I probably feel the most comfortable with Chapter 11 section 1. It's easy because you just follow the formulas for convert to rectangular and polar. Plotting the point is simple too. Pretty much all of Chapter 11 is easy for me which is very unusual.
CONVERT TO RECTANGULAR: formulas: x=rcos theta y=rsin theta
example: convert (4, 45 degrees) to rectangular x=4cos45=multiply 4x45. cos 45 is on the TRIG CHART! x=4(square root of 2/2)=4 y=4sin45=same things multiply 4 and 45 but TRIG CHART! x=4(square root of 2/3)=4 your answer is (4,4)
CONVERT TO POLAR: formulas: r=square root of x^2+y^2 tan theta=y/x
example: give the polar coordinates for (2,3) r=square root of 2^2+3^2 r=square root of 10 r=+-10 tan theta=y/x tan theta=3/2 take the inverse. =56.31 since tangent is positive you find the coordinates for quadrant 1 and 3. quadrant 1=56.31 to find quadrant 3 add 180 =236.31
I feel the most comfortable with chapter 10 section 1. It’s so easy and simple the only thing u just need to know is two formulas and trig chart (you should already know the trig chart since we been doing problems that the answer get replace to trig chart answers.)
You can’t use a calculator.
The two formulas are involving cosine and sine.
Cosine formulas is cos(alpha+/-beta)=cosalphacosbeta-/+sinalphasinbeta
Example cosine problem:
Find the exact value of cos 75degrees
Alpha 45 Beta 30
Cos(45degrees+30degrees)=cos45degreescos30degrees-sin45degreesin30degrees (Square root of 2/2)(Square root of 3/2)- (square root of 2/2)(1/2)= square root of 6/4-square root of 2/4 = square root of 6-square root of 2/4
Sin formulas is sin(alpha+/-beta)=sinalphacosbeta+/-cosalphasinbeta
Example sine problem:
Find the exact value of sine 15degrees Alpha 45degrees Beta 30 degrees
Sin(45degrees-30degrees)=sin45degreescos30degrees-cos45degreessin30degrees (Square root of 2/2)(square root of 3/2)-(square root of 2/2)(1/2) square root of 6/4-square root of 2/4 =square root of 6-square root of 2/4.
You just need to know those two formulas and trig chart and this is why chapter 10 section 1 is the most easiest and most comfortable section for me.
Along with Kaitlyn, I thought that Chapter 11 was the most understandable especially section 1. The first section dealt with two formulas – converting to polar and converting to rectangular. Converting to Polar: r = square root of x^2 + y^2 and tan theta = y/x Converting to Rectangular: x = r cos theta and y = r sin theta
EXAMPLE 1: Convert (1, square root of 3) to polar. A). r = square root of 1^2 + square root of 3^2 – r = +/- 2 B). tan theta = square root of 3 / 1 – tan theta = square root of 3 C). Take the inverse: theta = tan inverse square root of 3 D). Tan square root of 3 is in the trig chart so tan = 60 degrees E). Since tan is positive, find the angles in quadrant 3 by adding 180 degrees. F). Your answer is (square root of 3, 60 degrees) and (-square root of 3, 240 degrees)
EXAMPLE 2: Convert (2, 30 degrees) to rectangular. A). x= 2 cos 30 degrees and y = 2 sin 30 degrees B). Plug in cos 30 degrees and sin 30 degrees from your trig chart. C). x = 2(square root of 3/2) and y = 2(1/2) D). x = square root of 3 and y = 1 E). Your answer is (square root of 3, 1)
Just like Chapter 11, Chapter 10 consisted of strictly formulas. If you know them and understand what you’re doing, it’s simple. Sections 10-1 and 10-2 were the easiest from that Chapter. All you pretty much had to know was the formula to use and your trig chart. I would love to say that I mastered something from Chapter 9, but I kind of didn’t.. :/ that was the most difficult for me so far.
If I had to choose the trig concept that is the easiest to understand I would choose either chapter nine or chapter ten. Not only are those chapters the easiest, to me, they are actually pretty fun . In chapter nine both sections three and four are pretty simple and to the point. The only trouble one may have is remembering the formulas. Section three is explains the law of sines and section four explains the law of Cosines. In order to solve these problems you simply substitute and solve algebraicly.
Example:LAW OF SINES Sin(angle a)/a =sine(angle b)/b If angle A is 45 and angle b is 30 , and a is 4 and b is ? you would substitute to solve.
Sin(45) / 4 =sin(30) / b to solve, cross multi and divide. (sqrt2/2) /4 =(1/2) /b b is equal to the sqrt2.
The law of sines can only be used with non right triangles , and the angles must have a pair. Simple enough:)
So, I think the concept I understood the most is converting from polar to rectangular and rectangular to polar. Here are some examples of things I understand. Give the polar coordinates for (3,3). First, you must find r. r= square root of ( 3^2 + 3^2) r = square root of 18 Now that you have r, you need to find theta. theta= tan ^-1 (3/3) which is equal to one And tan of one is 45 according to the trig chart. So that makes your theta = 45 The answer you get is (square root of 18, 45 degrees) and (-square root of 18, 45 degrees).
Another example is.... converting to rectangular Find (-2,60 degrees) in rectangular x = -2 cos 60 y = -2 sin 60 x = -1 y = - square root of 3 Answer (-1, -square root of 3)
Out of chapters 9 through 11, I feel the most comfortable with 10. Section 10-1 is the easiest section. It involves the sine and cosine formula. They each have their own formulas:
Cos (alpha +/- beta) = cos alpha cos beta -/+ sin alpha sin beta Sin (alpha +/- beta) = sin alpha cos beta +/- cos alpha sin beta
All you have to do plug alpha and beta into the trig chart and add or subtract to find you angle. No calculators are allowed in this section.
Let’s try an example:
Example 1:
Find the exact value of
Sin 30 degrees Cos 15 degrees + Cos 30 degrees Sin 15 degrees Sin (30 degrees + 15 degrees) Sin 45 degrees = square root 2/2
Example 2:
Find the exact value of Sin 30 degrees
Sin (90 – 60) = Sin 90 degrees Cos 60 degrees – Cos 90 degrees Sin 60 degrees (1) (1/2) – (0) (square root 3/2) ½ - 0 = ½
This nine weeks in Mrs. Robinson’s Advanced Math class we have been focusing on chapters 9, 10, and 11. Although they started to get a little tricky at times, it was much easier to adjust to this class in the second nine weeks. One of the easiest things for me to catch on to was SOHCAHTOA in chapter nine. It was pretty simple and there were not too many complex problems that were given to us. They were all pretty basic. The formulas that we have to know are:
Sin theta = opp/hyp Cos theta = adj/hyp Tan theta = opp/adj Csc = hyp/opp Sec theta = hyp/adj Cot theta = adj/opp
SOHCAHTOA can also only be used with right triangles only.
Example 1 :
Solve for the triangle ABC. Angle C= 90 degrees, angle A = 35 degrees, and a = 21 degrees.
1. To find angle B, all you have to do is subtract the inside angles from 180. 180- 90- 35= 55 Angle B= 55 2. To find c you use sin: csin35 degrees = 21; c = 21/ sin35degrees; I don’t have a calculator with me right now, but all you have to do is plug in that problem exactly into the calculator and it will give you your answer.
The concept that i understood the most in chapters 9-11 is the law of sines and the law of cosines. These are formulas that you use when you want to solve a triangle. Law of sines is used when you have one pair of an angle and one side. You also must have another either side of angle. Law of cosines is what you use when no other formula works. You can use this formula when you have side, angle, side. The Law of sines formula is Sin (angle a) /a = sine (angle b) /b
Example 1:
Angle A is 65 and angle B is 25 , and a is 10, what is b?
sin (65 degrees) / 10 = sin (25 degrees) / b 5.19 = 24.2 / b ; find sin65/10 and sin25
5.19b = 24.2; multiply b by both sides
b = 4.6; divide by 5.19, and you get what you were looking for.
The law of sines is really simple if you know your formula. This formula can only be used if it is not a right triangle.
The easiest thing for me in chapters 9-11 is 9-2. In 9-2 you find the area of a triangle. This is simple because all you have to do is plug the numbers into the formula then plug it into your calculator to find the answer.
I probably feel the most comfortable with Chapter 11 section 1. It's easy because you just follow the formulas for convert to rectangular and polar. Plotting the point is simple too. Pretty much all of Chapter 11 is easy for me which is very unusual.
ReplyDeleteCONVERT TO RECTANGULAR:
formulas:
x=rcos theta y=rsin theta
example:
convert (4, 45 degrees) to rectangular
x=4cos45=multiply 4x45. cos 45 is on the TRIG CHART!
x=4(square root of 2/2)=4
y=4sin45=same things multiply 4 and 45 but TRIG CHART!
x=4(square root of 2/3)=4
your answer is (4,4)
CONVERT TO POLAR:
formulas:
r=square root of x^2+y^2 tan theta=y/x
example:
give the polar coordinates for (2,3)
r=square root of 2^2+3^2
r=square root of 10
r=+-10
tan theta=y/x
tan theta=3/2
take the inverse.
=56.31
since tangent is positive you find the coordinates for quadrant 1 and 3.
quadrant 1=56.31
to find quadrant 3 add 180
=236.31
your answer: 10,56.31
-5,236.31
I feel the most comfortable with chapter 10 section 1. It’s so easy and simple the only thing u just need to know is two formulas and trig chart (you should already know the trig chart since we been doing problems that the answer get replace to trig chart answers.)
ReplyDeleteYou can’t use a calculator.
The two formulas are involving cosine and sine.
Cosine formulas is cos(alpha+/-beta)=cosalphacosbeta-/+sinalphasinbeta
Example cosine problem:
Find the exact value of cos 75degrees
Alpha 45
Beta 30
Cos(45degrees+30degrees)=cos45degreescos30degrees-sin45degreesin30degrees
(Square root of 2/2)(Square root of 3/2)- (square root of 2/2)(1/2)= square root of 6/4-square root of 2/4 = square root of 6-square root of 2/4
Sin formulas is sin(alpha+/-beta)=sinalphacosbeta+/-cosalphasinbeta
Example sine problem:
Find the exact value of sine 15degrees
Alpha 45degrees
Beta 30 degrees
Sin(45degrees-30degrees)=sin45degreescos30degrees-cos45degreessin30degrees
(Square root of 2/2)(square root of 3/2)-(square root of 2/2)(1/2)
square root of 6/4-square root of 2/4 =square root of 6-square root of 2/4.
You just need to know those two formulas and trig chart and this is why chapter 10 section 1 is the most easiest and most comfortable section for me.
Along with Kaitlyn, I thought that Chapter 11 was the most understandable especially section 1. The first section dealt with two formulas – converting to polar and converting to rectangular.
ReplyDeleteConverting to Polar:
r = square root of x^2 + y^2 and tan theta = y/x
Converting to Rectangular:
x = r cos theta and y = r sin theta
EXAMPLE 1: Convert (1, square root of 3) to polar.
A). r = square root of 1^2 + square root of 3^2 – r = +/- 2
B). tan theta = square root of 3 / 1 – tan theta = square root of 3
C). Take the inverse: theta = tan inverse square root of 3
D). Tan square root of 3 is in the trig chart so tan = 60 degrees
E). Since tan is positive, find the angles in quadrant 3 by adding 180 degrees.
F). Your answer is (square root of 3, 60 degrees) and (-square root of 3, 240 degrees)
EXAMPLE 2: Convert (2, 30 degrees) to rectangular.
A). x= 2 cos 30 degrees and y = 2 sin 30 degrees
B). Plug in cos 30 degrees and sin 30 degrees from your trig chart.
C). x = 2(square root of 3/2) and y = 2(1/2)
D). x = square root of 3 and y = 1
E). Your answer is (square root of 3, 1)
Just like Chapter 11, Chapter 10 consisted of strictly formulas. If you know them and understand what you’re doing, it’s simple. Sections 10-1 and 10-2 were the easiest from that Chapter. All you pretty much had to know was the formula to use and your trig chart. I would love to say that I mastered something from Chapter 9, but I kind of didn’t.. :/ that was the most difficult for me so far.
If I had to choose the trig concept that is the easiest to understand I would choose either chapter nine or chapter ten. Not only are those chapters the easiest, to me, they are actually pretty fun . In chapter nine both sections three and four are pretty simple and to the point. The only trouble one may have is remembering the formulas. Section three is explains the law of sines and section four explains the law of Cosines. In order to solve these problems you simply substitute and solve algebraicly.
ReplyDeleteExample:LAW OF SINES
Sin(angle a)/a =sine(angle b)/b
If angle A is 45 and angle b is 30 , and a is 4 and b is ? you would substitute to solve.
Sin(45) / 4 =sin(30) / b to solve, cross multi and divide.
(sqrt2/2) /4 =(1/2) /b
b is equal to the sqrt2.
The law of sines can only be used with non right triangles
, and the angles must have a pair. Simple enough:)
So, I think the concept I understood the most is converting from polar to rectangular and rectangular to polar.
ReplyDeleteHere are some examples of things I understand. Give the polar coordinates for (3,3).
First, you must find r.
r= square root of ( 3^2 + 3^2)
r = square root of 18
Now that you have r, you need to find theta.
theta= tan ^-1 (3/3) which is equal to one
And tan of one is 45 according to the trig chart.
So that makes your theta = 45
The answer you get is (square root of 18, 45 degrees) and (-square root of 18, 45 degrees).
Another example is....
converting to rectangular
Find (-2,60 degrees) in rectangular
x = -2 cos 60
y = -2 sin 60
x = -1
y = - square root of 3
Answer
(-1, -square root of 3)
Out of chapters 9 through 11, I feel the most comfortable with 10. Section 10-1 is the easiest section. It involves the sine and cosine formula. They each have their own formulas:
ReplyDeleteCos (alpha +/- beta) = cos alpha cos beta -/+ sin alpha sin beta
Sin (alpha +/- beta) = sin alpha cos beta +/- cos alpha sin beta
All you have to do plug alpha and beta into the trig chart and add or subtract to find you angle. No calculators are allowed in this section.
Let’s try an example:
Example 1:
Find the exact value of
Sin 30 degrees Cos 15 degrees + Cos 30 degrees Sin 15 degrees
Sin (30 degrees + 15 degrees)
Sin 45 degrees = square root 2/2
Example 2:
Find the exact value of Sin 30 degrees
Sin (90 – 60) = Sin 90 degrees Cos 60 degrees – Cos 90 degrees Sin 60 degrees
(1) (1/2) – (0) (square root 3/2)
½ - 0 = ½
This nine weeks in Mrs. Robinson’s Advanced Math class we have been focusing on chapters 9, 10, and 11. Although they started to get a little tricky at times, it was much easier to adjust to this class in the second nine weeks. One of the easiest things for me to catch on to was SOHCAHTOA in chapter nine. It was pretty simple and there were not too many complex problems that were given to us. They were all pretty basic. The formulas that we have to know are:
ReplyDeleteSin theta = opp/hyp
Cos theta = adj/hyp
Tan theta = opp/adj
Csc = hyp/opp
Sec theta = hyp/adj
Cot theta = adj/opp
SOHCAHTOA can also only be used with right triangles only.
Example 1 :
Solve for the triangle ABC. Angle C= 90 degrees, angle A = 35 degrees, and a = 21 degrees.
1. To find angle B, all you have to do is subtract the inside angles from 180.
180- 90- 35= 55 Angle B= 55
2. To find c you use sin: csin35 degrees = 21; c = 21/ sin35degrees; I don’t have a calculator with me right now, but all you have to do is plug in that problem exactly into the calculator and it will give you your answer.
The concept that i understood the most in chapters 9-11 is the law of sines and the law of cosines. These are formulas that you use when you want to solve a triangle. Law of sines is used when you have one pair of an angle and one side. You also must have another either side of angle. Law of cosines is what you use when no other formula works. You can use this formula when you have side, angle, side.
ReplyDeleteThe Law of sines formula is Sin (angle a) /a = sine (angle b) /b
Example 1:
Angle A is 65 and angle B is 25 , and a is 10, what is b?
sin (65 degrees) / 10 = sin (25 degrees) / b
5.19 = 24.2 / b ; find sin65/10 and sin25
5.19b = 24.2; multiply b by both sides
b = 4.6; divide by 5.19, and you get what you were looking for.
The law of sines is really simple if you know your formula. This formula can only be used if it is not a right triangle.
The easiest thing for me in chapters 9-11 is 9-2. In 9-2 you find the area of a triangle. This is simple because all you have to do is plug the numbers into the formula then plug it into your calculator to find the answer.
ReplyDeleteExample:
A =1/2 ab sin(angle b/w)
a=2 b=4
A=1/2(2)(4)(sin 45 degrees)
A=2.82843
SIMPLE AS THAT!!