Graphing Sine & Cosine w/out a Calculator Pt 2

Graphing Sine & Cosine w/out a Calculator Pt 2


Ok, we have y equals the sine of 2x minus
pi. Now this is how I am going to have my kids do this so hopefully you appreciate this
and it works for you. We are going to list off all the information that we know about
this graph then make a t table and sketch it. Ok, the amplitude is set by “a” in the
beginning of your equation and mine is equal to one so the amplitude is one. The amplitude
it the absolute value of a. The period of sine, cosine, secant, and cosecant is 2pi
over divided by “b” which is in front of the x. It is 2pi divided by 2 so the period is
pi. The phase shift, which I am going to just write ps, the phase shift is your left or
right horizontal movement, is c divided by b so it is going to be pi over two. Because
we have a minus sign in our parenthesis which is part of our standard form, a*trigfunction(bx-c)
+ d. That minus sign is there in the original formula so the phase shift is the movement
to the right. The graph is going to be moved pi over two units to the right. Our vertical
shift, which I am going to write as VS, our vertical shift is from a plus or minus of
a constant at the end of our equation and there is none. Our vertical shift is zero.
Now if we are doing this without a calculator, we need to know what we are counting by, that
is going to be one quarter of the period. So we are going to count by one quarter of
the period of pi, so we get pi over four. I want to minimize the amount of mental math
I have to do, I am going to write my phase shift, my count value, and my c value all
with matching denominators… common denominators to alleviate the mental math. We have pi over
four, pi over two, and pi. The common denominator between all of these values is four. I am
going to write this phase shift as 2pi over four and when I write this equation in the
t table I am going to write c as 4p over four. Again count by, phase shift, and c should
all be written with a common denominator to make the mental math a little bit easier.
Here is my t table. X… and Y equals sine of 2x minus 4pi over 4. As long as you know
how to deal with fractions and you know just the quadrantal angles of the unit circle,
the thinking parts of this problem are done. We are going to start at the phase shift which
is 2p over four. Don’t bother reducing this, that is why/how I want to do all this mental
math easily. Now I want to count by one quarter of the period so two over four plus one over
four is three over four, so 3pi over 4, 4pi over 4, 5pi over 4, and I need five points
so 6pi over 4. Again we are counting by one quarter of the period. Now I am going to show
all the steps in one of these lines but after that I am going to do mental math so I can
get through two examples in fifteen minutes or at least attempt to. The sine of two times
2pi over four minus 4pi over four. Well, two times two is four so that becomes the sine
of 4pi over four minus 4pi over four. That comes out to be zero and as you are doing
this work, if you start at the phase shift and count by one quarter of the period, if
you do your fraction work correctly, you should you be getting your quadrantal angles zero,
pi over two, pi, three pi over two and back to two pi. If you do not get a trig function
giving you an answer of one, zero, or negative one you have done something wrong. So check
your work. That is zero and the sine of zero is zero. Now I am just going to verbally say
this. The sine of 2 times 3p/4 minus 4pi/4. Two times three is six over four. So six over
four minus four over four, six minus four is two and two over four is one half. So this
is the sine of pi/2 and the sine of pi over two is one. Plug in 4p/4. The sine of 2 times
4pi/4 minus 4p/4. That is eight and eight minus four is four over our common denominator
of four. Four pi divided by four is pi and the sine of pi is zero, so we are back to
0. The sine of 2 times 5pi/4 minus 4pi/4. Two times five is ten and ten minus four is
six over the common denominator of four. Six over four reduces down to three over two,
so this is the sine of 3pi/2 which is negative one. When you plug in the last one you are
going to get a value of zero. Here we have our t table set up and we are going to graph
two periods of this graph. Lay down the x axis which again is representing theta. We
have the values of 2pi/4, 3pi/4, 4pi/4, 5pi/4, and 6pi/4. Now those are the values that came
off my t table. They are all positive and I don’t know where zero is so I am going to
count backwards until I get to zero or I skip over it to a negative number. So 3, 2, 1,
ok 0, negative pi/4, negative 2p/4. Now that I know where zero is I am going to place my
y axis down here and mark off the values of one and negative one. Now I am going to start
graphing my sine function. Pi over two zero. Three pi over four and one. Pi, because four
divided by four is one, so pi zero. Five pi over four we are at negative one and at six
pi over four we are back to zero. Now here is what I have so far. Now do you remember
that the regular sine graph I did in the last video? The regular y equals the sine of x,
the parent function. It’s oscillation at zero zero. Well why is this one starting at 2pi
over four. Well it is because there is a phase shift of pi over two and 2pi/4 reduces down
to pi over two. That has pushed the graph to the right a little bit. What is the pattern
now? What if I wanted to graph two periods and not just one which is what I have got
here. Zero, negative one, zero, one, zero, and I have got these values on the x axis
to match my t table so the pattern is very easy now. 0,-1,0,1,0, ok so this point is
at pi over 4 negative one, back to zero, back to one, back to zero and then you see the
exact same pattern re-stamped basically from what I had over here. There is two periods
of y equals sine of two x minus pi. WHOOO!!! Let’s do another one. I think I have time.
I will just go until my fifteen minutes are up or do whatever I can manage to get done.
One more graphing problem. We have got y equals negative three cosine of one half x plus four.
No phase shift but we have a fractional b value which will increase the period. Amplitude
is the absolute value of a so it is going to be three. The period for sine, cosine,
secant, and cosecant is 2pi divided by b and when you pigtail that up, when you do 2 pi
divided one half that is the same as 2pi times two which is equal to 4pi. So since my b value
is less than one, it is actually going to stretch out the period and instead of oscillating
between zero and 2pi, it is going to take 4pi steps on the x axis to complete that period.
What else do we have. The phase shift, there is no plus or minus in my parenthesis so this
is zero. My vertical shift is my constant at the end, not inside my the math function
so it is four. We are going to count by a quarter of the period. What is the period?
The period is 4pi so we are going to count by one quarter of 4pi which is pi. Again,
if we are counting by a quarter of the period we are going to get those quadrantal angles
around the unit circle and get the easy values of zero, one, and negative one. T table. We
have x and y equals negative three cosine of one half x plus four. Books generally don’t
use parenthesis around this but I like to. This negative in front of the three is going
to reflect the graph over the cosine function over x axis. So instead of the function starting
at one it is going to start at negative one… maybe. Let’s see what the three does. We are
going to start our t table at the phase shift and count by a quarter of the period. And
we do not need to worry about common denominators because our denominator is one. We are starting
at the phase shift, which you are going to hear me say in all these videos now, and count
by a quarter of the period. So, pi, 2pi, 3pi, and we need five points so 4pi. Let’s plug
each one of these in. How much time do I have by the way? I better start moving! One half
of zero is zero and the cosine of zero, that is the x coordinate so that is one. So one
times negative three is negative three plus four. Negative three plus four is one. Plug
in pi. One half of pi is pi over two and the cosine of pi over two is zero, times negative
three is zero plus four which is four. Now are paying attention to what I am doing here?
I am plugging these in, I am working out, I am working my way back through the equation
until I get to leading coefficient and then I worry about the plus or minus at the end.
So 2pi times one half is 1pi. The cosine of ONE pi is negative one. This is negative one
here at 2pi and that times negative three is positive three plus four is seven. Let’s
plug in 3pi. The cosine of 3pi over 2 is, one, two, three pi over two, the cosine of
3p/2 is zero times negative three is zero plus four is back to four. Don’t…. Get the
five points before you start guessing what you think the pattern is by the way. And 4pi
times one half is 2pi. So the cosine of 2pi is again one. So the cosine of 2pi is one,
one times negative three is negative three and then plus four is back to one. So I have
all my points now and thankfully I have not run out of time. I have an x value of zero
so I don’t have to worry about where the y axis crosses. So we have zero, pi, 2pi, 3pi,
4pi, and this will give me one period. If I want I to go backwards or forwards I could
go 5pi, 6pi, 7pi, 8pi and so on and just start plotting my points. I don’t have to think
about the amplitude, the flip, the period change, or the vertical shift. All that is
just going to happen. We have zero one…we need a y value of 7 so 1,2,3,4,5,6,7. So zero
one, pi four, 2pi seven, 3p at four, then we are back down to one and that is one period
of cosine. And now I could do another pattern, another period and make if I like a second
period. You will probably be asked to graph two periods of these trig functions. Also,
in the minute that I have left here now that we have two periods of cosine. Instead of
starting at one, we sort of started at the bottom of the pattern, and in this case it
is not negative one, but it is at the bottom of the cycle. The amplitude. The overall height
is six from the top to bottom and if you take half of that you get the actual amplitude
of three. And by the way, as well, if you average these numbers of seven and one, if
you find the average of those numbers and get four. That happens to match the vertical
shift. So, if are ever given a graph and asked to find the vertical shift you average the
maximum and minimum values to find that vertical shift. ALL RIGHT. I am Mr. Tarrou. Thank you
for watching. GO DO YOUR HOMEWORK!!!

100 thoughts to “Graphing Sine & Cosine w/out a Calculator Pt 2”

  1. thanks so much for these videos. My professor didn't show us any kind of formula nor explain very well. I thought I could teach myself from our textbook but even the textbook makes it a thousand times harder than it is! 

  2. When would you know how to flip the graph? Because on the last problem you did, you didn't flip it. Or would it always be flipped when you graph it?

  3. You always pause and look over to your left, when you need to remember what to write.
    I'm curious. Do you have an overhead over there or something? Lol.

  4. When you found the average at the of the video why is that you divided by 2 to get the vertical shift??…or is it that you just simply divide by 2 everytime to get the vertical shift.

  5. You just saved my life. I wish you were my professor. You have no idea how much I would like to thank you. Thank you for teaching. 

  6. Ahhh!! you are magic. I have spent so much time stuck on this stuff and then in the first 2 minutes of this video you fixed all my problems. Thank you so much!

  7. Hey I just wanted to say that the graph of the second function is wrong (I think) it said cosine and you graphed it like  a sine function

  8. OMG this helped me understand so much!! This was way better than the Khan Academy video (which was not helpful at all).  Thank you!

  9. The alarm in the background haha! These videos are amazing! They are actually teaching me unlike my teacher. Thanks a bunch!

  10. Thank you #ProfRobBob for the time you spend creating videos (and replying to comments). Your channel is like the 'porridge that was just right' from Goldilocks and the Three Bears (in comparison to a few other math help YouTube channels I've seen).

  11. Thanks for your time on these videos and a huge thanks for taking the time to answer questions! I do have one quick question i'm a little confused on what the period is, i see tho that you keep using 2pie as the period. do i always use that? if not how do i determine what the period is?

  12. Ahh you are awesome RobBob! Math is not my strong suit, I am a pre-med student and have to go quite a ways in math and you have helped me so much in understanding what the heck is going on! I feel like it's so simple now! Thank you for taking the time to make these videos, you are inspiring many other students other than the ones you teach personally I'm sure! Keep it up! ~Samantha
    P.S. I love the enthusiasm, it made me smile big and loving what you do shows and makes a world of difference in our learning experiences! So thank you for that as well. Now, I have to go do my homework 😛

  13. I am having trouble figure out what to count by with the problem of f(x)=2sin(x(pi/3))+1. I have got to be doing something wrong but cannot figure out what it is. If my period is 2pi and I divide by four I get pi/2, and the phase shift is pi/3 what do I count by? pi/6? If I do that I do not seem to get my key points from the unit circle… Anyone have any suggestions?

  14. Thank you for taking your time to make these videos, ProfRobBob. The only thing I'm having trouble understanding was when you set up your t-chart for the first example. Why did you change sin( 2x – pi) to sin( 2x – 4pi/4)? Can you explain where the 4pi/4 came from?

  15. Guess who's going to pass Trigonometry course thanks to #ProfRobBob  . This totally helped me out! My final exam is tomorrow and I'm feeling ready. THANKS A LOT 🙂

  16. What happens when you don't have a "pi" in your period?  y= -sin pi x .  amp is 1, period is 2pi/pi, PS 0 VS 0, count is 1/2.

  17. I'm sooo glad that I found you, but I wished I had found you earlier. Anyways you're a math lifesaver, keep up the great work!

  18. Thank You so much for making this video, I never really understood phase shifts but this helped me so much. I have a test tomorrow and I know I can get a good grade now.

  19. I may be wrong but isn't pi the phase shift? When you do C/B doesn't this give the horizontal shift, not the phase shift?

  20. This is amazing, thanks for explaining it so well, especially during the T-Chart.  My teacher explained it in class, but I didn't really understand too well, but this cleared it up so much!Thanks again!   🙂

  21. Thanks a bunch from the bottom of my heart professor Rob Bob for the outstanding videos! Your videos literally help me learn the concepts way better than text book and lecture classes, no joke. I do like your catchy word Bam!

  22. I love your videos man!
    My teachers encourage me to use the calculator for everything. Luckily I never listen to her but instead seek the long way of doing things to gather a better understanding. Other students are so dependent on formulas and calculators that they have forgotten how to add fractions, crazy right.

  23. My fellow students are getting crushed in our Trig class, and I am excited to share these videos with them, as they have helped me understand the material tremendously well. Thanks, and keep up the great work!!

  24. Thank you so much, Mr. Tarrou for your videos. I've learned a lot from them. Your style of teaching works for me. I was so lost in class especially regarding graphing sine, cosine and tangent. BAM! 🙂

  25. Awesome videos! I am a musician trying to use math to create waveforms. Your videos have helped a lot. I have a question… What's the difference between sin(x+pi) and sin(x*pi)

  26. awesome lecture, just watching your video on graphing over and over until your method sticks for my trig exam 🙂 thanks for doing what you do

  27. Please explain to me why you arbitrarily create the count. When I graph y=sinx on a calculator, the line goes through pi and 2pi. You seem to arbitrarily divide the period by 4 to create a totally different  graph.

  28. It's bizarre and a testiment to your teaching that after all this time, these are still some of the best vids (you and Professor Leonard need to get together and do a collaboration series of videos!) What still gets me is that the Phase Shift is sort of "backwards" a positive phase shifts it to the left and a negative Phase shift pushes it right…which looks 'wrong'! My question though, is that do we always count by 1/4 of the period for the X axis?

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