Come up with your own trig graphing problem and walk someone through it step by step explaining each step and formula as though they have never taken Advanced Math.
HEY TAYOR HERE AND HERE IS MY ANSWER TO THIS PROMPT!!! My graphing problem will be a rose graphing problem so you need to know the following information. The rose formula is as followed. r= a sin ntheta or r= a cos ntheta The a can never be 0 and the n is an integer > 1. Also, if the n is an even number you need to multiply it by 2 in order for you to figure out how many petals the rose will have, and if it is an odd number you will just use that number in order for you to know how many petals the rose will have.
The following problems cannot be graphed on here but I will talk you through it.
ok say you have r= a cos 6theta. since we need to graph it you need to figure out how many petals it will have. So since it is an even number you will multiply 6 times 2 and you will get 12 as your answer. You will then evenly divide your petals up between the 4 quadrants and that is how you get your answer. I think I did this right...oh well. Till next time ja ne!!
This is Nathan with the response on the first holiday blog prompt.
I have decided to teach you about graphing circles in polar form. This is a very easy concept to grasp.
The two formulas you will need to know are: r=acos(theta) r=asin(theta)
When using, the first formula, r=acos(theta),"a" is the diameter of the circle that has its left-most edge at the pole.
Ex. r=7cos(theta) You will make the left-most edge start at 0, and extend the circle to 7 on the x-axis. As your "a" gets bigger, the circle gets bigger also.
For r=asin(theta),"a" is the diameter of the circle that has its bottom-most edge at the pole.
Ex. r=3sin(theta) You will mark your point of (0,3), start at 0 and draw a circle with those two points. 3 is the top-most point of the circle. Again, for this equation, as the "y" or "a" increases, the bigger the circle gets.
This is Lawrence with his first 5 out of 5 of the holiday blof prompts. i will be tour professor for this one blog so go get me a coke!!! haha jk jk. I decided to teach you all about graphing circles in polar form. This is a very easy concept to grasp, but since you don't know it yet its very hard for yall. (whisper voice:losers)
The two formulas you will need to know are: r=acos(theta) r=asin(theta)
When using, the first formula r=acos(theta)a is the diameter of the circle that has its left most edge at the pole.
Ex. r=4cos(theta) You will make the left-most edge start at 0 and extend the circle to 4 on the x-axis. As your a gets bigger the circle gets bigger too. fascinating isn't it.
For r=asin(theta)a is the diameter of the circle that has its bottom most edge at the pole.
Ex. r=3sin(theta) You will mark your point of 0,9 start at 0 and draw a circle with those two points. 9 is your top most point of the circle. for this equation as the y or a increases the bigger the circle gets.
now that im done teaching you, i will give you an exam that you better pass or you will fail my class. then after i grade your exams i will retire. okay thank you for taking my class and remember make it a great day or not the choice is yours.
hello, my name is mary, and i will be your teacher for today.
we will be working this problem cos(sin(4/5))(which requires graphing)
first off, you do not type this into your calculator, that's a no no. the trig function sin means Y/R and the trig function cos means X/R y/r and x/r are dealing with a coordinate plane. which is a regular graph with points and stuff. so, you see the sin(4/5) that means sin(y/r) oh, all these trig functions are related to triangles consisting of x,y, and r. this is kinda confusing i know, and its about to get worse. ok, so you know that sin(4/5) means that you have a triangle with the y side length as 4 and the hypotenuse side length at 5. these numbers lie on the axis of the graph. and then you need to find the x side length is you want to find the cos of the equation. so to do that, you just know in your brain that they are a pothagorean triple. there are a couple of these and you nneed to memorize them, then you know that the x side is 3 and the cos of the equation is 3/5. done!
Hi, My name is Nicala, and i will be your teacher for the day. Today i am going to teach you how to add trignometric formulas. I know what your thinking "I havent taking advanced math, so I dont know trig" but this will extremely easy as long as you how to add and you use part of the trig chart that i am going to explain to you. The Trig Chart sine 0=0 sine pi over 6= one half sine pi over 4= squared root of two over two sine pi over 3= squared root of three over two sine pi over 2= 1
cosine 0=1 cosine pi over 6= squared root of three over two cosine pi over 4= squared root of two over two cosine pi over 3= one half cosine pi over 2= 0
Now you are probably what is sine and cosine? and what is the trig chart? Sine and cosine are two of the six trig functions that you will use in advanced math. The trig chart is a chart of the trig functions. If you put sin 0 into your calculator (make sure its in degrees) you will get 0 like on the chart but if you put cosine pi over 6 you wont get the squared root of three over two but you the number will be equal to the squared root of three over two.
Okay so back to adding trigonemtric formulas. Directions:Find the exact value of sin 75. Now you know that sine 75 degrees is not on the trig chart so you need to find two on the trig that will be equal to it. you can use sin 45 and sin 30 and the formula you will be using is sin(alpha plus or minus beta)= sin(alpha)cosine(beta)plus or minus cosine(alpha)sin(beta alpha=30 beta=45 sin(30+45)=sin45cos30+cos45sin30 sin(75)=(squared root of two over two)(squared root of three over two)+(squared root of two over two)(one half)= squared root of six +squared root of two over four. Done.
Today i will be teaching you how to graph sin(2/3).
-The first thing you need to do is recognize your function sin. sin=y/r which means that 2 is on the y axis and 3 is your hypoteneuse.
-next you have to draw your triangle on the coordinate plane. to do this, you need to start from the origin and draw a line that goes upward and diagonal. then you have to draw a line that goes straight down reaching the x axis.
-once you have completed that, place the 2 on the y axis and place the 3 on the hypoteneuse of the triangle.
-now you want to find the x axis. all you have to do is use your hypoteneuse formula as if there were no coordinate plane there at all.
you will be given cotangent 3/4, find sine, cosine, tangent, secant, and cosecant. for this you will need to graph on the unit circle.
step 1: you graph on the unit circle where 3/4 would be. {quadrant 1} step 2: solve for the other side by using the formula a^2 + b^2 = c^2. {c = 5} step 3: know that sine = y/r, cosine = x/r, tangent = y/x, secant = r/x, and cosecant = r/y step 4: find what cosine would be using the sides you now have. {cosine = 3/5} step 5: find sine using the sides you now have {sine = 4/5} step 6: find tangent using prior knowledge of step 3 {tangent = 4/3} step 7: Find secant using step 3's information {secant = 5/3} step 8: find cosecant {cosecant = 5/4} step 9: list all answer in a neatly managed way {sine = 4/5, cosine = 3/5, tangent = 4/3, secant = 5/3, cosecant = 5/4}
Today I will be teaching you how to graph a simple triangle on the coordinate plane.
problem: sin 5/13 I'm sure at this moment you are probably freaking out because you dont remember what sin is from geometry. Sin is one of the six trig functions that you use all throughout advanced math.
now, on with the problem. sin is y/r, which means that 5 is the y axis and 13 is the hypotenuse. and since y is positive in this problem, the triangle will be in the first quadrant. okay, now that the basis of the triangle is drawn out, you have to figure out what the x axis is. because starting at the origin and creating a triangle from that, it means that it is a right triangle. the pythagorean theorem can be used. luckily, this grouping is a pythagorean triple and so the answer is ::: x=12
I'll be showing you how to rose curves which is produced from a polar equation in the form of:
r= a sin n(theta) or r=a cos n(theta)
* If n is an even integar, then the rose will have 2n petals. *If n is an odd integar, then the rose will have n petals. Now I'll be showing you Circles in Polar Form. r= a cos(theta) is a circle where "a" is the diameter of the circle tat has its left-most edge at the pole. Ex: r=12cos(theta) You will make the left-most edge start at 0, and extend the circle to 12 on the x-axis.
HEY TAYOR HERE AND HERE IS MY ANSWER TO THIS PROMPT!!!
ReplyDeleteMy graphing problem will be a rose graphing problem so you need to know the following information.
The rose formula is as followed.
r= a sin ntheta or r= a cos ntheta
The a can never be 0 and the n is an integer > 1.
Also, if the n is an even number you need to multiply it by 2 in order for you to figure out how many petals the rose will have, and if it is an odd number you will just use that number in order for you to know how many petals the rose will have.
The following problems cannot be graphed on here but I will talk you through it.
ok say you have r= a cos 6theta. since we need to graph it you need to figure out how many petals it will have. So since it is an even number you will multiply 6 times 2 and you will get 12 as your answer. You will then evenly divide your petals up between the 4 quadrants and that is how you get your answer.
I think I did this right...oh well. Till next time ja ne!!
This is Nathan with the response on the first holiday blog prompt.
ReplyDeleteI have decided to teach you about graphing circles in polar form. This is a very easy concept to grasp.
The two formulas you will need to know are:
r=acos(theta)
r=asin(theta)
When using, the first formula, r=acos(theta),"a" is the diameter of the circle that has its left-most edge at the pole.
Ex. r=7cos(theta)
You will make the left-most edge start at 0, and extend the circle to 7 on the x-axis. As your "a" gets bigger, the circle gets bigger also.
For r=asin(theta),"a" is the diameter of the circle that has its bottom-most edge at the pole.
Ex. r=3sin(theta)
You will mark your point of (0,3), start at 0 and draw a circle with those two points. 3 is the top-most point of the circle. Again, for this equation, as the "y" or "a" increases, the bigger the circle gets.
That's all for this response.
This is Lawrence with his first 5 out of 5 of the holiday blof prompts. i will be tour professor for this one blog so go get me a coke!!! haha jk jk.
ReplyDeleteI decided to teach you all about graphing circles in polar form. This is a very easy concept to grasp, but since you don't know it yet its very hard for yall. (whisper voice:losers)
The two formulas you will need to know are:
r=acos(theta)
r=asin(theta)
When using, the first formula r=acos(theta)a is the diameter of the circle that has its left most edge at the pole.
Ex. r=4cos(theta)
You will make the left-most edge start at 0 and extend the circle to 4 on the x-axis. As your a gets bigger the circle gets bigger too. fascinating isn't it.
For r=asin(theta)a is the diameter of the circle that has its bottom most edge at the pole.
Ex. r=3sin(theta)
You will mark your point of 0,9 start at 0 and draw a circle with those two points. 9 is your top most point of the circle. for this equation as the y or a increases the bigger the circle gets.
now that im done teaching you, i will give you an exam that you better pass or you will fail my class. then after i grade your exams i will retire. okay thank you for taking my class and remember make it a great day or not the choice is yours.
hello, my name is mary, and i will be your teacher for today.
ReplyDeletewe will be working this problem cos(sin(4/5))(which requires graphing)
first off, you do not type this into your calculator, that's a no no.
the trig function sin means Y/R
and the trig function cos means X/R
y/r and x/r are dealing with a coordinate plane.
which is a regular graph with points and stuff.
so, you see the sin(4/5) that means sin(y/r)
oh, all these trig functions are related to triangles consisting of x,y, and r.
this is kinda confusing i know, and its about to get worse.
ok, so you know that sin(4/5) means that you have a triangle with the y side length as 4 and the hypotenuse side length at 5. these numbers lie on the axis of the graph. and then you need to find the x side length is you want to find the cos of the equation. so to do that, you just know in your brain that they are a pothagorean triple. there are a couple of these and you nneed to memorize them, then you know that the x side is 3 and the cos of the equation is 3/5.
done!
Hi, My name is Nicala, and i will be your teacher for the day. Today i am going to teach you how to add trignometric formulas. I know what your thinking "I havent taking advanced math, so I dont know trig" but this will extremely easy as long as you how to add and you use part of the trig chart that i am going to explain to you.
ReplyDeleteThe Trig Chart
sine 0=0
sine pi over 6= one half
sine pi over 4= squared root of two over two
sine pi over 3= squared root of three over two
sine pi over 2= 1
cosine 0=1
cosine pi over 6= squared root of three over two
cosine pi over 4= squared root of two over two
cosine pi over 3= one half
cosine pi over 2= 0
Now you are probably what is sine and cosine? and what is the trig chart?
Sine and cosine are two of the six trig functions that you will use in advanced math.
The trig chart is a chart of the trig functions. If you put sin 0 into your calculator (make sure its in degrees) you will get 0 like on the chart but if you put cosine pi over 6 you wont get the squared root of three over two but you the number will be equal to the squared root of three over two.
Okay so back to adding trigonemtric formulas.
Directions:Find the exact value of sin 75.
Now you know that sine 75 degrees is not on the trig chart so you need to find two on the trig that will be equal to it.
you can use sin 45 and sin 30 and the formula you will be using is
sin(alpha plus or minus beta)= sin(alpha)cosine(beta)plus or minus cosine(alpha)sin(beta
alpha=30
beta=45
sin(30+45)=sin45cos30+cos45sin30
sin(75)=(squared root of two over two)(squared root of three over two)+(squared root of two over two)(one half)= squared root of six +squared root of two over four.
Done.
I'll be showing you how to rose curves which is produced from a polar equation in the form of:
ReplyDeleter= a sin n(theta) or r=a cos n(theta)
* If n is an even integar, then the rose will have 2n petals.
*If n is an odd integar, then the rose will have n petals.
Now I'll be showing you Circles in Polar Form.
r= a cos(theta) is a circle where "a" is the diameter of the circle tat has its left-most edge at the pole.
Ex: r=12cos(theta)
You will make the left-most edge start at 0, and extend the circle to 12 on the x-axis.
Today i will be teaching you how to graph sin(2/3).
ReplyDelete-The first thing you need to do is recognize your function sin. sin=y/r which means that 2 is on the y axis and 3 is your hypoteneuse.
-next you have to draw your triangle on the coordinate plane. to do this, you need to start from the origin and draw a line that goes upward and diagonal. then you have to draw a line that goes straight down reaching the x axis.
-once you have completed that, place the 2 on the y axis and place the 3 on the hypoteneuse of the triangle.
-now you want to find the x axis. all you have to do is use your hypoteneuse formula as if there were no coordinate plane there at all.
-answerrrr---> (squarerootof 5)
Charlie's...
ReplyDeleteyou will be given cotangent 3/4, find sine, cosine, tangent, secant, and cosecant.
for this you will need to graph on the unit circle.
step 1: you graph on the unit circle where 3/4 would be.
{quadrant 1}
step 2: solve for the other side by using the formula a^2 + b^2 = c^2.
{c = 5}
step 3: know that sine = y/r, cosine = x/r, tangent = y/x, secant = r/x, and cosecant = r/y
step 4: find what cosine would be using the sides you now have.
{cosine = 3/5}
step 5: find sine using the sides you now have
{sine = 4/5}
step 6: find tangent using prior knowledge of step 3
{tangent = 4/3}
step 7: Find secant using step 3's information
{secant = 5/3}
step 8: find cosecant
{cosecant = 5/4}
step 9: list all answer in a neatly managed way
{sine = 4/5, cosine = 3/5, tangent = 4/3, secant = 5/3, cosecant = 5/4}
Tori's...
ReplyDeleteToday I will be teaching you how to graph a simple triangle on the coordinate plane.
problem: sin 5/13
I'm sure at this moment you are probably freaking out because you dont remember what sin is from geometry. Sin is one of the six trig functions that you use all throughout advanced math.
now, on with the problem.
sin is y/r, which means that 5 is the y axis and 13 is the hypotenuse. and since y is positive in this problem, the triangle will be in the first quadrant.
okay, now that the basis of the triangle is drawn out, you have to figure out what the x axis is.
because starting at the origin and creating a triangle from that, it means that it is a right triangle. the pythagorean theorem can be used. luckily, this grouping is a pythagorean triple and so the answer is ::: x=12
Hello, my name is Fredrick Beethoven Fitzgerald, or Feroz for short, and I am about to teach you something I learned 3 minutes ago.
ReplyDeleteI find tan the easiest so we're gonna do tan 2/3.
Unless I'm teaching Kindergarten, you should know that tan = y/x, with 2 being on the y-axis and 3 being on the x-axis, hence y/x = 2/3.
So, since we have both and x and a y, does anyone know what we have to find? Anyone? No?
Well we have to find r, which is the hypotenuse. And I don't know why they call it r and not h, so don't even ask.
To find the hypotenuse, we use the Pythagorean theorem, which is a^2 + b^2 = c^2
So that gives us = 4 + 6 = c^2
= sqrt(10)
TaDa! Now draw it or something.
I'll be showing you how to rose curves which is produced from a polar equation in the form of:
ReplyDeleter= a sin n(theta) or r=a cos n(theta)
* If n is an even integar, then the rose will have 2n petals.
*If n is an odd integar, then the rose will have n petals.
Now I'll be showing you Circles in Polar Form.
r= a cos(theta) is a circle where "a" is the diameter of the circle tat has its left-most edge at the pole.
Ex: r=12cos(theta)
You will make the left-most edge start at 0, and extend the circle to 12 on the x-axis.