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!