degrees to radians

Since part two of our giant trig exam is tomorrow I figured I’d keep it simple and review how to convert degrees to radians and vice versa.

You are going to multiply 60 by pi/180.

All you really have to do is simplify 60/180 then add pi in.

You’re answer should therefore be 1pi/3, which is simply pi/3

Now for radians to degrees.

Convert 5pi/4 to degrees.

To convert radians to degrees the process is almost exactly the same

You multiply 5pi/4 times 180/pi.

The pi cancels leaving you with 5/4 times 180.

5 X 180= 900/4=225 degrees.

If you do not get a whole number, you need to convert to degrees minutes and seconds.

To convert to degrees minutes and seconds you multiply the number behind the decimal by 60. This number becomes minutes. If there is another set of numbers behind the decimal, multiply by 60 again. If you still don’t have a whole number after multiplying by sixty twice, you drop the number behind the decimal and the number in front of the decimal becomes seconds.

YOU WILL GET POINTS OFF IF YOU DO NOT CONVERT.

Good luck to everyone on the tests this week

--Sarah

7-1

One can measure angles in either degrees or radians. It really depends on whether the problem states it in degrees>0>degrees or rads>0>rads

1) Breaking a problem into degrees, minutes, and seconds

If one is converting degrees to minutes, all of the numbers that are behind the decimal have to multiplied by 60
-If one is converting degrees to seconds, all of the numbers that are behind the decimal have to multiplied by 60. 
-If one is to convert minutes and seconds back to degrees, then use the following equation: 

degrees+ (min/60)+ (sec/3600)

Ex 1: Convert 76.43 degrees to degrees, minutes, and seconds.

a) .43 X 60 = 25.8'
b) .8 X 60 = 48"
answer = 76 degrees 25' 48"

2) Converting from degrees to radians and radians to degrees

-To convert degrees to radians: degree(times)(pi/180)

-To convert from radians to degrees: radianspi (times)180/pi)

*pi will cancel out

Ex 2: Convert 235 degrees to radians.

a) 235 X (pi/180)

answer = (47/36)pi

-Sameer

10-2

Today I'm go to explain how to do chapter 10, section 2. This is the sum and difference for tangent. There's only two formulas for this secction. They're almost the same thing except the sign changes. 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) REMINDER: You do not plug in for formulas like the ones above, you replace. Okay, time for some examples!!!! Suppose tan alpha = 1/3 and tan beta = 1/2 Find tan(alpha+beta) = (1/3 + 1/2)/(1-(1/3)(1/2)) =( 2/6+3/6)/(1-1/6) = (5/6)/(5/6) = 1 Suppose tan alpha = 4/3 and tan beta = -1/2 Find tan(alpha+beta) = (4/3+(-1/2))/(1-4/3(-1/2) =(8/6+(-3/6))/(1-(-4/6) =(5/6)/(10/6) =1/2

8-4 Review

So this week we have the lovely trig exam, which I am just sooo excited for. SO since all we did was review and review AND review, I am going to just do my blog on 8-4.

8-4 is on the relationship among functions. This should be pretty easy. But before I can start with examples, i am going to give you a few notes that you need to know.

• The first thing you do is to do all the possible algebra to the problem.
• Once you do everything possible, you would first try to use your pythagorean identities.
• After that you would move everything to sin and cos.
• The next thing you would do is do algebra again.
• Once you do all of that you keep repeating steps 1 through 3 until your problem is completely simplified.

Now you need some formulas.

• cscx=1/sinx
• tanx=sinx/cosx
• cotx=cosx/sinx
• secx=1/cosx
• sin^2x+cos^2x=1
• 1+tan^2x=sec^2x
• 1+cot^2x=csc^2x

Well I guess now that you have all of what you need I'll do an example or two.

Example:  cos^2x+sin^2x

• No algebra can be done.
• So then you look for identies you can use, which this problem is one which means it will equal to 1.
• So you answer is going to be 1.

So that's it for this week. Later

Brad

Chapter 9 Review

Well, this week is the trig test so I thought it would be a good idea to review a trig chapter. This week I am going to review with you all the information that we learned in Chapter 9. Chapter 9 is all about triangles. There are a few formulas that you need to know for this chapter.

Notes:
• SOHCATOA: sin (theta) = opp. / hyp. cos (theta) = adj. / hyp. tan (theta) = opp./adj.
• To find the area of a right triangle, use the formula A=½ bh
• To find the area of a non right triangle, use the formula A= ½ (adj.) (adj.) sin(angle b/w)
• Law of Sines: (sin A / a) = (sin B / b) = (sin C / c)
• Law of Cosines: opp. leg^2 = (adj. leg^2) + (other adj. leg^2) – 2(adj. leg) (other adj. leg) cos (angle b/w)

Example: Find the area of non right triangle ABC when: AB=4, BC=6, and B=60 degrees
• ½ (4) (6) sin (60 degrees)
• 12 sin (60 degrees)
• A=10.392 u^2


-Braxton-

11-2

today im reteaching 11-2 which is complex numbers with polar and rectangular. the complex form of rectangular is z= x + yi. the complex form for polar is z= r cis (theta). you can also multiply these numbers and to do so for rectangular you would just do FOIL and get your answer. for polar you mulitply the r's and add the theta's. here is an example below.

EX:
express each complex number in polar form
1) -1 + i
you do same steps to convert from rectangular to polar.
(the square root of) (-1) ^2 + (1)^2 = (the square root of) 2
(theta) = tan (inverse) 1/-1
the quadrants that they are negative is 2nd and 4th
(theta) = tan (inverse) 45
convert to second and fourth quadrant
-45 +180= 135
-45 +360= 315
now you have to figure out which to use
(-1,1) is in second quadrant so you use 135
so your polar form is r cis (theta) = (the square root of) 2 cis 135 (degrees)

Review of Chapter 7

Okayy, so this week we reviewed for the trig test! So I am going to do my blog on a review of chapter 7. This is should be very easy because you should already know all of this. So lets get started! First, I am going to give you some formulas.

• K=1/2r^2 Ɵ
• K=1/2rs
• s=rƟ

In the last formula, r=distance between two objects, Ɵ=apparent size, and S=diameter of an object.

Okay, so now I am going to give you a few examples.

Example 1: A sector of a circle has a radius 8 cm and central angle 2 radians. Find its arc length and area.

In this problem,

• R(radius)=8cm
• Ɵ(central angle)=2
• K(area)=?
• S(arc length)=?

To solve this problem, you would use the equation k=1/2r^2Ɵ.

Therefore, k=1/2(8)^2(2) so k=64cm^2.

 

You then plug into k=1/2rs. Since you’re solving for s, it becomes k/1/2r=s.
Therefore s=64/4 so s=16cm.And those are your two answers!

 

Well, that's it for this weeeek. See ya later! Byeee.

 

--Halie!

11-1

11-1 Polar
This week I'm going to do a review on 11-1 where we learned how to convert from polar to rectangular and from rectangular to polar. In this section, we also use polar points in which we don't use (x,y) but (r,theta) instead.
-To convert from polar to rectangular: x = r(cos)(theta)    y = r(sin)(theta)
-To convert from rectangular to polar: r = square root of x^2 + y^2     theta = tan^-1(y/x)

Example 1: Give the polar coordinates for (5,0).
r = square root of 5^2 + 10^2
r = +/-5
theta = tan^-1(0/5)
theta = 0
final answer: (5,0) (-5,0)


Example 2: Give the rectangular coordinates for (3,30 degrees).
x = r(cos)(theta)
x = 3cos30
x = 3(sq. root of 3/2)
x = 3(sq. root of 3)/2
y = r(sin)(theta)
y = 3sin30
y = 3(1/2)
y = 3/2
final answer: (3(sq. root of 3)/2, 3/2)

12-5

Similar to what was learned in lesson 12-2 through 12-4, the only thing that changes is the number of variables present within the vector. It is pretty simple to do and not at all complicated with the addition of a variable.

Now there's is

- a x, y, and z

Ex 1: Simplify <2, 5, -4> + 3<2, 4, -1>

Distribute the THREE to the second vector

<2, 5, -4> + <6, 12, -3>

Now just add the terms corresponding in the first vector to the second vector (x1+x2, y1+y2, etc.)

The answer is <8, 17, -7>

-Sameer

vectors

This week we learned about vectors, a fairly easy thing to do. Things to know: 1. a vector is a slope 2. to add a vector you use the formula v+u = (a,b)+(c,d)=(a+c,b+d) 3. to subtract a vector you use the formula v-u = (a,b)-(c,d)=(a-c,b-d) 4. scalar multiplication is kv=k(a,b)=(ka,ka) 5. to find a vector from two points you do p2-p1 6. vector equation - (x,y)=(xo,yo)+t(ab) {xo and yo is the point and ab is the vector} 7. parametric equations - x=xo+at y=yo+bt 8. magnitude of a vector - |v|= square root of x^2+y^2 9. component form is (r cos theta, r sin theta) example 1. find a vector equation of the line through A(4,2) and B(3,4) 1. all you have to do is p2-p1 and write it in an equation 2. p2-p1 = -1,2 3. vector equation : (x,y)= (4,2)+t(-1,2) example 2. write example 1 in a parametric equation 1. x= xo+at 2. x= 4-t 3. y= yo+bt 4. y= 2+2t

blogs fir cory yay

So i remembered to do a blog tonight.  All i remember doing is vectors lol so ill talk about that

How to add vectors: v + u = (a, b) + (c, d) = (a + c, b + d)

To subtract vectors: v - u = (a, b) – (c, d) = (a - c, b – d)

Scalar multiplication: kv = k * (a, b) = (ka, kb)

To find a vector equation from two points, you do P2 - P1

Vector equation: (x, y) = (x0, y0) + t (a, b)

Parametric equations: x = x0 + at and y = y0 + bt

To find the magnitude of a vector, you do |v| = sq. root of (x^2 + y^2)

Component form is (r cos theta, r sin theta)

example: Find the equation for the vector from points A(3,4) and B(5,6)

P2-p1
(5-3, 6-4)

(2,2)

(x,y)= (3,4) + t(2,2)

yay for blogs

This week in preparation for the oh-so-lovely trig exam that will be starting Friday, we learned about vectors. They aren’t very exciting I can promise you that. Anyway, on with the blog.

Time for some notes:
• To add vectors: v + u = (a, b) + (c, d) = (a + c, b + d)
• To subtract vectors: v - u = (a, b) – (c, d) = (a - c, b – d)
• Scalar multiplication: kv = k * (a, b) = (ka, kb)
• To find a vector equation from two points, you do P2 - P1
• Vector equation: (x, y) = (x0, y0) + t (a, b)
• Parametric equations: x = x0 + at and y = y0 + bt
• To find the magnitude of a vector, you do |v| = sq. root of (x^2 + y^2)
• Component form is (r cos theta, r sin theta)

Example:
Simplify the following: U+V when U= <3,7> and V= <8,9>
Add the corresponding numbers U+V= <3+8,7+9>
Therefore, in the end, your answer is U+V= <11, 16>

--Sarah

vectors

so i guess this week i am going to do my blog on vectors, since its really easy. All you need to know is a few easy formulas and notes. Once you get know that you should be readyy to roll. woohooo. so now im going to give you what you need to know.

•  To add vectors: v + u = (a, b) + (c, d) = (a + c, b + d)
• To subtract vectors: v - u = (a, b) – (c, d) = (a - c, b – d)
• Scalar multiplication: kv = k * (a, b) = (ka, kb)
• To find a vector equation from two points, you do P2 - P1
• Vector equation: (x, y) = (x0, y0) + t (a, b)
• Parametric equations: x = x0 + at and y = y0 + bt
• To find the magnitude of a vector, you do |v| = sq. root of (x^2 + y^2)
• Component form is (r cos theta, r sin theta)

Now that you know all that you should be ready to do a few or maybe one example. yayy. Just remember to remember those formulas. It's very important!!!

Example: <2,3>-<5,1>

• 2-5=-3
• 3-1=2
• Your answer is going to be <-3,2>

so thats it. thanks for readingg. byeee peopleeee.

BRAD

Vectors

Okay, so this week I am going to teach you how to work problems with vectors. This is very easy as long as you remember the few formulas that you need to know. So now I am going to give you the formulas that you need to remember.

Formulas:

• To add vectors: v + u = (a, b) + (c, d) = (a + c, b + d)
• To subtract vectors: v - u = (a, b) – (c, d) = (a - c, b – d)
• Scalar multiplication: kv = k * (a, b) = (ka, kb)
• To find a vector equation from two points, you do P2 - P1
• Vector equation: (x, y) = (x0, y0) + t (a, b)
• Parametric equations: x = x0 + at and y = y0 + bt
• To find the magnitude of a vector, you do |v| = sq. root of (x^2 + y^2)
• Component form is (r cos theta, r sin theta)

NOTE: Vectors are the slope of a line.

Okay, so now I am going to give you a few examples to help you better understand.


Example: If g=(2,6) and f=(7,1) find g+f and g-f

• (2+7,6+1) = (9, 7)
• (2-7, 6-1) = (-5, 5)
• Your answers are (9,7) and (-5,5)

So it's as easy as that! I hope you now know how to work these types of problems. Well, I am going to bed. Byee!

--Halie!



Vectors

Things to know:

• Vectors are the slope of a line.
• They can be added or subtracted.
• Addition:  v + u = (a,b) + (c, d) = (a + c, b + d)
• Subtraction:  v - u = (a,b) - (c,d) = (a - c, b -d)
• Scalar Multiplication:  kv = k * (a,b) = (ka,kb)
• Find a vector equation from two points:  P2 - P1
• Vector equation:  (x,y) = (xo,yo) + t(a,b)
• Parametric equation:  x = xo + at and y = yo + bt
• Magnitude of a vector:  |v| = square root of (x^2 + y^2)
• Component form: (r cos theta, r sin theta)

Example:  A = (3, 10) and B = (9, 5), Find a + b

•  (3 + 9, 10 + 5) = (12,15)

Example: Find a - b

• (3 - 9, 10 - 5) = (-6,5)

Amber :)

General Formulas for Review

a = alpha
b = beta

cos(a +- b) = cos(a)cos(b) -+ sin(a)sin(b)
sin(a +- b) = sin(a)cos(b) +- cos(a)sin(b)
tan(a + b) = tan(a) + tab(b)/ 1 - tan(a)tan(b)
tan(a - b) = tan(a) - tab(b)/ 1 + tan(a)tan(b)

sin2a = 2sin(a)cos(a)
cos2a = cos^2(a) - sin^2(a)
          = 1 - 2sin^2(a)
          = 2cos^2(a) - 1

tan2a = 2tan(a)/ 1 - tan^2(a)

sin a/2 = +- square root of (1 - cos(a)/2)
cos a/2 = +- square root of (1 + cos(a)/2)
tan a/2 = +- square root of (1-cos(a)/ 1 + cos(a))
           = sin(a)/ 1 + cos(a)
           = 1 - cos(a)/ sin(a)

~ Parrish J. Masters Jr.

12-5

today im teaching you 12-5 which is the same as everything in 12-2 through 12-4 except it had three variables instead of 2. it has an x y and z. so all the formulas are the same except you just add a z to the equations and do it like that. i will provide an example below.

EX:
simplify:
(3,8,-2) + 2(4,-1,2)
you do the same first + first, second + second, third + third.
distrubute first
(3,8,-2) + (8,-2,4) = (11,6,2)

VECTORS

Today we will learn about vectors. Vectors are one of the most straight forward things we learned in trig. it's also the last thing in trig that we will learn for the year(i think) Sooo,

Here are some things you need to know:

1. vector: slope

2. vector addition: v+u = +=

3. vector subtraction: v-u = -=
4. scalar multiplication: kv=k=
5. to find a vector from two points: P2-P1
6. vector equation: (x,y)=(xo, yo)+t
7. parametric equation: x=xo+at and y=yo+bt
8. |v|(magnitude of a vector)=square root(x^2 + y^2)
9. component form:

That's all the notes we are going to learn for today. Now i'm going to show you some examples.

Example 1: <2,3>-<5,1>

After subtracting the first two and the last two, you get <-3,2>

Example 2: u=pq where p=(2,1) and q=(6,4). What is |u|?

You use the convert into a vector formula to get the vector.

Now you do the magnitude formula. so you get the Square root (4^2+3^2) which gives you 5.

That's about it for now, byeee,

Carley!

12-2 Vectors

This weekend I am going to teach you all about vectors. Vectors are the slope of a line. Vectors can be added and subtracted. There are a few things that you need to know about vectors before I show you all an example.
Notes:
• To add vectors: v + u = (a, b) + (c, d) = (a + c, b + d)
• To subtract vectors: v - u = (a, b) – (c, d) = (a - c, b – d)
• Scalar multiplication: kv = k * (a, b) = (ka, kb)
• To find a vector equation from two points, you do P2 - P1
• Vector equation: (x, y) = (x0, y0) + t (a, b)
• Parametric equations: x = x0 + at and y = y0 + bt
• To find the magnitude of a vector, you do |v| = sq. root of (x^2 + y^2)
• Component form is (r cos theta, r sin theta)

Example: If g=(2,6) and f=(7,1) find g+f and g-f
• (2+7,6+1) = (9, 7)
• (2-7, 6-1) = (-5, 5)

-Braxton-

12-1

12-2 Vectors
In this section on vectors, we learned how to do vector addition, subtraction, and scalar multiplication.  Also, we learned how to find a vector from two points, write a vector equation, and write a parametric equation.  We also covered how to find the absolute value of a vector and write an equation in component form.
- vector: slope
- vector addition: v+u = <a,b> + <c,d> = <a+c, b+d>
- vector subtraction: v-u = <a,b> - <c,d> = <a-c, b-d>
- scalar multiplication: kv = k<a,b> = <ka,kb>
- to find a vector from two points: P2 - P1
- vector equation: (x,y) = (xo, yo) + t<a+b> <---* (xo,yo) is the point, and (a,b) is the vector
- parametric equation: x = xo + at & y = yo + bt
- IvI = sq. root of x^2 + y^2 => magnitude of a vector
- component form: <rcos(theta),rsin(theta)>

Example 1: Given A(1,-2) and B(3,-2) find the a) component form b) absolute value of vector AB
a) P2 - P1
= (3-1,-2-1)
= <2,-4>


b) sq. root of 2^2 + (-4)^2
= sq. root of 20
= 2(sq. root of 5)


Example 2: If u = (1,1) and v = (2,4) find a) u+v b) u-v c) 2u-v
a) (1,1) + (2,4)
= <3,5>
b) (1,1) - (2,4)
= <-1,-3>
c) 2(1,1) - (2,4)
= (2,2) - (2,4)
= <0,-2>

OMG WE ONLY HAVE TO BLOG SIX MORE TIMES IN OUR LIVES!

Okay, so since i think we're going to be review trig for the rest of my life, i decided to go back to the beginning. We're going to review some reallllllly easy stuff!

We will be reviewing how to convert degrees into minutes and seconds, convert radians to degrees, and degrees to radians.

êThe first thing we’re going to learn is how to convert degrees into minutes and seconds.

1. The first thing you are going to do is take what is behind the decimal and multiply it by 60.
2. Next, to convert to seconds, you take what is behind the decimal in the minutes and multiply that by 60.

êêThe second thing we’re going to be learning is how to convert a degree into a radian.

1. Use the formula degree x π/180.

HINT: Type in the degree/180 into you’re calculator, put it into a fraction, and put Pibeside it.

êêêThe third and final thing we are going to be learning today is how to convert radians into degrees.

1. Use the formula Rads x 180/π=degree.

HINT: ∏ will cancel

Now, I am going to show you an example of each.

êExample 1: Convert 16.73º to minutes(‘) and seconds(“)

1. .73 x 60 = 43.8
2. .8 x 60 = 48

You’re answer would be 16º43’48”

êêExample 2: Convert 24º into Radians.

1. 24/180= 2/15 = 2π/15

You're answer would be 2π/15

êêêExample 3: Convert π/2 into degrees.

1. π/2 x 180/π =90º

You're answer would be 90º

YOUR WELCOME FOR THE COLORFULNESS!

10-3

Things you need to know:

• Sin2a = (2sina)(cosa)
• Cos2a = (cos^2a) - (sin^2a)
• Cos2a = 1 - 2sin^2a
• Cos2a = 2cos^2a - 1
• Tan2a = (2tana)/(1 - tan^2a)
• Sina/2 = +/- square root of ((1 - cosa)/2)
• Cosa/2 = +/- square root of ((1 + cosa)/2)
• Tana/2 = +/- square root of ((1 - cosa)/(1 + cosa))
• Tana/2 = +/- (sina/(1 + cosa))
• Tana/2 = +/- ((1 - cosa)/sina)

Example 1: Simplify square root of 1 - cos135/2

• First step is to find out what formula is being used.
• Formula that is being used: Sina/2 = +/- square root of ((1 - cosa)/2).
• Now you use Sina/2 as your formula.
• Plug in 135 into the formula Sina/2.
• Sina/2 = Sin135/2 = Sin67.5
• Final answer: Sin67.5 degrees

Example 2 : Simplify 2sin67.5 degrees(cos67.5 degrees)

• Sin2(67.5 degrees)
• Sin45 degrees
• Final answer: square root of 2/2

Amber :)

Review of 7-4

This weekend I am going to review with you all the information that we learned in chapter 7, section 4. In this section we learned how to find reference angles of trig functions. Reference angles must be between 0 and 90 degrees. The reason for this is so that you can find the exact value by using the trig chart, the unit circle, or a calculator. There are a few steps in the process of finding a reference angle of a trig function.

Steps:
1. First of all you have to find the quadrant that the angle is in and sketch it.
2. Then you must use the unit circle to determine whether the angle is positive or negative.
3. Finally, you have to subtract either 360 degrees or 180 degrees until the absolute value of theta is between 0 and 90 degrees.
4. If the problem asks for the exact value, then you would use the trig chart, the unit circle, or a calculator to find it.

Example: Find the exact value of sin 210 degrees
• Sin 210 degrees is in quadrant 3
• Sin is negative in quadrant 3
• 210-180=30
• -sin 30 degrees = ½

-Braxton-

Review of Polar to Rectangular!

I have almost nothing left to review or so it seems and we were off this week so obviously we have not learned any new and interesting mathematical skills so here I go with the reviewing and what not. I'm gonna talk about converting rectangular to polar and vice versa.

Convert (9,30 degrees) to rectangular

To do this, use the equation

x=9cos30 y=9sin30

You should get

(9 square root of 3/2,9/2)

Now, let's go over rectangular to polar.

Convert (3,3)

Use this equation:

x=square root of (3^2+3^2)=square root of 18 or +/-3square root of 2

theta=tan inverse (3/3) This equals one which is on your trig chart as 45 degrees. You'd then draw your coordinate plane and find that the other positive angle is 225 degrees. You should have two answers that are set up like this.

(3 square root of 2, 45 degrees)
(-3 square root of 2, 225 degrees)

That my friends is all there is to it. Easy right? Everyone get ready for a trig exam!

--Sarah

Finding all six trig functions

When given the point in ( , ) form, there are a few things to remember when finding all six trig functions:

* Don't assume…make sure you know which quadrant the point is supposed to be in.

* sin, cos, tan, csc, sec, and cot all need to be found.

* if it helps, graph the point on a coordinate plane. this will help when figuring it out

* Also, it helps to know the Pythagorean Triplets( 3-4-5, 5-12-13, 7-24-25, etc…) so you don't have to solve. (Good ACT tip if i may so!)

sin= y/r csc= r/y

cos= x/r sec= r/x

tan= y/x cot= x/y

Ex: 1 Find all six trig functions for the point (-3,4)

First I'm going to determine what quadrant this is supposed to be in by graphing it. It is in quadrant II. Therefore, x is negative and y is positive.

(by the way, r is always positive.)

sin= 4/5 csc= 5/4

cos= -3/5 sec= -5/3

tan= -4/3 cot= -3/4

-Sameer

2nd spring break blog

okay so today is last day of spring break which means last blog of spring break. today im going over chapter 9 which is triangles. ill be focusing on section 1 which is solving right triangles. to find sin of a right triangle you do opposite / hyptoenuse. to find cos you do adjacent / hypotneuse. to find tan you do opposite / adjacent. this is referring to the angle and the legs. whatever angle you are using you find sin cos or tan with the legs. the examples will explain this better below.


EX: in triangle ABC, angle A = 90 angle B = 25 and a= 18. find b and c
from this information you know angle C = 65 so now you can work the problem
first thing to do is solve for b or c first so lets go with b

Review of 10-3

Okay, so this week I am going to review problems in section 10-3. This is section is on how to solve problems using different formulas. These problems are very easy as long as you remember the formulas. So now I am going to list them for you.

Formulas:

• sin 2a= (2 sin a) (cos a)
• cos 2a= (cos^2 a) – (sin^2 a)
• cos 2a= 1- 2 sin^2 a
• cos 2a= 2 cos^2 a -1
• tan 2a= ( 2 tan a) / ( 1- tan^2 a)
• sin a/2= +/- sq. root of ((1 – cos a) / (2))
• cos a/2= +/- sq. root of (( 1+ cos a) / (2))
• tan a/2= +/- sq. root of (( 1- cos a) / ( 1+cos a))
• tan a/2= +/- (( sin a) / ( 1+ cos a))
• tan a/2= +/- (( 1- cos a) / (sin a))

Those are the formulas you are going to need to know for this section. So now that you know the formulas, I am going to work a few examples for you to better understand these problems.

Example: Simplify cos^2 15 degrees - sin^2 15 degrees

• cos 2(15 degrees)
• cos 30 degrees
• square root of 3 / 2 is your answer.

So that is how you work these types of problems. Hoped this helped you to remember! Byeee.

--Halie!

7-4

7-4 Review

This week I'm going to do a review on how to find reference angles. Reference angles are kind of like reduced fractions in that they must always be between 0 and 90. There are a few simple steps in finding a reference angle:

1) Find the original quadrant and sketch.

2) Determine if the angle is positive or negative using your unit circle methods.

3) Subtract 360 or 180 until the absolute value of "theta" is between 0 and 90.

Example 1: Find a reference angle for sin 128

- It's in quadrant 2, where sin is positive, which means the reference angle will be positive.

- 128 - 180 = -52; I make 52 positive.

- final answer = sin 52

Example 2: Evaluate cos 45

- All I have to do is go to my trig chart and find the value of cos 45.

- = sq root of 2/2

1st spring break blog

okay so today is 1 of 2 spring break blog. the next one will be done before monday. today i am teaching you ch 8-4 which is relationships amoung trig functions. this is the section with all of the identities and stuff. so there are 3 identities:

1) sin^2 x + cos^2 x = 1
2) 1 + tan^2 x = sec^2x
3) 1 + cot^2 x = csc^2 x

there are also 4 steps which are to be followed loosely:
1) Algebrea
2) Use pythagorean identites and switch everything to sin and cos
3) Algebra
4) Repeat

EX:
simpilify sin^2 x + cos^2 x / cot x sin x
algebra cant be done
the numerator is an identity so it becomes 1/ cot x sin x
now you switch to sin and cos and it becomes 1 / (cos/sin)(sin)
your sines cancel and its 1 / cos = sec x

review of chapter 7

Alright well Happy Easter everyone. I wanted to find something different to review that a lot of people probably don’t remember therefore I am going to be reviewing equations from chapter seven. These steps and formulas are most commonly used to solve different word problems. There are several formulas that you will have to remember in order to properly solve these word problems.

The first formula is as follows:
K=1/2r^2 Ɵ

In this equation, k is the area of a sector, r is the radius, and Ɵ is the central angle.

The second formula is follows:
K=1/2rs

In this equation, r is the radius, and arc is the length.

Another equation within this section is that for apparent size. That equation is as follows:
s=rƟ

In this formula, r=distance between two objects, Ɵ=apparent size, and S=diameter of an object.

EXAMPLE 1:
A sector of a circle has a radius 8 cm and central angle 2 radians. Find its arc length and area.

In this problem,
R(radius)=8cm
Ɵ(central angle)=2
K(area)=?
S(arc length)=?

To solve this problem, you would use the equation k=1/2r^2Ɵ.
Therefore, k=1/2(8)^2(2) so k=64cm^2.

You then plug into k=1/2rs. Since you’re solving for s, it becomes k/1/2r=s.

Therefore s=64/4 so s=16cm.

So, in the end the arc length (s) is 16cm and the area (k) is 64cm^2.
--Sarah

Trig Identities

Trig Identities:
sin^2x + cos^2x = 1
1 + tan^2x = sec^2x
1 + cot^2x = csc^2x

When simplifying these trig identities will be of vital use.
ex.)

1 - cos^2x/ 1 + cot^2x (Note: Both the numerator and the denomenator are part of a trig identiy. To simply themm basically (using the original trig identities) move parts of the equation around in order to get what is show in the problem to one side, then substitute what is already in the problem for what is on the other side of the equation.. I know no one probably understood what all that meant)
(For the numerator you would get this: sin^2x)

sin^2x/sec^2x (Keep simplifying... sec = 1/cos and you will have to use that)

sin^2x/1/cos2x (put sin^2x over one and sandwich)

Sin^2xcos^2x is your final answer...

~ Parrish J. Masters Jr.

Trig Review

So I am going to go over the Trig chart. There aren't actual notes on the trig chart so I am going to give you the trig chart. There are some notes that you need to know though.

Things you should know:

• Pi/6 = 30 degrees
• Pi/4 = 45 degrees
• Pi/3 = 60 degrees
• Pi/2 = 90 degrees

Trig Chart:

• Sin 0 = 0
• Sin pi/6 = 1/2
• Sin pi/4 = square root of 2/2
• Sin pi/3 = square root of 3/2
• Sin pi/2 = 1

• Cos 0 = 1
• Cos pi/6 = square root of 3/2
• Cos pi/4 = square root of 2/2
• Cos pi/3 = 1/2
• Cos pi/2 = 0

• Csc 0 = undefined
  
• Csc pi/6 = 2
• Csc pi/4 = square root of 2
• Csc pi/3 = 2 square root of 3/3
• Csc pi/2 = 1

• Sec 0 = 1
• Sec pi/6 = 2 square root of 3/3
• Sec pi/4 = square root of 2
• Sec pi/3 = 2
• Sec pi/2 = undefined

• Tan 0 = 0
• Tan pi/6 = square root of 3/3
• Tan pi/4 = 1
• Tan pi/3 = square root of 3
• Tan pi/2 = undefined

• Cot 0 = undefined
  
• Cot pi/6 = square root of 3
• Cot pi/4 = 1
• Cot pi/3 = square root of 3/3
• Cot pi/2 = 0

-Amber :)

A little more Trig Review!

Here's another trig section review, Dun, Dun, Dunnnnnnn:

I'm going to teach you to solve for ø. The steps are different than an algebra equation.

1. Isolate the trig function.

2. Take the inverse of the trig function (arc sin)sin–1 (arc cos)cos -1

3. Use the trig chart or calculator to find answer (only use positive value)

4. Use quadrants to find the right angle. Positive or negative with trig function.

There are at least two answers for each inverse.

To move quadrants:

• Q1àQ2 make negative add 180º
• Q1àQ3 Add 180º
• Q1àQ4 Make negative add 360º

Now I'm going to show you an example using the steps above.

Cos^-1 (.2)

1. The trig function is already isolated.

2. The inverse is already taken.

3. Since .2 is not on the trig chart, you have to change it into degrees. You would then get 11.46. This means that it is in the quadrant I.

4. Because Cos is x/r, it would be in quadrants I and III. Since we already know the degree in Quadrant I is 11.46 we have to find the degree in quadrant III, to do that you would at 180º. so 11.46º+180º=196.46º. Now we have to change the degree into degrees, minutes and seconds.We learned this in the first section. So for 11.46, you would get 11º27'36" and 196.46º would be 196º27'36".

Your final answer would be Cos^-1 (.2)=11º27'36" and 196º27'36"

I figured i'd use green since it's Easter and all,

--CARLEYYYY.

Review of the Trig Chart

Okay, so this week I am going to review the trig chart. You should remember this because it is one of the most important things you will learn in Advanced Math. So there are really no notes I can give you on this, so I am going to go straight into listing the trig chart for you.

sin 0= 0
sin pi/6=1/2
sin pi/4= square root of 2/2
sin pi/3= square root of 3/2
sin pi/2= 1 cos pi/2= 0

cos 0= 1
cos pi/6= square root of 3/2
cos pi/4= square root of 2/2
cos pi/3= 1/2
cos pi/2= 0

csc 0= undefined
csc pi/6= 2
csc pi/4= square root of 2
csc pi/3= 2 square root of 3/3
csc pi/2= 1

sec 0= 1
sec pi/6= 2 square root of 3/3
sec pi/4= square root of 2
sec pi/3= 2
sec pi/2= undefined

tan 0= 0

tan pi/6= square root of 3/3
tan pi/4= 1
tan pi/3= square root of 3
tan pi/2= undefined

cot 0= undefined
cot pi/6= square root of 3
cot pi/4= 1
cot pi/3= square root of 3/3
cot pi/2= 0

Well that's the trig chart! You better remember it, because it is very important. There are a few other things you may need to know.

Notes:

• pi/6=30 degrees
• pi/4=45 degrees
• pi/3=60 degrees
• pi/2=90 degrees

That's it for this week. Byeeee :)

--Halie!

Arithmetic and Geometric Sequences

This weekend I am going to review with you all the material that we learned in chapter 13, section 1. This section was an introduction to arithmetic and geometric sequences. In this section we learned how to determine whether a sequence was arithmetic or geometric, how to find the formula for each different sequence, and how the number of terms in a sequence or the indicated term in a sequence.

Notes:
• An arithmetic sequence is formed by adding the same number each time. The formula for an arithmetic sequence is t (n) = t1 + (n – 1) d
• A geometric sequence is formed by multiplying the same number each time. The formula for a geometric sequence is t (n) = t1 * r^(n-1)

Example: Identify whether the sequence is arithmetic or geometric, then find the formula for the nth term in the sequence. 2, 4, 6, 8, 10, …….
• It would be an arithmetic sequence
• t(n) = t1 + (n-1) d
• t(n) = 2 + (n-1) 2
• t(n) = 2 + 2n – 2
• t(n) = 2n



-Braxton-

Logarithms

Putting logs in exponential form, expandings logs and condensing them will all be taught today. This is all really simple once you know how to perform the proper steps in solving them.


Ex 1:put in exponential form
log base 2 8 = 3
you use the base which in this case is 2 and you raise that to what the whole thing equals, which is 3
2^3 = 8 : this is what your answer should look like.

Ex 2: expand

log (AB)^3
when expanding logs, you put everything terms of addition if multiplied and subtraction if divided and exponents go in front of the equation
3 log A + 3 log B

condense:
log 6 + log 5 - log 3
to condense you just do the expanding steps backwards.
log 6 * 5 / 3
log 30/3
log 10
since this is log 10 the base i also 10 which cancles everything

The answer is then 1

13-1

13-1 Review

This week I'm going to do a review on 13-1, where we learned how to identify arithmetic and geometric sequences, how to find the formula for each type of sequence, and how to find an indicated term or number of terms of a sequence.

-arithmetic sequence: a sequence that is generated by adding the same number each time

-formula: tn= t1 + (n - 1)d

-goemetric sequence: a sequence that is generated by multiplying the same number each time

-formula: tn = t1 x r^n-1

*to divide, we use fractions/2 = x 1/2

Example 1: Identify the following as arithmetic or geometric.

a) 4,6,8,...

= arithmetic

b) 2,4,8,16,...

= geometric

Example 2: Find the formula for the nth term.

a) 1,4,7,10...

- arithmetic

- tn = t1 + (n - 1)d

= tn = 1 + (n -1)3

= tn = 1 + 3n - 3

= 3n-2

b) 8,4,2,1,...

- geometric

- tn = t1 x r^n-1

= tn = 8 x 2^n-1

= tn = 8 x 2^n x 2^-1

= tn = 8 x 2^n/2

= 4 x 2^n

Example 3: Find the indicated term of the arithmetic sequence.

- t1 = 15, t2 = 21, t20 = ?

- 15 + d = 21

= d = 6

- tn = t1 + (n - 1)d

= t20 = 15 + (20 - 10)6

= 15 + 120 - 6

= 129

Example 4: Find the indicated term of the geometric sequence.

- t1 = 4, t3 = 36, t7 = ?

- 4 x r x r = 36

= 4r^2 = 36

= r = 3

- tn = t1 x r^n-1

= t7 = 4 x 3^7-1

= 4 x 3^6

= 2,916

MORE BLOGS

FIND ALL SIX TRIG FUNCTIONS

Given point (4,3) find all trig functions.
you draw the triangle and the other leg is 5

sin= 3/5
cos=4/5
tan=3/4
csc=5/3
sec= 5/4
cot=4/3


yay for blogs simple but whatever

BLOGS

Degrees to Radians and Vice Versa

To convert from degrees to radians:
divide the degree by 180 and add pie behind the coefficient.

To convert from radians to degree:
multiply the number in radians by 180 and take out the pie.

Examples:
Degrees to Radians

60 degrees x pie/180= pie/3

pie/2 x 180/pie (pies cancel)= 90 degrees

yay for blogs i know its late but sorry

Trig functionsss.

Today we are going to review how to solve a trig function. This requires 8 steps. I’m going to start of by telling you the steps. (For some reason, it kept crashing when I tried to insert a picture.)

1. Identify if there is a negative and which way it will start.

2. Find the Period.

3. Find the Amplitude

4. Write your five important points. (0, π/2, π, 3π/2, and 2π)

5. Divide every point by “B”

6. If anything is added or subtracted in parenthesis do the opposite for all 5 points

7. Sketch

8. Shift the graph up or down if anything is added or subtracted outside of parenthesis.

Now, we’re going to work and example:

Y=2 Sin 3x

1. There is no negative and it will start like this. (axis, up, axis, down, axis)

2. Use the formula 2π/B to find your Period. 2π/3

3. The Amplitude is the absolute value of the number before the function. Therefore, it is 3

4,5. 0x2=0

π/2x2=3π/2

πx2=π/3

3π/2x2=9π/2

2πx2=2π/3

6. Since there is nothing added or subtracted in parenthesis, you sketch the numbers we just found.

7. Sketch your points using the graph you found in step 1.

And that’s how you solve a trig function J

-Carley

This week we reviewed trig. I decided that I would blog about chapter ten formulas because there are so many and we all need to remember those babies. Here we go with the reviewing and the math and what not.

1. cos(alpha+/-beta)=cosalphacosbeta-/+sinalphasinbeta
2. sin (alpha+/-beta)=sinalphacosbeta+/-cosalphasinbeta
3. sinx+siny=2sinx+y/2 cosx-y/2
4. sinx-siny=2cosx+y/c sin x-y/2
5. cosx+cosy=2cosx+y/2cosx-y/2
6.cosx-cosy=-2sinx+y/2 sinx-y/2

EXAMPLE do you remeber how to d these?you bette :D
Find the exact value of sin75degrees.
sin(45degrees+30degrees)=sin45cos30-cos45sin30
45degrees is alpha and 20 degrees is beta. I simply plugged it in to equation 2 silly

THAT IS ALL I HAVE TO SAY ON THE SUBJECT.

--Sarah 😁

april fool's blog

since we just reviewed this week on ch 7 and ch8 i will go over some important stuff in ch 7. the main thing in ch 7 is converting degrees to radians and radians to degrees. now there are two formulas to do this. the first formula is degrees to radians. basically all you do is do the (degrees) * (pi)/180 or you can just do (degrees) / 180 and put pi next to it. to convert from radians to degrees is simple. the formula is radians * 180/(pi). in most cases pi will cancel and kleave you with a degree put sometimes it will not. examples will be listed below.

EX: convert 196 (degrees) to radians
196 / 180 (pi)
= 49(pi) / 45


b) convert 3(pi) / 4 to dergees
3(pi) / 4 * 180 / (pi)
pi's cancel
=3/4 * 180
= 135 (degrees)

Degrees - Rads and vice versa

To convert from degrees to radians, divide the degree by 180 and add pie behind the coefficient.
To convert from radians to degrees, multiply the number in radians by 180 and take out the pie.

Degrees to Rads:
180degrees: 180/180 = 1pie
360degrees: 360/180 = 2pie
Radians to Degrees:
1pie: (1pie)(180) = 180degrees
2pie: (2pie)(180) = 360degrees

P.S. Happy Birthday to me haah

7-1 Review

This weekend I am going to review with you all the information that we learned in Chapter 7 section 1. This lesson is about degrees, radians, and co-terminal angles. In this lesson we learned how to convert degrees to radians and radians to degrees. We also learned how to convert degrees to minutes and seconds and how to find co-terminal angles.

Notes:

• When you convert from degrees to radians, you use this formula: degrees * pi/180
• When you convert from radians to degrees, you use this formula: radians * 180/pi
• To convert from degrees to minutes and seconds, you take what is behind the decimal and multiply it by 60. Then you would do it again to find seconds. If there is still a decimal, then you drop the numbers after the decimal
• To convert from minutes and seconds back to degrees follow this formula:
Degrees + min/60 + sec/3600
• When you find a co-terminal angle, you add/subtract 360 degrees or 2 pi

Example: convert 60 degrees to radians
• 60 * pi/180
• 60 pi/180
• Pi/3


-Braxton-

8-4 Review

Things you should know:

• Do all the possible algebra in the problem first.
• After you do everything possible, you should use the Pythagorean identities.
• After you do that you should move everything to sin and cos.
• After that you should do algebra again.
• Once you do all of those steps, you repeat steps 1-3 until the problem is simplified

• Pythagorean Identities:

1. sin^2x + cos^2x = 1
2. 1 + tan^2x = sec^2x
3. 1 + cot^2x = csc^2x

• Changing to sin to cos:

1. cscx = 1/sinx
2. tanx = sinx/cosx
3. cotx = cosx/sinx
4. secx = 1/cosx

Example 1: cos^2x - 1

• You can not do any algebra in this problem.
• If you noticed, this problem is part of your Pythagorean identities
• Since we know sin^2x + cos^2x = 1 then we know cos^2x - 1 = sin^2x.
• Final answer: sin^2x.

Example 2: 1 + cot^2x

• This is the same thing as the last example.
• You can not do any algebra in this problem.
• This problem is part of your Pythagorean identities.
• 1 + cot^2x = csc^2x
• Final answer: csc^2x

-Amber :)

7-1

Since we've been doing trig review I'm gonna go over 7-1. This section is about converting degrees to radians, radians to degrees, and how to find co-terminal angles.

When you are converting from degrees to radians you use the formula degrees*pi/180

When you are converting from radians to degrees you use the formula radians*180/pi

When you convert from radians to degrees and you have to find minutes, you take what's behind the decimal and multiply it by 60. If you have to find seconds then you use what's behind the decimal after you find minutes and multiply it by 60. If there is a decimal, just drop what's behind it.

To convert back to degrees, you use this formula: degrees+min/60+sec/3600

To find a co-terminal angle, you add or subtract 360 if in degrees, or add or subtract 2pi if in radians.

example 1:

Convert 200 degrees to radians

200*pi/180

10 pi/9

example 2:

convert 5 pi/9 to degrees

5 pi/9 * 180/ pi

the pis cancel

100 degrees

Review of 8-4

Okay so this week I am going to review section 8-4 which is on the relationship among functions! This was confusing when I first learned it, but it is actually very easy once you remember all the relationships. So first I am going to give you a few notes that you are going to need.

Notes:

• The first thing you do is to do all the possible algebra to the problem.
• Once you do everything possible, you would first try to use your pythagorean identities.
• After that you would move everything to sin and cos.
• The next thing you would do is do algebra again.
• Once you do all of that you keep repeating steps 1 through 3 until your problem is completely simplified.

Okay, so now that you know all that, I am going to give you a few extra things you are going to need to know.

Pythagorean Identities:

• sin^2x+cos^2x=1
• 1+tan^2x=sec^2x
• 1+cot^2x=csc^2x
  

Changing to Sin and Cos

• cscx=1/sinx
• tanx=sinx/cosx
• cotx=cosx/sinx
• secx=1/cosx

Now that you know everything you need to know, I will work a few examples for you. This is hopefully help you to better understand how to work these problems.

Example 1: 1+tan^2x

• There is no possible algebra that can be done to this problem.
• This problem is one of your pythagorean identities, which is going to make your problem equal to secx^2. That is then going to be your answer.
• So you answer is secx^2.

Example 2: cos^2x+sin^2x

• There is no possible algebra that can be done to this problem.
• You are first going to look for identities, which you can tell that this problem is one of the
• pythagorean identities which means it will equal to 1.
• So you answer is going to be 1.

And that is it for this weeek. See youu next weeek. Byeee.

--Halie!