# Graphing Trig Functions (Phase Shift)

(D4M) — Here is the video transcript for this video.

00:00
what does phase shift mean left right or up or down
in this episode we concentrate on graphing trig functions that have a
phase shift this approach has the additional benefit
of understanding the similarities between sine and cosine functions
let’s do some math [Music] hi everyone welcome back in this episode
graphing trig functions we’re going to concentrate on the phase shift
and so let’s go ahead and get started here what is a phase shift so
now actually before we get started i want to mention very very briefly that
this episode is part of the series trigonometry is fun step-by-step
tutorials for beginners and in the link is below in the
description and so you can get all caught up on all the episodes
um and so yeah in this episode we’re going to concentrate on the phase shift

00:01
so first off um let’s let me remind you that the absolute value of a is
gives the amplitude the amplitude and so we’re looking at
graphs that look like this a sine bx plus c plus d and
the same for cosine and secant and cosecant and tangent and cotangent so
i’ll just write it out for sine so a is the number in front here
and if you have sine or cosine then the absolute value of a gives the amplitude
in fact i’ll just focus on uh sine and cosine right here

00:02
now we could have a plus or minus c here and we could have a plus d okay um so
the period um the 2 pi over b is gives us a period in the last episode
we talked about um when c was zero when d was zero and the b could be positive
or negative in particular we talked about how the sine function is an odd
function and the cosine function is an even function and so you know
we can call this one here standard form standard form and let’s just form
and we can talk about shifted form so a sine and then we’ll factor out a b here
and say b and then x plus or minus c over b
and let’s just put brackets around here and then plus d
and then we’ll have the same for cosine b

00:03
and we factor out a b here so b and then x plus or minus c over b
and then plus d now in terms of this video here um
in terms of this video here the d is going to be 0 and we’re going to have
and we’re going to focus on the d which is going to be called the vertical shift
um in another episode so d will be 0 for today in this episode
and look so look for that episode coming up episode
uh called vertical translations and so [Music] then the fourth one here is

00:04
the um c is called the phase angle the phase angle and um plus or minus um
c over b is called the uh phase shift so or sometimes we’ll call it the
horizontal shift and so this is going to move opposite in sign
so if this is a positive c over b then we’re going to move to the left if it’s
a negative c over b we’re going to move to the
right and so this is the outline here um and this is called a shifted form
so these are all the same like this is the same as this one
it just has a different form to it so there’s because there’s two forms
there’s two ways we can sketch this so we’re going to focus on moving from
standard form to the sketch today to the sketch in this episode and then

00:05
look for an upcoming episode where the title is graphing trick functions
but we’ll have horizontal shifts and so today i’m we’re going to focus on
uh the standard form and we’re going to use the word freight phase shift
but because you know these have two different forms one for sine and two
different forms for cosine there’s two different approaches so we’re gonna
we’re gonna cover both approaches in this series
and and today we’re gonna cover the phase shift approach which is gonna
start with standard form right here but again in this episode the d is zero so
and then we’ll do some episodes where um you know you can have any approach
you want to take in it and we’ll kind of tie it all together in a review video
okay so that’s what we’re going to do in today and so let’s go ahead and get

00:06
started with our first example so here we go an example please there we go
so let’s get this out of the way here so let’s look at a concrete example and
understand more clearly what i was just saying now here the a is the
amplitude so the absolute value of a which in this example is the absolute
value of -3 which is just three amplitude is always positive amplitude is three
and the period is so two pi over b which is two pi right here
so i’ll just write like this 2 pi over b which is b
b is in this case 2 pi also so 2 pi right there so the period is 1.
normally cosine you will make a period say from zero to two pi
and so now here the period is going to be just zero to one

00:07
um and so what we’re going to do is in order to deal with the phase shift
that we have going on right here so this is the plus c
right and so if we think about this right here in the form a cosine b x plus c
plus d here in this case the c is positive 4 pi
and so that’s going to be the shift right there now we could say 4 pi over
2 pi which gives me a 2 positive 2 so that means we’re going to shift left 2.
um so that’s one way to remember it however i think it’s
a lot easier if we just take this part right here the 2 pi x plus 4 pi
and set that equal to 0 and also set it equal to

00:08
4x so set take the whole angle right here and set it equal to
so it’s 2 pi x plus 4 pi sorry 4 pi plus 4 pi equals 2 pi now
do you know why this is equal to zero the angle right here and it’s also equal
to two pi well the reason why is because that’s the normal period or not normal
but that’s a period usually i call that the first period
but that’s the period of cosine is zero to two pi
so i’m looking for the new period right we know the period is one
but what are the tick marks for it i mean where where does it cross the
x-axis and you know where does it have its maximum and minimum
so we can solve this right here so this is two pi x equals minus four pi
and then we’ll divide by two pi and in fact we see x is just minus two
and we’ll do the same thing over here two pi x

00:09
so two pi minus four pi is minus two pi cancels the two pi’s x is minus one
so let’s just abbreviate that so we can save some space so x is two
and x is minus one and so these are our new tick marks so
normally cosine you know zero to zero to two pi so now i’m gonna go from minus
two to one and basically what we’re doing is we’re
shifting the graph to the left so here we go i’m gonna sketch the graph now and
we’re gonna go from -2 right here and this will be -1 right here
and you know that’s not a great scale right there that i have got way too much
positive x so let’s just redo that and let’s say this is minus two
and let’s say this is minus one and so now um you know this is a

00:10
negative right here for the cosine right here so um
it’s going to start down here at -3 right here because that’s the
amplitude so it’s going to come in here like this
and then halfway it’s going to go up and then
it’s going to come back down halfway it’s going to reach all the way to the peak
and then it’ll start repeating itself at the minus 1 right here
so this part right here is halfway right here which is minus 3 over two
and the height here is three and this is minus one and
um so i need these two right here so this will be minus seven over four
and this one right here will be minus five over four
so we have those tick marks right there so this one right here is minus five
over four minus seven over four you can think of

00:11
minus two as minus eight over four and this will be a minus one right here
and the minimum right here is happening at minus three
so the minimum right here is minus one minus three but then we have it again at
minus two minus three and so on and so here’s a
you know reasonable sketch of the graph for this function right here
and so normally the cosine would have started right here
and go down and come back up right but we’ve shifted this over
and you know how much have we shifted it over
this is where the c over b that i was talking about earlier comes in
so the c is what um four pi and the b is two pi
so it’s been shifted to the left two so instead of starting here at zero like
this it’s been shifted over and now it starts at -2

00:12
but also because of the minus sign it’s been reflected
so it’s actually not this graph that’s been shifted over it’s the reflection
so i’ll sketch the reflection right here right so something like that
here’s the reflection y equals um minus three cosine x
so there’s that graph right there and it’s been shifted to the left two units
shifted to the left two units there all right and so here you know here’s
the sketch uh that’s just to help illustrate what i’m talking about here
in terms of shifting it to the left two units right there all right let’s look
at some more examples now and the first one is going to be this one right here
so let’s see here we have sine of 2x plus pi and so

00:13
what we’re going to start off with is saying 2x plus pi equals 0
and 2x plus pi equals 2 pi and let’s move the pi over so we get minus pi
and then we’ll divide by my two and then here we get 2x equals pi
so x equals pi over two and notice the period is two pi over b
and so if you’re still just learning this let me just make sure we’re
we’re good on this this is a sine bx plus c
plus d but right now the d is just 0. so the b is two the period is pi
you can see this distance between these two right here is pi
right positive pi over two and then we have a negative negative pi over two

00:14
and then we just have the regular sign graph whatever the graph looks like for
example this one’s not reflected at all so let’s try to sketch this graph here
and it’s it’s um starting right here at minus pi over two
and sine normally starts like right here goes up like that
i’ll just sketch dot it in real quick there’s the normal sine graph right
there but uh we’re shifting it over now we’re starting over right here so
we have this um tick marks to go through this one’s pi over two
now what’s halfway is zero so the graph is going to come up
and then back down halfway through it and then it’s going to go back down come
back up and so this will be zero this will be minus pi over four
this will be pi over four right so halfway right here where it

00:15
reaches its height of one and then right here where it reaches its um
halfway right here where it reaches a height of one and then halfway between
these two which is pi over four it reaches there it’s minimum at minus one
right there so we’ve shifted the graph over so what graph do you like to see
that it got shifted over so there’s that and that
there’s the normal sine graph y equals sine x
and you know that’s a 2 pi right there right so it’s been shifted over
but also not only shifted over but the period has changed because we
have a 2x here so the period is changed and we shifted it
over so this one here in orange isn’t really the um
one that was shifting we’re shifting this one right here sine 2x
and so this one right here has pi and then pi over 2

00:16
and so this one right here has been shifted over and so by how much so c over b
right and so this is pi over two and since it’s positive pi over two it’s
going to shift to the left minus pi over two so it’s gonna start to
go like that and the period is pi so so the graph is in black all the rest
of is just help understand it all right let’s do another one
let’s do another cosine um and so this one right here the period isn’t changing
amplitude is three and what is the phase shift right so
let’s do the amplitude so the amplitude is three amplitude
and then we’re going to take two pi over b
which in this case is just one so two pi this is period and then c over b

00:17
which is going to be pi over two over one pi over two is the phase shift
i’ll just abbreviate that phase shift it’s positive so i’m going to be moving
it to the left so normally cosine will look like this i’ll put the
cosine right here um it’s going to look like this then come back up
and the period will be two pi and then halfway will be pi
and halfway will be pi over two and halfway will be three power two
so there’s the cosine graph in fact let’s make that the 3 cosine graph
because we have a 3 in front so the height here is at 3
and this one right here is that minus three and so we want to take this graph
right here and shift it using the phase shift we’re going to shift to the left
pi over two now we didn’t have to sketch that first

00:18
um so what we can do is we can just simply say x plus pi over two equals zero
in fact let’s do that over here x plus pi over two equals zero
and then x plus pi over two equals two pi so this is the we’re graphing cosine
right and so the period is zero to two pi so where is it going to start
at minus two pi and where is it going to end so it’s two pi minus pi over two
and that’s just better known as three pi over two
and so there’s where we’re starting let’s sketch the graph now
so we’ve shifted by minus pi over two and remember this is positive cosine
here so it’s going to start up here and it’s going to come down
and then it’s going to go back up and then start

00:19
stop right there it’s going to start repeating and
you know it’s not 2 pi anymore it’s been shifted to the left this is 3 pi over 2
this is minus pi over 2. this is zero and
this right here is halfway between these two right here right so what’s halfway
well we can just add pi over twos this is a pi over two distance because that’s
minus pi over two so it’s one pi over two two pi over two and then this will
be three pi over twos um [Applause] one pi over 2 2 pi over 2 which is just
1 pi over 2. this should be just pi and this is a height of three

00:20
and minus three um so let’s see here it’s going to go to 0 and then -3
and then back to 0 and the period here [Music] is 2 pi so this should be
pi over 2 is 4 pi over 2 minus pi over 2 which yes 3 pi over 2.
um so that should be 3 pi over 2 this is 1 pi over 2 and then another pi over 2
that should be pi over 2 and this should be pi all right there we go
all right yep and so there we go okay so yeah it just keeps repeating

00:21
and so there’s number two there we have a phase shift so we just take the
regular graph with of amplitude 3 for cosine and we shift it left by pi over 2.
all right let’s do one more now this has a minus and this has a minus
here so let’s see what’s going on okay so the amplitude is
absolute value of minus one so just one the period is um 2 pi over b so 2 pi
and the phase shift is c over b um so it’s just c so minus 3 pi over 4
over 1 if you want this is a negative so now we’re going to
shift it to the right um and the reason why is because it’s
positive 3 pi over 4 that gets us back to 0.

00:22
and so if you’re not clear about that well we’ll just go ahead and solve these
two equations so we’ll take the whole angle that we’re put inputting into sine
and remember sine you know starts at 0 then goes to 2 pi for one full cycle
so we have x minus x minus 3 pi over 4 equals 0 so x equals positive 3 pi over 4
and then the other one is x minus 3 pi over 4 equals 2 pi
and then we’ll get the right tick mark here so 2 pi plus 3 pi over 4
and so think of 2 pi let’s bump that over so x equals two pi
plus three pi over four and then x equals so think of this two pi
is eight pi over four so this would be eleven pi over four all right and so

00:23
what we’re going to have here is um we’re going to start at 3 pi over 4 and
we’re going to end at 11 pi over 4 but we’re still going to have the same
period right here all right so let me sketch the graph right here
and so this is 3 pi over 4 about right here and
this right here has got a negative in front of it
so we’re going to start and we’re going to go down
and let’s get a good shape in here and so this is 3 pi over 4
and then we’ve reached one full cycle right here at 11 pi over four
and so what’s halfway between these two right here right so
to find out halfway it’s going to be what 14 pi over four or seven pi over two
that’s halfway right just add them up you get 14 pi over 4 right so 7 pi over 2

00:24
and then chop that in half right there you know so that would be 10 pi over uh
no that would be 3 pi over 4 plus 14 pi over 4 or said differently 17 pi
yeah so that’s 17 pi over 4 right that’s 3 pi over 4 plus um 14 pi over 4
so that would be 17 pi over 4 right here 17 pi over 4 and then
find the midpoint here would be 18 pi uh no would be 14 pi plus 11 pi
over four so that would be 25 pi over four there we go so
let’s see here what did we say that was 25 pi over four

00:25
and that’s 17 pi over four know if you can read that 17 pi over four
okay and so the reason why because that’s 3 pi over 4
and then 7 pi over 4 is going to be 14 pi over 4 and so that’s 17 pi over 4
in the same way for all of them right there in any case this is going to be
minus 1 right here and the height here is a 1. and there we go we got all the
we got the zeros right there and right there and right there for the
uh period for the one cycle there and we
have the maximum and the minimum we have the minimum and the maximum and
so and then we have the nice curvy shape to it that that’s appropriate
all right so there’s number three right there
all right let’s do a cosine another cosine here

00:26
all right and so this one right here is the phase shift is minus pi over four
because the b is one and the a is one and so yeah we’re just going to be
shifting it to the right by pi over four so i’ll take the regular cosine graph
which looks like this right here and this is a 2 pi and a pi and a pi over 2
and a 3 pi over 2. and so this is just y equals cosine x period is 2 pi
of course the height here is 1 and the minimum right here is that is -1
and then we reach the maximum again right there
so that’s just the regular cosine now we’re shifted to pi over four but to the
right so c over b is minus pi over four and then over one so
we’re going to shift it to the right now you know we just need to take each of

00:27
these tick marks right here 0 1 pi over 4 pi 3 pi over 2 2 pi so
we’re going to take each of them and then pi and then 3 pi over 2 and then 2 pi
and then i’m going to add the pi over 4 to it to each of these
to get the new tick marks because the phase shift is saying shift
it and that’s negative so we’re going to shift it to the right
and the way the reason why we shift it to the right is because if you plug in
pi over 4 into here you get out zero and so that’s where we’re starting right
here so normal cosine you plug in zero you get out one this
one you have to plug in pi over four to get the zero to get out the one
all right so what are these tick marks here pi over four and right so this is
um you know think of it in terms of 4 right so this
would be 2 pi over 4 so this will be 3 pi over 4

00:28
and think of this as 4 pi over 4 so that would be 5 pi over 4.
and think of this as 6 pi over 4 so 7 pi over 4 and then think of this as 8 pi
over 4 so this would be 13 pi over 4 or if you wish you could have seen the
pattern here in any case there is the new tick marks
so i’m going to construct the graph right here
and it’s been shifted to the right so it’s the same shape
so i’m going to come in here like this and i got my new tick marks right here
so pi over 4 3 pi over 4 five pi over four seven pi over four
and then thirteen pi over four those are
my new tick marks and of course we still have the one
and the minus one right there so there’s one way to get the tick marks

00:29
is just by actually viewing the graph um with only the period first
and then do the phase shift right there so we’ll do another video where we take
this approach and emphasize it more alright let’s go back and do another one now
let’s do a cosecant now so let me move out of the way slightly here
let’s move up here all right so let’s do a cosecant now
so actually the graph that i’m going to sketch first before i do the cosecant is
the sine so i’m going to sketch one half sine of 2x minus pi over 4.
i’m going to sketch this graph first and then when i finish sketching this graph
then i’ll come back and sketch the cosecant there and i’ll do it in a
different color i’ll do it in an orange so let me graph this one right here
first so you know the the amplitude is one half for this one

00:30
the period is going to be uh two pi over the b which is two so the period is pi
and the phase shift right here is c over b so it’s going to be c over b
which is minus pi over four over the b which is two so it’s minus pi over eight
so we’re going to shift to the right by pi over eight and
we’re going to have a new period of just pi so what we can do here is we can
you know take this starting and ending points here so 2x minus pi over 4 equals
0 and 2x minus pi over 4 that’s the whole angle right there
and that’s equal to 2 pi because normally cosecant has a period of 2 pi

00:31
just like sine does and so here we go so this is 2x equals pi over 4
and then divide by 2 we get pi over eight so as you can see it’s
going to be shifted to to the right to pi over eight
now this one right here is uh two x and this will be two pi plus pi over two
so 2x equals think of this as 4 pi over 2 so this will be 5 pi over 2
and then we’ll divide it into again 5 pi over 4. all right so
and this is just like a 2 pi and then it’s been shifted by pi over 8.
right think of the 2 pi as 16 pi over 8 plus pi over 8. so that’s 5 pi over 2

00:32
or 2 5 pi over 4 and that is 16 pi over eight but the period is only pi
okay i see so pi plus pi over eight which is
eight pi over eight plus pi over eight which is nine pi over eight the period is
pi so we’re going to do pi plus pi over eight um eight plus
one is nine pi over eight 5 8 something doesn’t quit quite look right
because here we’re gonna start at pi over eight and we’re gonna go to five pi

00:33
over four let’s see if i did this right so this will be um
what the heck could i do here that’s over four
um and so this will be eight pi over four so that’s 11 pi over four
and then we’ll chop that in half and say 11 pi over eight
which should be the same thing as pi plus pi over eight because the period is
gonna it’s gonna start repeating when you get to pi because that’s the period
and we’re shifting it to the right and so that should be eight pi over eight
so it should be nine pi over eight and so what did i do again
this is eight pi plus pi that’s nine pi over four and then chop that in half
nine pi over eight all right now everything’s making sense to me
all right sorry about that all right so long story short this is

00:34
where it start stops uh beginning of the cycle and the ending of the cycle there
and so we can start to sketch the graph and i’m just going to bump these up so i
have space here so let’s say this is pi over eight and this one is
nine pi over eight here all right and so let’s sketch the graph here now and
this is um we’re still graphing the sign here
and then we’ll come back in and do the cosecant here
so the sign here is going to start at the power rate and this is a positive
sign here so let’s go to pi over eight right here and we’re going to come up
and then we’re going to come down and then we’re going to get one cycle
there and that’s going to be 9 pi over 8 and what’s halfway right here
so halfway is um you know just add these
up and divide by two right so 10 pi over eight over 16

00:35
you know chop that in half so 10 pi over 16
and then chop that down into 2 5 pi over 8 so 5 pi over 8 right here and
then we can find the height right here adding them up will be 6 pi over 8 so 6
pi over 16 to get halfway and then 3 pi over 8 so 3 pi over 8
and then to get this one right here we can you know find the midpoint here
which will be 14 pi over eight and then double it to
get cut it in half so 16 so seven pi over eight so seven pi over eight right
here there’s the tick marks right there and the height here is at one half
and the minimum is right here value one half so now to get the cosecant graph
i’m going to need my isotope so i’m not going to ask where is this graph right

00:36
here where are the zeros of this graph right here so right here at
x equals pi over eight and then right here at x equals five pi over eight
and then right here at x equals nine pi over eight there we go
so it says pi over eight and i should just label them up here so 5 pi over 8
and then 5 wait this one’s 5 pi over 8 this one’s pi over eight
and this one is nine pi over eight and now we can sketch the
cosecant graph i’m gonna do that in orange so it reaches its low point there for
this branch at the one half and three pi over eight

00:37
and we got this branch over here and i think the isotopes are important
because it shows behavior so we have the upper branch and the lower branch for
one cycle and then it would just repeat another upper branch another lower
branch another upper branch another lower branch so i think the best way to
sketch this cos cosecant graph is to first sketch the sine graph right there
and then you can erase it if you want because you don’t really need it it’s
just there to help guide us where does it bend right here where does it bend
right here all right so there’s number five all right number six
let’s do a secant one now so i’m going to take the same um a similar approach
instead of doing secant we’re going to do the um cosine so let’s
say right here 2 cosine of pi x minus 2 pi so i’m going to graph this

00:38
one right here first so the period is going to be what so the period is
2 pi over the b which is pi so the period is 2 and the phase shift is
and the in the amplitude is 2. um but if it’s a secant graph i don’t
specifically say what the amplitude is simply because secant doesn’t have an
amplitude cosine does but any case the phase shift is c over b
and so it’s going to be a minus 2 pi over the b which is just pi
so minus two so it’s negative so i’m going to shift it to the right pi over two
and we can find the tick marks here by starting with the pi over x
pi x minus two x and so that’s the whole angle right
there and cosine is going to start at zero to get our first cycle and then pi
x minus two x equals two pi and so one full cycle of cosine is

00:39
between 0 and 2 pi and so let’s solve these and now that said minus 2 pi so that
should be 2 pi right here and again a pi sorry
okay so now we have pi x equals positive 2 pi
and then we’ll just cancel the pi’s and so pi x equals
2 pi and then move 2 pi over we get 4 pi and then cancel the pi’s so two to four
is the starting tick mark and that’s the where the cycle ends
and notice the distance between them is two so that makes sense because the
period is two right there all right so let’s sketch the graph here
and we’ll keep in mind that we’re going to do cosine and then we’re going to
come in and do the secant here so i want to put this right here
um maybe a little bit over here actually and let’s put it right here so let’s

00:40
call this a 2 and let’s call this a 4 and the height is going to be a height
of 2. and now this is cosine with a positive in front positive 2.
so i’m going to start up here at a positive 2 right here is a
maximum right there and i’m going to come through here
and then i’m going to come back over here and i should always put that tick
mark last before i shape it now it’s shaped now let’s put the tick mark
and then we just put an x now halfway is a three
so that’s where it’s going to hit the minimum right there
and then halfway right here is just you know add them up divide by two so five
over two and then add these up seven so seven over two
seven over two right there and so yeah this is the graph for this
one right here y equals two well there’s a two here and a minus two here
and then pi x minus two pi so there it is it’s been shifted over to the right

00:41
two because the phase shift is minus two so it’s been shifted to the right two
and here’s y and the period is only two so one period
is two right here so it’s a really good graph there now we need our isotopes to
graph the secant because secant is one over cosine so where is the cosine zero
so right here at five halves so here’s a vertical isotope right here
and i think we can slip it in right here five over two
and then it’s also zero right here at seven over two
so there’s the two vertical isotopes for the secant graph and now we can fill in
the secant graph right here in the orange
so we’re gonna have a lower branch right here
and then we’re going to have a partial upper branch right here
and a partial upper branch right here there we go so there’s the secant

00:42
right there there’s graph number six all right
so let’s look at this right here now by the way if you’re enjoying this video
let me know use the comments below and let me know that you liked it
if not let me know what you didn’t like all right so now let’s graph another uh
secant cosecant sorry cosecant um and we have a minus sign in front and
we have a 2 here so we’re changing the period it’s going to go twice as fast
right so the period is going to be 2 pi 2 pi over pi
sorry 2 pi over 2 which is pi and the phase shift is minus two pi over three
over the two or so differently minus two pi over three times a half
two’s cancel and we’re going to get minus pi over 3.

00:43
so there’s the phase shift 2’s cancel we’re just going to get minus pi over 3
and so it’s going to be we’re going to be
shifting it to the right but pi over 3. um so i’ll show you one way to do that
there’s there’s two different ways i’ll show you one way but also it’s going to
be a minus right so the auxiliary function that i’m
graphing is y equals minus sine of the angle 2x minus 2 pi over 3.
so i’m going to sketch this one right here first and black and then we’ll come
in and sketch the cosecant in orange so to sketch this one right here i
notice there’s a minus here all right so let me uh do that right here
so i’m going to be going down and i’m not going to start here at 0
because um it’s been shifted here so you know normally when we sketch the

00:44
graph of of sine it’ll be like this that’s that’s minus sign
and so what’s halfway and what’s this so the period is pi
so let’s say here we have 2x and so this is pi and so halfway is pi over 2
and halfway again is pi over 4 and then pi over 4 2 pi over 4 3 pi over 4
so that’s how i would sketch this graph right here minus sine 2x
and then we have a height of 1 and minus 1. so that’s minus sine 2x
but we have a phase shift of minus pi over 3 so i’m going to take each of
these tick marks right here and i want to add a pi over 3 because we’re
shifting to the right i want to add a pi over 3 to them
so we can do that over here we can say 0 plus pi over 3 so

00:45
0 and then and then pi over 4 and then pi over 2 and then 3 pi over 4
and then pi so i just listed the tick marks that i
have from sketching minus sine 2x minus sine 2x the period has been changed and
it’s negative so it’s going to start going down right here like that so pi
over four pi over two three pi over four pi and i’m gonna add the phase shift to
them um pi over three plus pi over three i’m gonna get my new tick marks here
so this will be pi over three and so think of this as over twelve
so this will be three plus four so seven pi over twelve
and let’s think of this one as um over a six
so this will be three plus two so this will be five pi over six

00:46
and then let’s think of this one as over twelve so this will be nine plus four
so thirteen pi over twelve and then think of this one right here um over three
and so this will be four pi over three okay so
these will be the new tick marks here and so that’s how i got those tick marks
there um however maybe you took a different approach
maybe we did as we’ve been doing before in this episode where we take the 2x
minus um what is that 2 2 pi over 3 set that equal to 0
and the 2x minus 2 pi over 3 equals to 2 pi and that’s because sine
of an angle it’s going to start at 0 and then the sine of the angle must be 2 pi
and that will be one full complete cycle there
so this will be 2x equals and then we’ll move the 2 pi over 3 over

00:47
and then we’ll divide by the 2 on both sides we get pi over 3
which was this first one here and then we get 2 pi 3 and so we get 2x equals
2 pi plus two pi over three and so let’s think of this one as being over
three so six pi so eight pi over six and then we’ll chop that down and say
four pi over six and then chop it down one more time two pi over three um
two two pi over six which is pi over three wait no something’s not right there
um so two pi and then added two pi over three
and so this will be six pi over three plus two pi over three
and so that is eight pi over three and so chop it in half that’ll be four

00:48
pi over three all right and so that one should agree right there
um this approach um we had to sketch um an auxiliary to our auxiliary graph
but i got the tick marks pretty pretty quickly there
um but anyways we can do this approach here too
so there’s where it starts and that’s why we’re shifting to the right by pi
over three and this is where it ends um and so i’ll just roll these up and
say pi over three well i got them over here so i’m good
all right so now let’s sketch this graph
right here finally this let’s sketch the auxiliary graph right here and
let’s do that right here let’s do this let’s say about right here
and this is a minus sign so we’re going to start going down
and there’s my first tick mark of pi over 3 and i’m going to be going down

00:49
[Music] and then back up and then down and this tick mark right here is the 4
pi over 3 and then halfway so i can find the
midpoint of them but we already have it right here it’s going to be 5 pi over 6
and then halfway i can find the midpoint between
these two but it’s already right here 13 pi over 12
and then i can find the midpoint between
these two but it’s already right here at 7 pi over 12.
and that’s going to be the minimum and the maximum
and this is going to happen at a a one and a minus one right here
and now we can go now that we have this sketched for one cycle
from my from pi over three to four pi over three
now we can sketch the isotopes and get the cosecant graph going so here’s where

00:50
sine is zero of this function right here zero so right here at
x equals pi over three and it’s zero right here at x equals five pi over six
and a zero right here at x equals 4 pi over 3
and so now we’re ready for the shape of the cosecant graph the final graph
everything else was made just to so we can see this right here
and there’s an upper branch right there and here’s a lower branch right here
there we go and that’s -1 and that’s a 1 right there
and so there we have the growth and it’s getting closer and closer to this
isotope and here as we travel this way we’re getting closer and closer to the
isotope all right and so there’s uh number seven right there
all right let’s do one more let’s uh do a secant and then let’s see
pause the video now and see if you can get it all on your own

00:51
and give that a try of course you may be doing that all
along but any case i like to just you know mention it pause the video
all right now actually this has a minus sign in front of here so
hopefully you took this approach here um we’re going to do secant of
i’m going to put a square bracket here secant of minus 2 um
i’m going to put a minus here with a 2x and a minus pi over 3.
so i’m taking secant of all this now the reason why i want to factor out
the minus sign from these two and by the way think of the plus as a
minus times a minus so i’m going to factor out one of those
minuses and one of those minuses gets left behind or said differently
a negative times a negative is a positive

00:52
but in any case if i factor out that now what does secant do to the minus sign
remember secant is an even function so this will be secant of 2x
minus pi over 3. so this is the function right here that i want to sketch
and before i do that i want to sketch the cosine so we okay so
let’s go over here and say i’m going to sketch the graph of y equals cosine of
2x and then minus pi over 3. so the period is going to be 2 pi over 2 or just pi
and the phase shift is minus pi over three over two
we’re said differently minus pi over six that says minus pi over six there

00:53
um and so we’re going to be shifting to the right pi over six so the right
by pi over six and so what we can do is we can first graph this one right here
cosine of 2x and then we can sketch cosine 2x minus pi over 3
and then once we have that graph we’ll be able to superimpose this graph right
here on top of it so let’s do that so let’s grab this one first so the
period is pi so it’s just going to look like a normal cosine graph
because it’s positive in front of here and it’s just a one the amplitude of
this right here is just one so i’m just going to sketch a nice cosine in here
and the period here is pi and so i can chop this in half is pi over two

00:54
and then half again is pi over four and then add up my pi over four so i get
three pi over fours here and so this is just y equals cosine of two x right here
so the point is is that it might be easier to get the tick marks by shifting
this by pi over six because i can just add a pi over six to each of these tick
marks so i can do that say i’ll just get out of the way let’s say over here
so i’m going to take my tick marks that i have here 0 pi over 4 pi over 2
3 pi over 4 and then pi and then i want to add a pi over 6 to all of them
so pi over 6 plus pi over 6 plus pi over 6.
and let’s add some fractions there pi over 6 is this one

00:55
now think of this as over 12 and so that one i’ll multiply by 3 so 3
pi plus 2 pi is five pi and this is a two and a six
so think of this as over a six and so i need a three pi plus a two pi
so it’s five pi over six and four part four and six so think of this
as over twelve and so i’m going to need a nine pi plus a two pi so it’s 11 pi
and then think of this one as over six so this will be six pi plus pi so seven
pi over six and so there’s our new tick marks for this one right here
so i think we can sketch it in here and instead of starting here at 0 1
now we’re going to start at pi over 6 so pi over 6 is about right here

00:56
and we’re going to keep the same shape here very good and the height here is one
and minimum right here is minus one and this right here is seven pi over six
it should be just pi plus another pi over six
and the midpoint is right here five pi over six
and so midpoint between these two and this one right here where we have a
0 is going to be 5 pi over 12 and the midpoint right here is going to
be the 11 pi over 12. and so now we have this sketch right here right here
that says pi over three so this is minus c over b in the b z two okay

00:57
so now we’re ready for our isotopes we got one right here
and we got one right here at 11 pi over 12 and we can label these up here
5 pi over 12. and we can label this one right here 11 pi over 12.
and then i’ll use orange to sketch in the graph right here
of the secant 2x minus pi over 3 and so we have a lower branch right here
and we have a partial upper branch right here
and another partial upper branch right here and we got all these points here
labeled that points labeled this one this one
this one and this one but we only really
need the orange we don’t really need the black graph it’s just there to help us
so we got um um two halves of an upper branch and a
lower branch there to get one full cycle there

00:58
and so that took care of this graph right here which is our last one right there
so i want to say thank you for watching and
the question now becomes now that we’ve done phase shift we’ve done period for
sines and cosines and secants and cosecants so now the question is what
are the graphs of tangents and cotangents look like and that video
starts right now