HEY TAYLOR HERE AND HERE IS MY ANSWER TO THIS WEEKS BLOG!!!!!!!!!!
I find the formula for sin in section 10.1 as the one as I understand the most.
The formula that you need to know for tis.
sin (alpha+/- beta) = sinalphaxcosbeta +/- cosalphasinbeta
Alpha and Beta can come from your trig chart.
Find the exact value of sin 135 degrees.
sin ( 90+45)= sin 90 degrees x cos 45 degrees + cos 90 degrees x sin 45 degrees First, you need to figure out what angles in the trig chat equal 135 degrees and if you need to subtract or add to get to it. Since 90 degrees plus 45 degrees equals 135 degrees, you will use the addition formula for this equation. Then you simply fill out your formula with the information that you were given.
1 x square root of 2/2 + 0 x square root of 2/2 Then you find out the trig functions by using your trig chart.
square root of 2/2 + 0 Then you multiply the 2 sides by themselves.
sin 135 degrees= square root of 2/2 Finally you add them up and you get this as your answer.
Simplify sin 45 degrees x cos 60 degrees + cos 45 degrees x sin 60 degrees.
sin 45 degrees x cos 60 degrees + cos 45 degrees x sin 60 degrees First, write out your equation.
sin (45 degrees+ 60 degrees) Then, since you know that you’re using sin’s formula and addition thanks to the formula. You will put it like this.
sin 105 degrees Then you add the two degrees together and you get this. Since 105 degrees is not on the trig chart you will leave it as it is.
That is what I feel most comfortable with doing from chapters 9-11. SO UNTIL NEXT TIME JA NE!!!!(goodbye in Japanese)
I would have to agree with Helen, the 10-1 formulas were comforatble, but I still think that the 11-1 Polar formulas were my most comfortable formulas and/or section.
Formulas: To convert to Rectangular-x=rcos(theta) y=rsin(theta) To convert to Polar- r=[square root of(x^2 + y^2) ] & tan(theta)=y/x
( []=square root of )
Ex.
Give the rectangular coordinate(s) for the point (3,30 degrees)
The trig concept that i feel the most comfortable with would probably have to be from chapter 10-1 because all you have to do is know your formulas (which are easy) and plug your numbers into it. You don't even need a calculator to do this, you just need to know your trig chart.
Lesson 10-1 is the on i feel the most comfortable with. Formula: sin(A+/-B)=sinA cosB +/- cosA sinB cos(A+/-B)=cosA cosB -/+ sinA sinB Ex: simplify sin30 cos45 + cos30 sin45 =sin(A+B) = sin(30+45) = sin 75
What's easiest? These triangles we've been learning about for like 198339847 years! lol, and the only thing easier than law of sines is sohcahtoa, but you knowww. so about that law of sines..
You can only use law of sines when you have a non right triangle and if you have a pair, an angle and an opposite leg. and it really helps when you have a calculator and draw it out.
Ex. In triangle ABC, AB= 25; A=110; and B=20. Find AC and BC. Draw the triangle. You will find the missing angle by 180-110-20=50. So, sin50/25=sin110/BC. Cross multiply to get: BC=25sin110/sin50BC=30.667 m Now to find AC: sin50/25=sin20/AC AC=25sin20/sin50 AC=11.162 m EX.2 In triangle RST, angle S=126 degrees, s=12, t=7. Determine whether angle T exists. If so, find all possible measures of angle T. sin126/12=sinT/7T=sin^-1(7sin126/12)T=28.159 degrees
Now you must find another angle. Sin is related to Y and is positive in Q I and II.-28.159+180=151.841
Now you test the angles in the triangle. 151.841 + 126 exceeds 180. So the only angle that can work, is 28.159.
the trig concept that i was most comfortable with was the sections one and two in chapters 10. Formulas sin (alpha+/- beta) = sinalphaxcosbeta +/- cosalphasinbeta cos(alpha+/-beta)=cosalphacosbeta -/+ sinalpha sinbeta tan(alpha+/-beta)= tanalpha+/tanbeta 1-tanalphatanbeta
find the exact value of cos 15. cosine(45-30)=cos45cos30+sin45sin30. cosine(45-30)squared root of two over two times squared root of three over two plus squared root of two over two times one half =squared root of six over two plus the squared root of two over two
Okay, I don't know what chapter it was, but the easiest thing, to me, was the one when we had the law of sines in it.
Example number one of law of sines: [You are given the length of 2 sides of a triangle and one angle.] A: 82 degrees a: 2 inches b: 12 inches [You then plug numbers into the formula sintheta/opposite leg = sintheta/opposite leg, so...] sin82/2 = sintheta/12 [now you cross multiply. therefore the answer is...] 5.94 degrees
Another example for the law of sines: [For this type of problem, you are given two angles of a triangle and then just one side length] A: 37 degrees B: 33 degrees b: 4 inches [So, you plug this into the same formula used in the first example] sin37/a = sin33/4 [Then, once again, you cross multiply. So the answer is...] 4.42 inches
The section that I feel most comfortable with is section 10-4. All it talks about is solving trig equations. Basically it gives you an equation and you just have to simplify.
Ex 1: First you want to simplify: 3cos(2X)+cos(X)=2 -Cos2X is an identity and can be broken up into 2cos^2(X-1). 3(2cos^2(X-1))+cos(X)=2 6cos^2X-3+cosX-2=0 6cos^2X+cosX-5=0 -Then you set up as a quadratic and solve for X: 6X^2+X-5 (6X^2+6X)-(5X-5) 6X(X+1)-5(X+1) (6X-5)(X+1) X=(5/6), -1 -Last, you plug in for cos and solve: CosX=-1; cosX=5/6 X=cos inverse(-1); X=cos inverse(5/6) X=180 degrees; X=33.557 degrees & 326.443 degrees
Ex 2: First you simplify by trying to get things all into the same terms like sin: cos2X=1-sinX -cos2X is an identity and can be broken up into 1-2sin^2X. This puts things in terms of sin. 1-2sin^2X=1-sinX 1-2sin^2X-1+sinX 2sin^2X+sinX=0 -Set up as a quadratic and solve for X 2X^2+X=0 X(2X-1)=0 X=0, (1/2) -Plug in for sin and solve SinX=0; sinX=(1/2) X=sin inverse (0); X=sin inverse(1/2) X=0 degrees; X=30 degrees
HEY TAYLOR HERE AND HERE IS MY ANSWER TO THIS WEEKS BLOG!!!!!!!!!!
ReplyDeleteI find the formula for sin in section 10.1 as the one as I understand the most.
The formula that you need to know for tis.
sin (alpha+/- beta) = sinalphaxcosbeta +/- cosalphasinbeta
Alpha and Beta can come from your trig chart.
Find the exact value of sin 135 degrees.
sin ( 90+45)= sin 90 degrees x cos 45 degrees + cos 90 degrees x sin 45 degrees First, you need to figure out what angles in the trig chat equal 135 degrees and if you need to subtract or add to get to it. Since 90 degrees plus 45 degrees equals 135 degrees, you will use the addition formula for this equation. Then you simply fill out your formula with the information that you were given.
1 x square root of 2/2 + 0 x square root of 2/2 Then you find out the trig functions by using your trig chart.
square root of 2/2 + 0 Then you multiply the 2 sides by themselves.
sin 135 degrees= square root of 2/2 Finally you add them up and you get this as your answer.
Simplify sin 45 degrees x cos 60 degrees + cos 45 degrees x sin 60 degrees.
sin 45 degrees x cos 60 degrees + cos 45 degrees x sin 60 degrees First, write out your equation.
sin (45 degrees+ 60 degrees) Then, since you know that you’re using sin’s formula and addition thanks to the formula. You will put it like this.
sin 105 degrees Then you add the two degrees together and you get this. Since 105 degrees is not on the trig chart you will leave it as it is.
That is what I feel most comfortable with doing from chapters 9-11.
SO UNTIL NEXT TIME JA NE!!!!(goodbye in Japanese)
The concept that I feel most comfortable with is section 10-1 because all you have to do is plug it into the formula.
ReplyDeleteFormula:
sin(A+/-B)=sinA cosB +/- cosA sinB
cos(A+/-B)=cosA cosB -/+ sinA sinB
Examples:
1) Find the exact value of sin 75
A= 30 B=45
sin(A+B)=sin30 cos45 + cos30 sin45
=(1/2)(√2/2) + (√3/2)(√2/2)
= √2+√6/4
2) simplify sin30 cos45 + cos30 sin45
=sin(A+B)
= sin(30+45)
= sin 75
I would have to agree with Helen, the 10-1 formulas were comforatble, but I still think that the 11-1 Polar formulas were my most comfortable formulas and/or section.
ReplyDeleteFormulas:
To convert to Rectangular-x=rcos(theta) y=rsin(theta)
To convert to Polar- r=[square root of(x^2 + y^2) ] & tan(theta)=y/x
( []=square root of )
Ex.
Give the rectangular coordinate(s) for the point (3,30 degrees)
x=3cos30 degrees y=3sin30 degrees=3/2
=3(pi3/2)=(3[3]/2, 3/2)
This comment has been removed by the author.
ReplyDeleteThe trig concept that i feel the most comfortable with would probably have to be from chapter 10-1 because all you have to do is know your formulas (which are easy) and plug your numbers into it. You don't even need a calculator to do this, you just need to know your trig chart.
ReplyDeleteFormulas:
1) sin(A+-B)=sinAcosB+-cosAsinB
2) cos(A+-B)=cosAcosB-+sinAsinB
Examples: find sin75
sin(45+30)=sin45cos30+cos45sin30
=(squarerootof 2/2)(squarerootof 3/2) + (squarerootof 2/2)(1/2)
=(squarerootof 6/4) + (squarerootof 2/4)
=(squarerootof 6)+(squarerootof 2)/4 <-----answer
Examples: Simplify sin50cos20-cos50sin20
=sin(50-20)
=sin(30)
=1/2 <---- answer
Lesson 10-1 is the on i feel the most comfortable with.
ReplyDeleteFormula:
sin(A+/-B)=sinA cosB +/- cosA sinB
cos(A+/-B)=cosA cosB -/+ sinA sinB
Ex: simplify sin30 cos45 + cos30 sin45
=sin(A+B)
= sin(30+45)
= sin 75
What's easiest? These triangles we've been learning about for like 198339847 years! lol, and the only thing easier than law of sines is sohcahtoa, but you knowww. so about that law of sines..
ReplyDeleteYou can only use law of sines when you have a non right triangle and if you have a pair, an angle and an opposite leg. and it really helps when you have a calculator and draw it out.
Ex. In triangle ABC, AB= 25; A=110; and B=20. Find AC and BC.
Draw the triangle. You will find the missing angle by 180-110-20=50.
So, sin50/25=sin110/BC.
Cross multiply to get:
BC=25sin110/sin50BC=30.667 m
Now to find AC: sin50/25=sin20/AC
AC=25sin20/sin50
AC=11.162 m
EX.2
In triangle RST, angle S=126 degrees, s=12, t=7. Determine whether angle T exists. If so, find all possible measures of angle T.
sin126/12=sinT/7T=sin^-1(7sin126/12)T=28.159 degrees
Now you must find another angle. Sin is related to Y and is positive in Q I and II.-28.159+180=151.841
Now you test the angles in the triangle. 151.841 + 126 exceeds 180. So the only angle that can work, is 28.159.
Feroz with blog prompt #4.
ReplyDeleteI'm really comfortable with 10-1, but since most of these comments cover that, I guess I'll do 10-3. (Double/Half-Angle)
This is probably the only section I did homework for and it kinda paid off.
Sin2a = 2sinacosa
Cos2a = cos^2a - sin^2a
or = 1 - sin^2a
Tan2a = 2tana/1 - tan^2a
sin a/2 = 1 - cosa/2
cos a/2 = 1 + cosa/2
*Half-Angle formulas are under a square root
**There were other formulas but those are the ones I know off the top of my head.
Anyway, those poorly written formulas probably didn't make sense, so I'll do an example.
ex. If sina = 5/13, find cosa and 2sina
cosa would be 12/13, since 5, 12, and 13 are a triple.
So, using the 2sina formula:
2(5/13)(12/13) = 120/169
And those are the answers.
the trig concept that i was most comfortable with was the sections one and two in chapters 10.
ReplyDeleteFormulas
sin (alpha+/- beta) = sinalphaxcosbeta +/- cosalphasinbeta
cos(alpha+/-beta)=cosalphacosbeta -/+ sinalpha sinbeta
tan(alpha+/-beta)= tanalpha+/tanbeta
1-tanalphatanbeta
find the exact value of cos 15.
cosine(45-30)=cos45cos30+sin45sin30.
cosine(45-30)squared root of two over two times squared root of three over two plus squared root of two over two times one half
=squared root of six over two plus the squared root of two over two
This is Nathan with the week 4 blog prompt response. Hopefully I will pick a chapter in between nine and eleven this time.
ReplyDeleteI think that I will go over chapter 10, with the sin and cos formulas.
sin(alpha+-beta)=sin(alpha)cos(beta)+-cos(alpha)sin(beta)
cos(alpha+-beta)=cos(alpha)cos(beta)-+sin(alpha)cos(beta)
Find the exact value of cos 75.
cos(45+30)=cos45cos30-sin45sin30
=√2/2(√3/2)-√2/2(1/2)
=√6-√2/4
Find the exact value of sin 90
sin(60+30)=sin60cos30+cos60sin30
=1/2(1/2)+√3/2(√3/2)
=1/4+√6/4
=1+√6/4
This comment has been removed by the author.
ReplyDeleteCharlie's Blog...
ReplyDeleteOkay, I don't know what chapter it was, but the easiest thing, to me, was the one when we had the law of sines in it.
Example number one of law of sines:
[You are given the length of 2 sides of a triangle and one angle.]
A: 82 degrees a: 2 inches b: 12 inches
[You then plug numbers into the formula sintheta/opposite leg = sintheta/opposite leg, so...]
sin82/2 = sintheta/12
[now you cross multiply. therefore the answer is...]
5.94 degrees
Another example for the law of sines:
[For this type of problem, you are given two angles of a triangle and then just one side length]
A: 37 degrees B: 33 degrees b: 4 inches
[So, you plug this into the same formula used in the first example]
sin37/a = sin33/4
[Then, once again, you cross multiply. So the answer is...]
4.42 inches
The section that I feel most comfortable with is section 10-4. All it talks about is solving trig equations. Basically it gives you an equation and you just have to simplify.
ReplyDeleteEx 1: First you want to simplify:
3cos(2X)+cos(X)=2
-Cos2X is an identity and can be broken up into 2cos^2(X-1).
3(2cos^2(X-1))+cos(X)=2
6cos^2X-3+cosX-2=0
6cos^2X+cosX-5=0
-Then you set up as a quadratic and solve for X:
6X^2+X-5
(6X^2+6X)-(5X-5)
6X(X+1)-5(X+1)
(6X-5)(X+1)
X=(5/6), -1
-Last, you plug in for cos and solve:
CosX=-1; cosX=5/6
X=cos inverse(-1); X=cos inverse(5/6)
X=180 degrees; X=33.557 degrees & 326.443 degrees
Ex 2: First you simplify by trying to get things all into the same terms like sin:
cos2X=1-sinX
-cos2X is an identity and can be broken up into 1-2sin^2X. This puts things in terms of sin.
1-2sin^2X=1-sinX
1-2sin^2X-1+sinX
2sin^2X+sinX=0
-Set up as a quadratic and solve for X
2X^2+X=0
X(2X-1)=0
X=0, (1/2)
-Plug in for sin and solve
SinX=0; sinX=(1/2)
X=sin inverse (0); X=sin inverse(1/2)
X=0 degrees; X=30 degrees