(sinx-1)
(2sinx+1) (sinx-1)
2sinx+1=0 sinx-1=0
x=sin^-1(-1/2) x=sin^-1(1)
x=30(plugged in as positive) x=90

Sin is negative is Q III and IV
So, x=210 degrees, and 330 degrees
210 x π/180=7π/6
330 x π/180=11π/6
90=π/2

So, finally, your answers are 7π/6, 11π/6, and π/2.

I understood section 8-5 pretty well, but the graphs in the sections before still confuse me a little. So, overall, chapter 8 was not that bad. It's late on Sunday and more than half the class has not done the blog. WOW

Kaitlyn's Blog #5

In this blog, I am going to explain 8-5.

This section is pretty much the same as 8-1 except you might have to use identities to get your trig function on the side by itself. You also have to see if the trig function has the same inverse or not.


THINGS YOU CAN'T DO!!!
1)you cant divide by a trig function when solving to cancel
2)you also cant cancel from inside the trig function


Example:

2 sin (squared) (theta) + 1 = 0
2 sin (squared) (theta) = -1
sin (squared) (theta) = -1/2
sin (theta) = +/- (square root) -1/2



(theta) = sin-1 ((square root) 1/2)
(theta) = 45


Q1: 45
Q2: 135
Q3: 225
Q4: 315

And that's how you do 8-5..

Dylan's SHWEET Blog :)

8-5

In this section we learned about using identities to get get the same trig. func: if you can't, try to make it tan or cot by sin/cos or cos/sin.**

*You follow inverse rules for these problems.*

THINGS YOU CAN'T DO!!!-
1. divide by a trig. func. when solving to cancel
2. cancel from the inside of a trig. func.
EX:
sin(2x)/2 => you can't cancel the 2

sin 2x=sin x => you can't divide and cancel

Now for real problems:

2sin^2Θ-1=0 (if you get confused, replace the sin with a normal equation: 2x^2-1=0)
2sin^2Θ=1
√sin^2Θ=√1/2
sinΘ= +/- √1/2
Θ=45°, (now we have to find the other quadrants) 135°,225°, 315°
the angles are the answers

now for another:

2sinΘ= cosΘ
2(sinΘ/cosΘ)=1
2tanΘ=1
tanΘ=1/2
Θ=tan^-1(1/2) (inverse rules :D )
now to find the quadrants, I & III
Θ=26.565°, 206.565°

the end :)

Lawrence's cool blog

I will be telling you what i learned in section 8-5

first, you have to check and see if your problem has the same inverse, if it doesn't then you have to use the identities that you learn in section 8-4 to get the same trig function.

Hint: ( YOU CAN'T TRY TO MAKE TANGENT OR COTANGENT BY SINE OVER COSINE OR COSINE OVER SINE)


THINGS YOU CANT DO!!!!!!!!!!!!!!!!!!!!! NEVER IN YOUR LIFE!
1) you can't divide by a trig function when solving to cancel
2) you also can't cancel from inside the trig function 91

example:
2sin(squared) theta-1=0

2sin(squared) theta=1

sin(squared) theta= 1/2

sin theta= +or- square root of 1/2

then you find the degree of all the quadrants.

gotta remember
1) Q2: make negative add 180
2) Q3: add 180
3) Q4: make negative add 360

you take the inverse of sin and you get 45 degrees in Q1
Q2 is 135 degrees
Q3 is 225 degrees
Q4 is 315 degrees
(final answers)

that is all i learned in section 8-5. hope it helps some of you out.

4th Blog...

I'm pretty much lost on what material is in what section right now but I know how to find an equation from a graph, shifts and everything. I've even got a picture as an example.


I apologize for the picture quality but its good enough.

The process of finding an equation from a graph is basically the same as having to sketch a draft from an equation.

Everything you need to sketch a graph can be found inside of the graph. You need an A and a B and need to know whether it is positive or negative or sine or cosine, and whether or not it shifts. Everyone of these things can be deduced just from looking at the picture.

First you check to see if its a sine or cosine function and if its positive or negative. In this case, it looks like it would be cosine but its really a positive sine function with a vertical shift.

Its positive because the graph opens upward after passing the y axis and its sine because it doesn't cross the axis at its amplitude.

That leads us to the next step which is finding the Amplitude and Period of the graph.

To find the Amplitude just count how many points are from the point that crosses the y axis and the highest point on the graph, here its 2, not 3.

Now the Period is a little tricky. You need to count to the fifth point on the graph and multiply that point by 2pi.

Its kind of hard to explain how to count the points without physically showing it but I'll try. You start with the point that crosses the y axis, that's your first point, next you count the point that is highest on the graph, the next is going to be parallel to the first and then you count the lowest point on the graph, and so on and so forth until you have a total of five points.

Here its going to be 4.

Then you just multiply 4 x 2pi and that will give you B. Which comes out to be 1/2 pi or pi/2 they are the same thing.

Alright so far we have found that: our graph is a positive sine function with a vertical shift
our Amplitude is 2
and our B is pi/2

NOW WE CAN WRITE OUR EQUATION!! :O

Okay so we have everything we need, so lets get started.

We know its positive and its sine so we can start with: y = sin

Our Amplitude is 2 sooooo it then becomes: y = 2 sin

Next our B is pi/2: y = 2 sin pi/2x

Now we have to represent the vertical shift. It shifts up by one point so it will be represented by a - 1 in the equation. You always put the opposite sign of the way it shifts.

When your done its going to look a little like this: y = 2 sin (pi/2x - 1)

Hope it helped.

My apologies if it didn't...

Helen's Blog #5

I'll be going over section 8-5.

First you would have to see if your problem has the same inverse, if it doesn't then you would use the identities to get to the same trig function. If you can't you would have to try to make it tan or cot by (sin/cos) or (cos/sin).

Then you would follow the inverse rules.

THINGS YOU CAN'T DO!:

1: Divide by a trig function when you are solving to cancel.
2: Cancel from the inside of a trig function.

Sin (2x)

2 * You can't cancel the 2

Sin ²x= Sin * You can't divide and cancel

Examples:

2 Sin ² θ-1 =0

2 Sin ² θ=1
Sin ² θ= ½
Sin θ= +/- Sq. root of ½
θ= Inverse Sin (Sq. root of ½)

QI: 45°
QII: 135°
QIII: 225°
QIV: 315°

Taylor's (real) 4th Blog

The following information is based on section 7.2 of my Advanced Math notes.

These are the formulas that you will need to work the following problems and equations.

S=RTheta(This is a circle with a line going horizontally through it)

k= 1/2r^2theta

k= 1/2rs

S stands for the arc length of the equation/formula.

R stands for the radius of the equation/formula.

Theta stands for the central angle of a problem.

K stands for the area of a sector.

Yet, sometimes you will not just be ask to find the arc length .

They will sometimes call s, r, andTheta other things in some word problems.

Theta is sometimes called the apparent size.

S is sometimes called the diameter of the object.

R is some times called the distance between the objects.

IMPORTANT NOTE THETA MUST BE IN RADIA NOT DEGREES!

Now lets apply what we just learned into the following problems.

The arc length of an object is 6cm and the central angle is .0045 radians. Find the area and the radius of the object.

Theta= .0045S= 6 cmK= ?R= ?

First identify what you know from the problem and what you need to find out.

6=Rx.0045

Next, chose one of the formulas and fill in what you can. Lets find R first since it iis easier to find. Then fill in the appropriate places with the right numbers.

5.9955 cm=R

Then subtract .0045 from both sides to find out what R is. Make sure to put the length of measurements in the appropriate places.

K=1/2x5.9955x6

Now lets use one of the K equations to find the area of the object. So once more fill in the appropriate areas with the right numbers.

K=17.9865 cm^2

Finally, you just need to to multiply the appropriate numbers and this is what you should get as your answer.

Ok lets try this problem next.

The distance between the moon and Earth is 4x10^3 mi and the apparent size of the moon is 5000 radians. What is the diameter?

Theta=5000 radiansS= ?R= 4x10^3 mi

First, write down what you do know from the problem and what you need to find.

S=4x10^3x5000

Then place the information into a problem so that you can work it out.

S= 20000000 mi

Then you multiply 4x10^3 and 5000 and you get your answer.

That is what we learned in lesson 7.2.

Nicala's Blog

in section eight-five

you have to check to see if your problem is has the same inverse if it does not the you have to use the identities that you learn in eight-four to get the same trig function.
(hints: you can not try to make it or tan or cot by sin over cos or cos over sine.

after you you follow the inverse rules

THINGS YOU CAN NOT DO!
you can not divide by a trig function when solving to cancel.
you also can not cancel from the inside of a trig function.

example

4 sin(squared)(theta)=1

you have to move the 4 to the other side equal side

sin(squared)(theta)=-1/4

then you have to take off the squared so you squared root both sides of the problem

sin(theta)=plus or minus squared root 1/4

you move sin over and make the sin an inverse

theta=sin(inverse)(squared root 1/4)
when you put in your calculator it equals to = 25 degrees.

then you graph it on the y-axis and x-axis

Week 3 Prompt

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

Malorie's blog

8-2



this section deals with graphs with a period change.



To graph, you must know what graphs of sine and cosine look like.

(sorry i don't have pictures like some people . . . DYLAN.)



A graph of sine would go through the origin (0,0)



A graph of cosine will not go through the origin.



*hint: if it is positive, it will go through the y-axsis above 0

if it is negative, it will go through the y-axsis below 0.*





To begin graphing with a period change: y=Acos/sinBx



A) you must find the amplitude and period.



the amplitude is the distance from 0 to the highest point.

A= Amp

the period is how long it take a function to repeat itself.

2Pi/B= Period



B) You must find the points on the graph.

You take the five numbers below and divide by B

1. 0

2. pi/2

3. pi

4. 3pi/2

5. 2pi



C) Sketch the graph





Example:



1) y=4sin4x



amp: 4

period: 2pi/4= pi/2



1. o/ 4= 0

2. pi/2 * 1/4= pi/8

3. pi * 1/4= pi/4

4. 3pi/2 * 1/4=3pi/8

5. 2pi *1/4 =pi/2





since this is sine, the line on the graph will go through 0, and go up to four, then, go down to negative 4, and continously go on.

you will then lable where the graph goes through 0, then when it's at it's highest point, then when it passes the x-axsis, then it's lowest point, and lastly, where it begins to repeat itself at.



(sorry if the graph is hard to explain)




and now you know how to graph with a period change (:

Feroz's Epic Blog #3

I already did 8-1 and I don't get 8-3, so I guess I'll do 8-2.

This section is about finding the Amplitude and Period of graphs. Be it sine or cosine.

Before we start I should probably explain what they are. Amplitude is the highest point on the graph and Period is how often the graph repeats itself.

To find the Amplitude, you must find the absolute value of A, which would be the number in front of the trig function.

To find the Period, you must divide B, the number in front of the x, by 2π.

Now, you must find the first five points of the graph. This requires the quotient of B from these points:

1. 0

2. π/2

3. π

4. 3π/2

5. 2π

Some more important things to know. Sine graphs start from or go through the origin and cosine graphs start from a certain point.

Let's do an example:

y = 4sin2πx

The absolute value of 4 (A) is 4, so 4 is the amplitude.

To find the period, divide 2π/2π, which equals 1.

The next step is to divide those five points I gave you earlier by B, which is 2π.

1. 0/2π = 0

2. π/2/2π = 1/4

3. π/2π = 1/2

4. 3π/2/2π = 3/4

5. 2π/2π = 1

Btw, an easy way to divide fractions is to multiply by the reciprocal of the number you are dividing by.

Your fifth point should be the same as your period, as you are dividing 2π by the same number.

I can't do graphs on here but the rest should be in your notes.

So, yeah. That's it.

Dylan's Fantastic Sarcastic BLOG :D...

In 8-1 through 8-3 we've been learning about: moving trig. func., graphing with period change, and finding the shift in a period change.









8-1 - Moving trig. func.
To solve for Θ you get the trig. func. by itself and then take an inverse.

An inverse has 2 answers with some exceptions, find where the angle is based on trig. func. and if the number is positive or negative.

Steps :
Take inverse of the positive number to find the quadrant 1 angle.

To Get To:
Q2 make the degree negative and add 180 degrees
Q3 add 180 degrees
Q4 make the degree negative and add 360 degrees

Example:
sin x=.6 for x
x=sin^-1(.6)
x=36.870°
now find the next angle
x=36.870°, 143.13°

8-2 - Graphing Trig. Func.
y=sin x and y=cos x, when graphed, are either started at the origin or the point on the y axis.

forms:
y=A sin Bx or y=A cos Bx

Amplitude (peak height from origin)= absolute value of A

Period(repetition of func.)= 2pi/B

To graph with period change:
a. take all 5 points
b. divide by b
c. sketch

5 points:
1.0
2.pi/2
3.pi
4.3pi/2
5.2pi

Example
y=-2 sin pi(x)
|-2|=Amp.=2 2pi/pi=Per.=2
refer to image in begging for example of graphs of sin and cos

8-3 - Finding Shift In Period Change
Shifts are harder to find than normal graphing.

y=A sin (Bx +/- C) +/- D
or
y=A cos (Bx +/- C) +/- D

To graph:

Follow previous steps for period change with 5 points.

Add or subtract C from those 5 points. (If the equation is + C you subtract and vice versa.)

Move all y values up D or down D.

Amplitude is different for values now
Amp=max.-min./2
EX.5-(-1)/2=|3|=3=Amp.

Example

y=4 sin (2x + pi) -1
Amp.=4
Per.=pi
1.o=0
2.pi/2=pi/4
3.pi=pi/2
4.3pi/2=3pi/4
5.2pi=pi

now subract C

1.o=0=-pi
2.pi/2=pi/4=-3pi/4
3.pi=pi/2=-pi/2
4.3pi/2=3pi/4=-pi/4
5.2pi=pi=0



Overall these lessons were epic... :)

Dylan

The notes from sections 8-1 and 8-2.

In section 8-1 we are solving for (theta), to solve for (theta) you have to get the trig function by itself and then take the inverse of it.

- an inverse has two answers with some exceptions, find some exceptions, findwhere the angle is based on the trig function and if the number is positive or negative.

Steps:

1. take the inerse of the positive number to find the Q1 angle to get this

Q2: make negative degrees and add 180 degrees.

Q3: Add 180 degrees.

Q4: Make negative degrees and add 360 degrees.

Ex 1:

Solve sin x= -6 for x

x= sin-1 (.6)

x= 36.870, 143.130 degrees

8.2

In this section you are graphing the period changes of sine and cosine.

- y= Asin Bx or y= A cos Bx

- absolute value of {A} = amplitude

-2(pie)= Period

Steps:

Take all 5 points (0, (pi)/2, (pi), 3(pi)/2, 2(pi))

- divide by B

-Sketch the graph

Ex

y=4sin3x

amp: l4l = 4
Pi/6period: 2pi/3 =2pi/3

Now by dividing 0, pi/2, pi, 3pi/2, & 2pi by 3, you get these 5 points.
0/3= 0
pi/2/3= pi/6
pi/3= pi/3
3pi/2/3= pi/2
2pi/3=2pi/3

My 3rd post?

This is supposed to be on 8-1 through 8-3 but I don't really know much from 8-1 because I missed most of that section because I was on vacation. The rest of it I think I know fairly well though. Unfortunately I left my notebook in my locker at school so this blog might not be too good...

Anyway, I'm lost on section 8-1 so I guess I'm just going to talk about section 8-2.

8-2

This section was finding the Amplitude and Period of sine and cosine graphs.

First you need to know the formulas y=AsinBx and y=AcosBx

A stands for the Amplitude and B for the Period.

Amplitude represents the highest point on the graph and Period is how often the graph repeats itself.

You have to find the absolute value of A and divide B by 2pi.

The result of that will give you your Amplitude and Period.

Next you need to find the first five points of the graph and to do that you need to find the quotient of B from these points:
1.) o
2.)pi/2
3.)pi
4.)3pi/2
5.)2pi
And then just sketch your graph and you are done.

Sine graphs start from or pass through the origin and cosine functions start from a particular point.

Example: y=2sin4pix

Amp=2 Per=1/2 ( 2pi/4pi pi's cancel and 2/4 simplifies to 1/2 )

1.) 0/4pi = 0
2.) pi/2 / 4pi = 1/8
3.) pi/ 4pi = 1/4
4.) 3pi/2 / 4pi = 3/8pi
5.) 2pi/ 4pi = 1/2

* Your fifth point and you period should match, that is a way to check if you found everything correctly

After that all you have left to do is graph, the sign of the Amplitude (+ or -) determines which way the graph opens up.

Then you're all done, yippee!

Kaitlyn's Blog #4

I am going to explain Chapter 8, Sections 1 and 2. This chapter was actually pretty easy for me. I understood it more quickly than i understood chapter 7.



8-1



In this section, we are solving for (theta) by getting the trig function by itself and then taking the inverse of it. An inverse must have 2 answers, with exceptions. Here are the formulas to getting the numbers in the different quadrants:

Q2: make negative and add 180

Q3: add 180

Q4: make negative and add 360



Ex: 2sinx=1 -- simplifies to sinx=1/2

-Take the inverse of this -- x=sin^-1 (1/2) = 30



- on the unit circle, this would be in the 1st quadrant. Since sin is positive in the first and second quadrant, you want to get 30 in those quadrants (30 is already in the first so just solve for the second). -30+180=150



- Your answers are: 30 and 150





8-2

In this section, you are graphing the period changes of sine and cosine

- y=AsinBx or y=AcosBx

- A=amplitude

- 2(pi)/B=period

Steps to graphing with a period change:

- Take all 5 points (0, (pi)/2, (pi), 3(pi)/2, 2(pi))

-Divide by B

-Sketch your graph

Ex: y=10sin5x

amp: 10 period: 2(pi)/5

1. 0/5= 0
2. (pi)/2/5= (pi)/10
3. (pi)/5= (pi)/5
4. 3(pi)/2/5= 3(pi)/10
5. 2(pi)/5= 2(pi)/5

(Then you would graph these points. The graph will come out to be wavy and since it is sin, it will pass through the origin.)

Mary's 3rd?! 4th?!? blog.

On our blogs, we are supposed to explain what we learned, but i'm having major trouble with everything this week. I took the time to re-do homework from the week to try and better understand what we're doing.



Let's start off in 8-1



This section is about solving for (theta) which we were told to use as 'x'. Then after you get the trig function and the theta by themselves, you take the trig function inverse of the number. Then you find the quadrants where the trig function is positive, and those two numbers will be your answer.

ex. 4cos(theta)=2
cos(theta)=1/2
cos(-1)(1/2)=(theta)
plug that in your calculator
60!
cos 60 is positive
so what other quadrant is cos positive
c'mon.. you know this...
its quadrant 4!
so how do we get to q4?
let's refer back to the notes.
i actually don't have an example b/c i didn't copy one from charlie, so i'm hoping this is correct.
to go into quadrant 4, you make negative, and add 360.
so we get -60+360=300degrees
so your answer is 60 and 300, i think.
please correct me if i'm wrong.

Moving on to 8-1 and 8-2 overlap.
honestly, all i know from that section is that if the problem says find "inclination" that you have to use m =tan(alpha)
and that if it says find directon angle that you use tan2(alpha)=b/a-c

Let's go on to 8-3!

here are the fomulas we were given to work this section
y=Asin(Bx+/-C) +/-D
y=Acos(Bx+/-C)+/-D

to graph these, here are the steps
1.follow the steps for graphing with a period change.
a)take all 5 points
B)divide by B
C)sketch
2.add or subtract c from those 5 points
3.move all y values up D or down D

now that i go over the notes, and i'm gonna work some more problems, this really isn't that hard.
the example is really long, but it follows these steps exactly.. do it on your own... haha.

Lawrence's Blog

the notes i will go over are a combination of sections 8-1 and 8-2.

these are the formulas B-Rob gave us

m=tan(alpha) for a line
tan2(alpha)= B/a-c if A does not =C for a conic
(alpha)= pi/4 if A=C for a conic
ax^2 +bxy+ cy^2+ dx+ey+f=0

example.

to the nearest degree, find the inclination of the line 2x+5y=15

you have to solve for y
5y=-2x+15

then you divide by 5
y=-2/5x+3

now you are able to use one of the formulas b-rob gave us.

the formula that worked in this problem is:
m=tan(alpha)

m is -2/5. so you get:
-2/5=tan(alpha)

now you take the inverse of tan
(alpha)= tan^-1(2/5)= 21.801

now your back into section 8-1 where you have to find what quadrants tan is positive in. tan is y/x. so since you have to find where y and x are positive you look on the chart and its in Q2 and Q4.

now you can continue solving the problem.

Q2: make negative and add by 180 degrees
-21.801+18= 159.801 degrees

Q4: make negative and add by 360 degrees
-21.801+360= 338.199 degrees.

that is your two final answers. i hope yall understood what i did and i also hope it helped those who dont understand it as well. it took me awhile to understand it. i just had to keep working problems then i finally got it.

Charlie's 4th Blog

This week we went through sections 8-2, an ‘overlap’ of 8-1 & 8-2, and 8-3.

In 8-2, it gives you functions y=AsinBx or y=AcosBx. Where you take A and find the absolute value of it. The absolute value of A equals the amplitude (aplitude- the distance from 0 to the highest point; ½ the vertical distance). Then you divide 2pi by B. This equation gives you the period (period- how long it takes a function to repeat itself). Then after finding these two things, amplitude and period, you graph the function with a period change following these steps: 1) take all 5 points 2) divide by B 3) sketch. The five points step 1 is talking about are 0, pi/2, pi, 3pi/2, and 2pi.

*remember* sine starts at 0, and cosine starts at the point.



EXAMPLE:
y=4sin3x
amp: l4l = 4
Pi/6period: 2pi/3 =2pi/3

Now by dividing 0, pi/2, pi, 3pi/2, & 2pi by 3, you get these 5 points.
0/3= 0
pi/2/3= pi/6
pi/3= pi/3
3pi/2/3= pi/2
2pi/3=2pi/3

Then graph/sketch.

Nathan's Blog

The notes that I will go over are a combination of section 8-1 and 8-2.

These are the formulas:

m=tan (alpha) for a line

tan 2(alpha)=B/A-C if A does not = C for a conic

(alpha)=π/4 if A = C for a conic

Ax^2+Bxy+Cy^2+Dx+Ey+F=0

Ex. Find the inclination of the line to the nearest degree: 6y=2x+12

Now, you have to find slope by solving for y:

y=1/3x+2

m=1/3

1/3=tan (alpha)

(alpha)=tan^-1 (1/3)

(alpha)=18.435

Now you have to find which quadrants tan is positive in: Q I and III

You already have Q I.

To get to quadrant III, you have to add 180.

18.345+180=198.345

(alpha)=18.345 degrees and 198.345 degrees.

I understood most of the concepts from chapter 8, but you have to pay close attention to fully understand it. I am probably going to have to work more problems in the book. A special tribute goes out to all of the victims and families of the 9/11 tragedy. GOD BLESS AMERICA

Helen's Blog #4?

I'll be showing you what I learned on section 8-1 and 8-2.

8-1 ( Solving for Theta)

1: To solve for theta you get the function by itself, then you take the inverse.

2: An inverse has 2 answers with some exceptions, Find where the angle is based on trig function and it the number is positive or negative.

3: Steps to take the inverse of the positive number to find the quadrant angle.

STEPS:

To get to :
Q2: Make negative, then add 180°
Q3: Add 180°
Q4: Make negative, then add 360°

Example:

3 Cos(theta) + 9=7

1: You would have to subtract 9 on the other side.
3Cos(theta)= -2

2: You would divide 3 on the other side.
Cos(theta)=-2/3

3: You would have to get the inverse cosine of -2/3.
theta=Cos^{−1(-2/3)
= 48.1897}°

4: Moving to QII & QIII
QII: Make negative, then add 180°( -48.1897+180= 131.810°)

QIII: Add 180° ( 48.1897+180= 228.190°)

Answer : 131.810°,228.190°

8-2

To Graph with a Period Change

• y= A sin Bx or y= A cos Bx

• |a|= Amplitude
• 2/ B= Period
  

STEPS:

1: Take all 5 points
2: Divide by B.
3: Sketch

Example:

y=2 sin 2x

Amp: 2
Period: 2/2=

1: 0 0/2=0
2:/2 *1/2=/4
3:/2
4:3/2 * 1/2=3/4
5:2/2=

Last you would graph it.

Taylor's (real) 3rd Blog

These are the notes on what we learned in section 7.5

You will need to know the following so that you can work out the equations.

Tan Theta=y/x
Cot Theta = x/y
Sec Theta = r/x
Csc Theta = r/y

Ok now that we know what Tan, Cot, Sec, and Csc are lets use them to solve the following equations.

Since Sin Theta = 4/6 0<> 90 degrees find the meaning of the other five trig functions.

(REMEMBER I DON’T KNOW HOW TO PUT A GRAPH ON THE COMPUTER SO I WILL JUST TYPE IT OUT.)

Ok, we know the since Sin is positive then the y needs to be positive and that can only happen on the second and first quadrant. Also since 4/6 is supposed to be less than 90 degrees we need to be in the first quadrant. Ok now we need to find x. You can draw a triangle and put in the info that you know to help you.

A^2+4^2=6^2

We will use the Pythagorean theorem to help us find x.

A^2+16=36
a^2=20

We will then use the exponents to get theses answers and then we will subtract 16 from 36 and get 20 then we will find the square root of a and 20 and the answer will be the square root of 20.

Now we just fill in the blanks.

Cos x/r= square root of 20/6

Tan y/x= (since we can’t have a square root at the bottom we will multiply it by the square root of 20 and reduce and get this) square root of 20/5

Cot x/y= square root of 20/4

Sec r/x=(since we can’t have a square root at the bottom we will multiply it by the square root of 20 and reduce and get this) 3 times the square root of 20/10

Csc r/y = 6/4 (reduced is) 3/2

Now lets try to find the reference angle of tan.

Find the reference angle for Tan 65 degrees.

First we find what quadrant its in. Which is the first because of the degrees which means tan is positive.

Tan 65 degrees

Since 65 degrees is already in-between 0 and 90 degrees we don’t need to subtract and since it’s not on the Trig chart that means that Tan 65 degrees is your answer.

That is what I learned in 7.5. Until next time!

Nicala's Blog

in chapter eight section two we learn about ampitiude, period, and graphing them coordinating plane.

amp(amptitude)is the distance from zero to the highest point in the vertical graph (aka y-axis).



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





the forms for this problems are: y=AsinBx or y=AcosBx


[A]=amp


two (pie) divided B=period





to graph with a period changes

1. divde by five points



2. divide by B



3. sketch the graph

example: y=-4cos 3x

amp is 4 period 2(pie) divided by B is equaled to 2(pie) divided over 3

then you have to divide 0, (pie) divide 2, (pie), three(pie) divided two and two(pie) by three and then sketch your graph.

Section 7.5 summary

Some of the formulas you might need to know to work the problems.

Tangent(theta)= y/x

Secant(theta)=r/x

Cosecant(theta)=r/y

Cotangent(theta)=x/y

Now here are some examples you can use to understand.

If theta is a second quadrant angle and tan(theta)= -3/4, find the values of the other five trigonometric functions.

Since theta is a second quadrant angle and tan(theta)=y/x= -3/4, you can draw a graph. I cant draw a graph on here but you should be able to visualize what im saying.

cos(theta)=x/r= -4/5

sin(theta)= y/r= 3/5

sec(theta)= 1/cos(theta)= -5/4

csc(theta)= 1/ sin(theta)= 5/3

cot(theta)= 1/ tan( theta)= -4/3

so that is some of the stuff i learned from graphing sine and cosine in section 7.5 last week.

Taylor's 4th Blog

These are the notes on what we learned in section 7.5

You will need to know the following so that you can work out the equations.

Tan Theta=y/x
Cot Theta = x/y
Sec Theta = r/x
Csc Theta = r/y

Ok now that we know what Tan, Cot, Sec, and Csc are lets use them to solve the following equations.

Since Sin Theta = 4/6 0<> 90 degrees find the meaning of the other five trig functions.

(REMEMBER I DON’T KNOW HOW TO PUT A GRAPH ON THE COMPUTER SO I WILL JUST TYPE IT OUT.)

Ok, we know the since Sin is positive then the y needs to be positive and that can only happen on the second and first quadrant. Also since 4/6 is supposed to be less than 90 degrees we need to be in the first quadrant. Ok now we need to find x. You can draw a triangle and put in the info that you know to help you.

A^2+4^2=6^2

We will use the Pythagorean theorem to help us find x.

A^2+16=36
a^2=20

We will then use the exponents to get theses answers and then we will subtract 16 from 36 and get 20 then we will square the a and 20 and the answer will be the square root of 20.

Now we just fill in the blanks.

Cos x/r= square root of 20/6

Tan y/x= (since we can’t have a square root at the bottom we will multiply it by the square root of 20 and reduce and get this) square root of 20/5

Cot x/y= square root of 20/4

Sec r/x=(since we can’t have a square root at the bottom we will multiply it by the square root of 20 and reduce and get this) 3 times the square root of 20/10

Csc r/y = 6/4 (reduced is) 3/2

Now lets try to find the reference angle of tan.

Find the reference angle for Tan 65 degrees.

First we find what quadrant its in. Which is the first because of the degrees which means tan is positive.

Tan 65 degrees

Since 65 degrees is already in-between 0 and 90 degrees we don’t need to subtract and since it’s not on the Trig chart that means that Tan 65 degrees is your answer.

That is what I learned in 7.5. Until next time!

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!**

Charlie's Blog #3

This week we basically reviewed for the Chapter 7 test. Which was degree's to rads, rads to degrees, coterminal angles, minutes & seconds, the central angle stuff, sin & cos, reference angles, tan sec cot & csc, inverses, and the trig chart. But at the end of the week we took Chapter 8 notes.


To solve for theta you get the trig. function by itself &then take an inverse.

An inverse has 2 answers with some exceptions. Find where the angle is based on the trig. function &if the number is positive or negative.

For 8-1 theres easy steps to follow..
1.) Take the inverse of the positive number to find the quadrant 1 angle
2.) to get to:
-quadrant 2= make negative degree & add 180 degrees
-quadrant 3= add 180 degrees
-quadrant 4= make negative and add 360


**For example:

Cos x=.9
x=cos^-1(.9)
x=.451 degrees
inverse of .451; -.451 + 360
x=1.103 degrees; 359.549 degrees

In this problem cos is positive over positive which deals with the x axis. So this means we have to find the first and the fourth quadrant. To get the first we took the inverse of .451 and to get the fourth quadrant we made .451 negative and added 360, this gives us both answers.