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# Peerless Cucumber Learning Trigonometry

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This post is a continuation of another thread learning Trigonometry with Melody. The thread was getting a bit too long.

Thank you Melody! This has been a massive help and I anticipate learning more with you.

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Melody's Previous Question: --------------------------------------------------------------------------------------------------------------------------------------------------------------------------

y = 4-3 sin (x/4) Graph: https://www.desmos.com/calculator/r10dftnolq

y = a sin (kx) Graph: https://www.desmos.com/calculator/g3sl7begnr

 y = a sin (kx) y = 4-3 sin (x/4) Amplitude 1 3 Midline y = 0 y = 4 y-intercept 0 4 wavelength 2pi 8pi

Equation y = 4-3 sin (x/4):

Question: "What is the coefficient of x, I mean what number is x multipled by?"

I believe it's being divided by pi, correct?

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I hope these are correct, if not, I apologize, sometimes I struggle when adding new concepts and need a little more guidance.

Feb 6, 2020
edited by PeerlessCucumber  Feb 6, 2020
edited by PeerlessCucumber  Feb 6, 2020
edited by PeerlessCucumber  Feb 6, 2020
edited by PeerlessCucumber  Feb 6, 2020

#1
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Thanks for your lovely comments. I am very pleased to help you. y=4-3sin(x/4)  is spot on

the coefficient of x is   1/4    and

the wavelength will be   2pi / (1/4) = 8pi

I have altered your graph a little so that the x axis is in radians and I have made the step 2pi.

I have made the step equal to one quarter of the wavelgth, then it will always look nice and be easy to read from.

Here is my new graph version

https://www.desmos.com/calculator/hizjhkehuc

See if you can do this too for your future graphs.

Questions for you to answser:

NOW, without looking at the desmos graph I want you to tell me what first maximum point is.

That is worded in very lay terms ... I want the maximum point with the smallest positive x value.

And I want you to tell me what the first minimum point is too.

(These are co-ordinates with both an x and a y value)

I suggest you sketch it freehand to answer these two questions.

You can check on the desmos graph that i have just given you.

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Now for y=ksin(ax)

You have treated your graph like it is fixed, it isn't, see how you can drag the a and the k into different numbers?

You have to pretend that k and a are just numbers and think about what that would mean.

the amplitude is given by the number that sin is multiplied by so the amplitude is 'k'

the wavelength is 2pi/ (the number in front of x, the x pronumeral)

in this case the pronumeral of x is 'a'

so the wavelenth is   2pi/a

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y=fcos(gx) -t     (where f, g and t are all positive constants.)

find

amplitude, y intercept, midline and wavelength.

Feb 6, 2020
#2
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First Question:

The first highest point should be seven. The first lowest point should be one, correct?

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Second Question:

 y = f cos (gx) -t Amplitude f y-intercept t Midline y = t Wavelength (gx)*pi

With what I'm understanding, we're using these as just placeholders, right? Thus, I can't give any specifics, or am I all wrong?

You have the right idea, but you have made a couple of mistakes.

put some numbers in yourself and see if you can spot your errors.  :)

Amplitude is the only one that is completely correct.

PeerlessCucumber  Feb 6, 2020
edited by Melody  Feb 6, 2020
#3
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I plugged in some numbers with the graph example, which confirms the amplitude, of course; however, I'm struggling with the others still. The wavelength I'm not sure about because I thought the wavelength was dictated by dividing the (gx) and pi? Or was it 2pi? Secondly, by switching the number on 't' affects where on the graph it sits, up or down, and so how am I suppose to determine a specific midline or y-intercept?

PeerlessCucumber  Feb 6, 2020
#4
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y = 4-3 sin (x/4)

"The first highest point should be seven. The first lowest point should be one, correct?"

Correct except I wanted the actual points, so i want the x value as well as the y value at the highest point.

Sketch it freehand and see what you get, then check with desmos.

second response:

y = f cos (gx) -t

We are both in full agreement that the amplitude is f.  So that is good. im going to substitute some constants to help with the other features.

let    f=4, g=2 and t = 3

then you would have

y=4cos (2x) - 3

Sketch freehand, you should be able to. Then check with desmos.

 y=4cos(2x)-3 y = f cos (gx) -t amplitude 4 f midline y intercept (remember this is a cos graph) is the y intercept at the top middle or bottom? wavelength   $$\lambda$$ A maximum point A minimum point

By the way, I do understand that these new concepts are very hard for you.  And I am happy to keep helping you to understand.

You need to understand that when you get each concept you need to practice, practice, practice, or else it will not sink in.

You can practice by making up your own similar questions, doing them yourself, and then checking your answers with Desmos.

Melody  Feb 6, 2020
edited by Melody  Feb 6, 2020
#5
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Melody:  I have put my response in the middle of your answers.

First Question

(please include the question so I don't have to go look for it)

y = 4-3 sin (x/4)

The first lowest point should be (2pi, 1) and the first highest point should be (6pi, 7) At least, I think this is correct.

YES, that is perfect, I hope you worked it out with a rough sketch BEFORE you used Desmos.

https://www.desmos.com/calculator/hizjhkehuc     I like your graph --------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Second Question

 y = 4 cos (2x) -3 you did this one perfectly y = f cos (gx) - t y = f cos (gx) - t   melody's correction Midline y = - 3 y = - t yes Amplitude 4 f yes y-intercept 1  (-3+4=1) g     x -t+f it is easier to get this as you hand sketch the graph Is the y-intercept at the top,   middle, or bottom? Top Top yes Wavelength π (2pi/2=pi) 2π (?)   x 2pπ/g A Maximum Point (π, 1) (2π, 0) (?) (0,-t+f), (2π/g, -t+f) don't worry about these last 2, they were too hard anyway. A Minimum Point (π/2, -7) (π, -2)  (?) (π/g,-t-f)

I'm not sure how to fill in wavelength, the maximum point, and the minimum point. If you input this into the desmos graph, it gives a rough example, but considering it can be moved due to all these just being variables, I'm just answering with what the baseline graph would look like.

Good, Once you get the wavelength consistently correct that will work well.

anyway those last two questions, max and min with letters was a bit hard, don't get hung up on those.

Please correct me, I think I'm starting to make sense of it more consistently somewhat though.

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Note: I do relatively well usually once I make sense of the concept, suddenly something will just click and I understand it. I avoid over-practicing until I feel I have a general sense of what is wrong and right with the material, so that I don't form bad habits or get myself confused. This has been a really great experience having someone work with me one-on-one. I appreciate your patience and willingness a great deal, even if I might be a bit slow.

PeerlessCucumber  Feb 6, 2020
edited by Melody  Feb 6, 2020
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y=-cos(8x)

This is how I would go about understanding and building this equation.

I would always do a rough graph whether I was asked to or not.

y=cosx    looks like this     you must memorize this But mine has a cos(8x)......

y=cos(8x)

but     my wavelength is     2pi/8      so the middle must be  pi/8

mmm so now I have  so the 2pi on the x axis must be replaced with  2pi/8,    and the middle will be half that, pi/8

I sketch that.

then it is multiplied by -1 so I turn it upside down.

Here I have built it.

The black one is the finished one Mmm, it would be so much easier if you were in front of me. It would be much easier to explain.

Maybe I might half give up on that approach.

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Graph this.  Do not use desmos except at the very end to check

y=3sin4x - 2

midline=?    draw it with a dotted line.

what is the y value for the highest point, put a light line across to indicate the top.

what is the y value for the lowest point, put a light line across to indicate the bottom.

Will the y intercept be in the middle or at the top or bottom?  put it in.

What is the wavelength?

Divide the wavelength by 4 and put the 4 points on the x axis. (so you will have 1 whole wavelength.)

Now graph it, or at least try to. See how you go.

Feb 6, 2020
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 y = 3 sin 4 x - 2 Midline y = -2 A Maximum Point (pi/8, 1) A Minimum Point (3pi/8, -5) Will the y-intercept be top or bottom?       Wavelength It intercepts at the midline?       pi/2

I would love to upload my terrible little sketch graphs, but it's always a pain in the butt to upload them to my computer.

Hopefully these are correct, I tried double-checking them with desmos.

I think I'm starting to make sense of it a little bit better?

https://www.desmos.com/calculator/kzrq9wtnkp

PeerlessCucumber  Feb 7, 2020
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Yes you appear to be getting the hang of it very nicely.

so

y=7 sin3x -6

1) midline   dashed line

2)topline (my term, not formal)           very light line

3) bottom line (my term, not formal)     very light line

4) wavelength

5) A quarter of the wavelength (this will help with choosing the scale on the x axis)

6) What would be sensible points for the x axis?

7) Will it start on the midline, above or under, why?

8) Will is start going down or going up, why?

9) y intercept

10) plot 4 (or 5) points in advance just by knowing what the properties of the graph will be.

(I mean maxima, minima, and points on the midline)

11) Graph it from x=0 to x= wavelength

12) what is the first maxima? (read from graph)

13) what is the first minima?

14) What points are on the midline?

Feb 7, 2020
edited by Melody  Feb 7, 2020
#9
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 y = 7 sin 3 x -6 Midline y = -6                  correct Topline 1 (?)                   correct Bottom line -13 (?)                  correct Wavelength 2π/3                    correct Quarter of the Wavelength π/3 (?)                  wrong              $$\frac{2\pi}{3}\div4 = \frac{\pi}{6}$$ What would be sensible points for the x-axis? As in where it intersects?        every quarter like, ... 0, pi/6, 2pi/6 3pi/6, 4pi/6, 5pi/6  .. these fractions should be simplified though Will it start on the midline, above or under, why? It will start on the midline because it is a sine graph.   correct Will is start going down or going up, why? It will start upwards because of the positive number in front.  correct y-intercept -6     correct Plot 4 (or 5) Points in Advance    (I mean maxima, minima, and points on the midline.) (4π/3, 6) & (3π/2, 1) & (5π/2, -15) &        (8π/3, -6) (0,-6) (2pi/3,-6)(pi/3,-6)(pi/6,1)(3pi/6,-13) But it is easier if you do it straight onto the rough graph. Graph it from x=0 to x= wavelength I'm not sure what this means.   I mean graph it by hand for one whole wavelength, starting at the y axis. What is the first maxima? (π/6, 1)          yes What is the first minima? (π/2, -13)         yes What points are on the midline? (4π/3, 6) & (8π/3, -6)       NO you can fix this

Sorry that I couldn't answer all of these fully or properly I assume, still getting the hang of it.

https://www.desmos.com/calculator/hocp6dfrvm

I just improved the scale of the graph

https://www.desmos.com/calculator/q10ry2yrjp

Also if you hit the region circle on the side you will see one full wavelength.

You can see also how it is divided nicely into 4 quarters.

PeerlessCucumber  Feb 9, 2020
edited by Melody  Feb 9, 2020
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ok, new one. :)   (make sure you fully understand the last one first though)

y=  -2cos4x +3

(Remember to think about how the - sign will affect the curve. )

1) midline   dashed line

2) topline (my term, not formal)           very light line

3) bottom line (my term, not formal)     very light line

4) wavelength

5) A quarter of the wavelength (this will help with choosing the scale on the x-axis)

6) What would be sensible points for the x-axis?

Easiest to see this from your rough sketch that you are already making.

7) Will it start on the midline, above or under, why?

So far you have just done this from memory of curve shape which is great.

But, but you can just substitute.

when x=0    y=-2cos(0)+3 = ?

8) Will is start going down or going up, why?

9) y intercept

10) plot 4 (or 5) points in advance just by knowing what the properties of the graph will be.

(I mean maxima, minima, and points on the midline)

11) Graph it from x=0 to x= wavelength

12) what is the first maxima? (read from graph)

13) what is the first minima?

14) What points are on the midline?

Your table looked great but it would have been more time consuming.

You can just copy my questions and answer after them if you want.

You can check your sketch and answers with Desmos.  Make the x axis units equal to 1/4 of the wavelength.

Feb 9, 2020
#11
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Equation: y = -2 cos 4 x + 3

Desmos Graph: https://www.desmos.com/calculator/e5jilhrvfk

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- Questions -

1) Midline: y = 3                    yes

2) Topline: 5                           y=5   It is a line!

3) Bottom Line: 1                      y=1

4) Wavelength: π/2                          2pi/4 = pi/2      GOOD

5) Quarter Of The Wavelength: π/4     X     one quarter of pi/2   IS NOT pi/4           $$\frac{1}{4}*\frac{\pi}{2}=\frac{\pi}{8}$$

6) What would be sensible points for the x-axis? π/4, 3π/4, 5π/4, 7π/4, 9π/4 (?)

pi/8, 2pi/8=pi/4, 3pi/8, 4pi/8=pi/2   etc

7) Will it start on the midline, above or under, why? ​It will start below the midline because it is a cos graph.

NO   Sin graphs start in the middle because sin0=0

+cos0=1    -cos0=-1      This equation has the minus sign so it will start (cross the y axis) at the bottom.

8) Will is start going down or going up, why?​ It has begun at its lowest point as a cos graph and will begin upwards.

Up because it is is already at the bottom.

9) y-intercept: 1     YES    (this is the bottom!)

10) Plot 4 (or 5) points in advance just by knowing what the properties of the graph will be.

I mean maxima, minima, and points on the midline: (π/8, 3), (3π/8, 3), (5π/8, 3), (7π/8, 3), (9π/8, 3) [Or] (π/4, 5), (3π/4, 5) [Etc]      (0,1)

11) Graph it from x = 0 to x = Wavelength:

12) What is the first maxima? (π/4, 5)

13) What is the first minima? (π/2, 1)

14) What are points on the midline? (π/8, 3), (3π/8, 3), (5π/8, 3), (7π/8, 3), (9π/8, 3) (?)

Good but I also wanted the ones at the top and bottom (0,-1), pi/4,5), (pi/2,-1)

Look at the graph.

Hopefully these are all correct!     You are getting there :)

You need to do a freehand sketch as you answer the questions. They will complement each other and make it easier.

Here is a graph with a better step on the x axis. I have made the x step equal to one quarter of a wave.

https://www.desmos.com/calculator/lh4pzfxeja

I have only added the shaded green so that you can easily see one wavelength.

See how it starts at the bottom.

when x=0   y=-2cos(4*0)+3 = -2*1+3 = 1   (which is at the bottom of the wave) PeerlessCucumber  Feb 11, 2020
edited by Melody  Feb 11, 2020
edited by Melody  Feb 11, 2020
#12
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ok, another new one. :)   (make sure you fully understand the last one first though)

y=  0.5sin(2x) -1

1) midline   dashed line

2) topline (my term, not formal)           very light line    Hint:  it is a line, the equation needs any equal sign.

3) bottom line (my term, not formal)     very light line

4) wavelength (show a little working)

5) A quarter of the wavelength (this will help with choosing the scale on the x-axis)

NOTE: At this point you should have a rough sketch started

6) What would be sensible points for the x-axis?

Easiest to see this from your rough sketch that you are already making.

7) Will it start on the midline, above or under, why?

So far you have just done this from memory of curve shape which is great.

But, but you can just substitute.

when x=0    y=-2cos(0)+3 = ?

8) Will is start going down or going up, why?

9) y intercept

10) plot 4 (or 5) points in advance just by knowing what the properties of the graph will be.

(I mean maxima, minima, and points on the midline)

11) Graph it from x=0 to x= wavelength

12) what is the first maxima? (read from graph)

13) what is the first minima?

14) What points are on the midline?