In this video, we're going to be solving whole collection of
trigonometric equations now be cause it's the technique of
solving the equation and in ensuring that we get enough
solutions, that's important and not actually looking up the
angle. All of these are designed around certain special angles,
so I'm just going to list at the very beginning here the special
angles and their sines, cosines, and tangents that are going to
form. The basis of what we're doing.
So the special angles that we're going to have a look at our
zero. 30 4560 and 90 there in degrees.
If we're thinking about radians, then there's zero.
Pie by 6.
Pie by 4.
Pie by three.
And Π by 2.
Trig ratios we're going to be looking at are the sign.
The cosine. On the tangent of each of these.
Sign of 0 is 0.
The sign of 30 is 1/2.
Sign of 45 is one over Route 2.
The sign of 60 is Route 3 over 2 and the sign of 90 is one.
Cosine of 0 is one.
Cosine of 30 is Route 3 over 2 cosine of 45 is one over Route
2, the cosine of 60 is 1/2, and the cosine of 90 is 0.
The town of 0 is 0 the town of 30 is one over
Route 3 that Anna 45 is 110 of 60 is Route 3 and
the town of 90 degrees is infinite, it's undefined.
It's these that we're going to be looking at and working with.
Let's look at our first
equation then. We're going to begin with some very simple
ones. So we take sign of X is equal to nought .5. Now
invariably when we get an equation we get a range of
values along with it.
So in this case will take X is between North and 360. So what
we're looking for is all the values of X.
Husain gives us N .5.
Let's sketch a graph of sine X over this range.
And sign looks like that with 90.
180
270 and 360 and ranging between one 4 sign 90 and minus
one for the sign of 270.
Sign of X is nought .5. So we go there.
And there.
So there's our first angle, and there's our second angle.
We know the first one is 30 degrees because sign of 30 is
1/2, so our first angle is 30 degrees. This curve is symmetric
and so because were 30 degrees in from there, this one's got to
be 30 degrees back from there.
That would make it
150. There are no more answers because within this range as we
go along this line.
It doesn't cross the curve at any other points.
Let's have a look at a cosine cause
of X is minus nought .5 and the
range for this X between North and 360.
So again, let's have a look at a graph of the function.
Involved in the equation, the cosine graph.
Looks like that. One and
minus one. This is 90.
180
270 and then here at the end, 360.
Minus 9.5.
Gain across there at minus 9.5 and down to their and
down to their.
Now the one thing we do know is that the cause of 60 is plus N
.5, and so that's there. So we know there is 60. Now again,
this curve is symmetric, so if that one is 30 back that way
this one must be 30 further on. So I'll first angle must be 120
degrees. This one's got to be in a similar position as this bit
of the curve is again symmetric.
So that's 270 and we need to come back 30 degrees, so
that's 240. Now we're going to have a look at an example where
we've got what we call on multiple angle. So instead of
just being cause of X or sign of X, it's going to be something
like sign of 2X or cause of three X. So let's begin with
sign of. 2X is equal to Route 3 over 2
and again will take X to be between North and
360.
Now we've got 2X here.
So if we've got 2X and X is between Norton 360, then the
total range that we're going to be looking at is not to 722.
X is going to come between 0 and 720, and the sign function
is periodic. It repeats itself every 360 degrees, so I'm going
to need 2 copies of the sine
curve. As the first one going up to 360 and now I need a second
copy there going on till.
720
OK, so sign 2 X equals root, 3 over 2, but we know that the
sign of 60 is Route 3 over 2. So if we put in Route 3 over 2 it's
there, then it's going to be these along here as well. So
what have we got? Well, the first one here we know is 60.
This point we know is 180 so that one's got to be the same
distance. Back in due to the symmetry 120, so we do know that
2X will be 60 or 120, but we also now we've got these other
points on here, so let's just count on where we are. There's
the 1st loop of the sign function, the first copy, its
periodic and repeats itself again. So now we need to know
where are these well.
This is an exact copy of that, so this must be 60
further on. In other words, at 420, and this must be another
120 further on. In other words, at 480. So we've got two
more answers. And it's X that we actually want, not
2X. So this is 3060.
210 and finally
240. Let's have a look at that with a tangent
function. This time tan or three X is equal to.
Minus one and will take X to be
between North and 180.
So we draw a graph of the tangent function.
So we go up.
We've got that there. That's 90.
This is 180 and this is 270 now. It's 3X. X is between
Norton 180, so 3X can be between North and 3 * 180 which is
540. So I need to get copies of this using the periodicity of
the tangent function right up to 540. So let's put in some more.
That's 360. On there.
That's 450.
This one here will be 540 and that's as near
or as far as we need to go. Tanner 3X
is minus one, so here's
minus one. And we go across here picking off all the ones that we
need. So we've got one there.
There there. These are our values, so 3X is equal 12.
Now we know that the angle whose tangent is one is 45,
which is there. So again this and this are symmetric bits of
curve, so this must be 45 further on. In other words 130.
5.
This one here has got to be 45 further on, so that will be
315. This one here has got to be 45 further on, so that
will be 495, but it's X that we want not 3X, so let's divide
throughout by three, so freezing to that is 45 threes into that
is 105 and threes into that is 165. Those are our three answers
for that one 45 degrees.
105 degrees under
165. Let's take cause of X over 2 this time. So
instead of multiplying by two or by three, were now dividing by
two. Let's see what difference this might make equals minus 1/2
and will take X to be between North and 360. So let's
draw the graph.
All calls X between North and 360, so there we've
got it 360 there.
180 there, we've got 90 and 270 there in their minus. 1/2 now
that's going to be.
Their cross and then these are the ones that we are after.
So let's work with that. X over 2 is equal tool. Now where are
we? Well, we know that the angle whose cosine is 1/2 is in fact
60 degrees, which is here 30 in from there. So that must be 30
further on. In other words, 120 and this one must be 30 back. In
other words, 240. So now we multiply it by.
Two, we get 240 and 480, but of course this one is outside
the given range. The range is not to 360, so we do not
need that answer, just want the
240. Now we've been working with a range of North 360, or in one
case not to 180, so let's change the range now so it's a
symmetric range in the Y axis, so the range is now going to run
from minus 180 to plus 180
degrees. So we'll begin with sign of X equals 1X is to
be between 180 degrees but greater than minus 180 degrees.
Let's sketch the graph of sign in that range. So we want to
complete copy of it. It's going to look like that.
Now we know that the angle who sign is one is 90
degrees and so we know that's one there and that's
90 there and we can see that there is only the one
solution it meets the curve once and once only, so
that's 90 degrees.
Once and once only, that is within the defined range. Let's
take another one.
So now we use a multiple angle cause 2 X equals 1/2
and will take X to be between minus 180 degrees and
plus 180 degrees. So let's sketch the graph. Let's remember
that if X is between minus 180 and plus one 80, then
2X will be between minus 360.
And plus 360.
So what we need to do is use the periodicity of the cosine
function to sketch it.
In the range. So there's the knocked 360 bit and
then we want.
To minus 360. So I just label up the points. Here is 90.
180
Two 7360 and then back this way minus 90 -
180. Minus 270 and
minus 360. Now cause 2X is 1/2, so here's a half.
Membrane that this goes between plus one and minus one and if we
draw a line across to see where it meets the curve.
Then we can see it meets it in four places. There, there there
and there we know that the angle where it meets here is 60
degrees. So our first value is 2 X equals 60 degrees.
By symmetry, this one back here has got to be minus 60.
What about this one here? Well, again, symmetry says that we are
60 from here, so we've got to be 60 back from there, so this
must be 300 and our symmetry of the curve says that this one
must be minus 300, and so we have X is 30 degrees minus 30
degrees, 150 degrees and minus
150 degrees. Working with the tangent function tan, two
X equals Route 3 and again will place X between
180 degrees and minus 180 degrees. We want to sketch
the function for tangent and we want to be aware
that we've got 2X.
So since X is between minus 118 + 182, X is got to be between
minus 360 and plus 360.
So if we take the bit between.
North And 360.
Which is that bit of the curve we need a copy of that between
minus 360 and 0 because again the tangent function is
periodic, so we need this bit.
That
And we need that and it's Mark off this axis so we know where
we are. This is 90.
180
270 and 360. So this must be minus 90 -
180 - 270 and minus 360.
Now 2X is Route 3, the angle whose tangent is Route 3. We
know is 60, so we go across here at Route 3 and we meet the curve
there and there.
And we come back this way. We meet it there and we meet there.
So our answers are down here.
Working with this one, first we know that that is 60, so 2X is
equal to 60 and so that that one is 60 degrees on from that
point. Symmetry says there for this one is also 60 degrees on
from there. In other words, it's
240. Let's work our way backwards. This one must be 60
degrees on from minus 180, so it must be at minus 120. This one
is 60 degrees on.
From minus 360 and so therefore it must be minus 300.
And so if we divide throughout by two, we have 31120 -
60 and minus 150 degrees. We want to put degree signs on
all of these, so there are four solutions there.
Trick equations often come up as a result of having expressions
or other equations which are rather more complicated than
that and depends upon
identity's. So I'm going to have a look at a couple
of equations. These equations both dependa pawn two identity's
that is expressions involving trig functions that are true for
all values of X.
So the first one is sine squared of X plus cost
squared of X is one. This is true for all values of X.
The second one we derive from
this one. How we derive it doesn't matter at the moment,
but what it tells us is that sex squared X is equal to 1 + 10
squared X. So these are the two identity's that I'm going to be
using. Sine squared X plus cost squared X is one and sex squared
of X is 1 + 10 squared of
X OK. So how do we go about using one of those to do
an equation like this? Cos
squared X? Plus cause of X is equal
to sine squared of X&X is between 180
and 0 degrees.
Well.
We've got a cost squared, A cause and a sine squared.
If we were to use our identity sine squared plus cost squared
is one to replace the sine squared. Here I'd have a
quadratic in terms of Cos X, and if I got a quadratic then I know
I can solve it either by Factorizing or by using the
formula. So let me write down sign squared X plus cost
squared. X is equal to 1, from which we can see.
Sine squared X is equal to 1 minus Cos squared of X,
so I can take this and plug it into their. So my
equation now becomes cost squared X Plus X is equal
to 1 minus Cos squared X.
I want to get this as a quadratic square term linear
term. Constant term equals 0, so I begin by adding cost squared
to both sides.
So adding on a cost squared there makes 2 Cos squared X plus
cause X equals 1. 'cause I added cost square to get rid of that
one. Now I need to take one away from both sides to cost squared
X Plus X minus one equals 0.
Now this is just a quadratic equation, so the first question
I've got to ask is does it factorize? So let's see if we
can get it to factorize.
I'll put two calls X in there and cause X in there because
that 2 cause X times that cause X gives Me 2 cost squared and I
put a one under one there 'cause one times by one gives me one
and now I know to get a minus sign. One's got to be minus and
one's got to be plus now I want plus cause X so if I make this
one plus I'll have two cause X times by one.
Is to cause X if I make this one minus I'll have minus Cos X from
there. Taking those two together, +2 cause X minus Cos X
is going to give me the plus Kozaks in there, so that equals
0. Now, if not equal 0, I'm multiplying 2 numbers together.
This one 2 cause X minus one and this one cause X plus one, so
one of them or both of them have got to be equal to 0.
So 2 calls X minus one is 0.
All cause of X Plus One is 0, so this one tells me that
cause of X is equal to 1/2.
And this one tells Maine that cause of X is equal to minus
one, and both of these are possibilities. So I've got to
solve both equations to get the total solution to the original
equation. So let's begin with this cause of X is equal to 1/2.
And if you remember the range of values was nought to 180
degrees, so let me sketch cause of X between North and
180 degrees, and it looks like that zero 9180.
We go across there at half and come down there and there is
only one answer in the range, so that's X is equal to 60 degrees.
But this one again let's sketch cause of X between North and
180. There and there between minus one and plus one and we
want cause of X equal to minus one just at one point there and
so therefore X is equal to 180 degrees. So those are our two
answers to the full equation that we had.
So it's now have a look at
three. 10 squared X is equal to two sex squared X
Plus One and this time will
take X. To be between North and 180 degrees. Now, the identity
that we want is obviously the one, the second one of the two
that we had before. In other words, the one that tells us
that sex squared X is equal to 1 + 10 squared X and we want to be
able to take this 1 + 10 squared and put it into their. So we've
got 3. 10 squared X is equal to 2
* 1 + 10 squared X Plus one. Multiply out
this bracket. 310 squared X is 2 + 210 squared
X plus one.
We can combine the two and the one that will give us 3.
And we can take the 210 squared X away from the three times
squared X there. That will give us 10 squared X. Now we take the
square root of both sides so we have 10X is equal to plus Route
3 or minus Route 3.
And we need to look at each of
these separately. So.
Time X equals Route
3. And Tan X equals minus Route 3.
Access to be between North and 180, so let's have a sketch of
the graph of tan between those values, so there is 90.
And there is 180 the angle whose tangent is Route 3, we know.
Is there at 60 so we know that X is equal to
60 degrees? Here we've got minus Route 3, so
again, little sketch.
Between North and 180 range over which were working here, we've
got minus Route 3 go across there and down to their and
symmetry says it's got to be the same as this one. Over here it's
got to be the same either side. So in fact if that was 60 there
this must be 120 here, so X is equal to 120 degrees.
So far we've been working in degrees, but it makes little
difference if we're actually working in radians and let's
just have a look at one or two examples where in fact the range
of values that we've got is in radians. So if we take Tan, X is
minus one and we take X to be between plus or minus pie.
Another way of looking at that would be if we were in
degrees. It will be between plus and minus 180. Let's
sketch the graph of tangent within that range.
Up to there.
That's π by 2.
Up to their which is π.
Minus Π by 2.
Their minus
pie. Ton of X is minus one, so somewhere
across here it's going to meet the curve and we can
see that means it here.
And here giving us these solutions at these points. Well,
we know that the angle whose tangent is plus one is π by 4.
So this must be pie by 4 further on, and so we have X is equal to
pie by 2 + π by 4. That will be 3/4 of Π or three π by 4, and
this one here must be.
Minus Π by 4 back there, so minus π by 4.
Let's take one with a multiple angle.
So we'll have a look cause of two X is equal to Route
3 over 2.
I will take X between North
and 2π. Now if X is between North and 2π, and
we've got 2X.
And that means that 2X can be between North and four π.
So again, we've got to make use of the periodicity.
Of the graph of cosine to get a second copy of it.
So there's the first copy between North and 2π, and now we
want a second copy that goes from 2π up till four π.
We can mark these off that one will be pie by two.
Pie.
Three π by 2.
This one will be 5 Pi by two. This one three Pi and this
one Seven π by 2.
So where are we with this cost? 2 X equals. Well, in fact we
know cost to access Route 3 over 2. We know that the angle that
gives us the cosine that is Route 3 over 2 is π by 6. So
I'll first one is π Phi six, root 3 over 2. Up here we go
across we meet the curve we come down. We know that this one here
is π by 6.
Let's keep going across the curves and see where we come to,
what we come to one here which is π by 6 short of 2π. So
let me write it down as 2π - Π by 6, and then again we come
to one here. Symmetry suggests it should be pie by 6 further
on, so that's 2π + π by 6, and then this one here.
Is symmetry would suggest his pie by 6 short of four Pi,
so four π - π by 6. So let's do that arithmetic 2X
is π by 6.
Now, how many sixths are there in two? Well, the answer. Is
there a 12 of them and we're going to take one of them away,
so that's eleven π by 6. We're going to now add a 6th on, so
that's 13 Pi by 6.
How many 6th are there in four or there are 24 of them? We're
going to take one away, so that's 23. Pi over 6. Now we
want X, so we divide each of these by 2π by 1211 Pi by
12:13, pie by 12, and 20, three π by 12, and there are our
four solutions. Let's have a look at one where we've got the
X divided by two rather than multiplied by two. So the sign
of X over 2 is minus Route 3 over 2.
And let's take X to be between pie and minus π. So
will sketch the graph of sign between those limited, so it's
there. And their π zero and minus pie.
Where looking for minus three over 2. Now the one thing we do
know is that the angle who sign is 3 over 2 is π by 3.
But we want minus Route 3 over 2, so that's down there.
We go across.
And we meet the curve these two
points. Now this curve is symmetric with this one.
So if we know that.
Plus Route 3 over 2.
This one was Pi by three. Then we know that this one must be
minus π by 3.
This one is π by three back, so it's at 2π by three, so this one
must be minus 2π by three, and so we have X over 2 is equal
to minus 2π by three and minus, π by three, but it's X that we
want, so we multiply up X equals minus four Pi by three and minus
2π by 3.
Let's just check on these values. How do they fit with the
given range? Well, this 1 - 2π by three is in that given range.
This one is outside, so we don't want that one.
A final example here, working with the idea again of using
those identities and will take 2 cost squared X.
Plus sign X is equal to 1.
And we'll take X between North and 2π.
We've got causes and signs, so the identity that we're
going to want to help us will be sine squared plus
cost. Squared X equals 1.
Cost squared here.
Cost squared here. Let's use this identity to tell us that
cost squared X is equal to 1 minus sign squared X and make
the replacement up here for cost
squared. Because that as we will see when we do it.
Leads to a quadratic in sign X, so it's multiply this out 2 -
2 sine squared X plus sign X is equal to 1 and I want
it as a quadratic, so I want positive square term and then
the linear term and then the constant term. So I need to add.
This to both sides of 0 equals 2 sine squared X. Adding it to
both sides. Now I need to take this away minus sign X from both
sides and I need to take the two away from both sides. So one
takeaway two is minus one.
And now does this factorize? It's clearly a quadratic. Let's
look to see if we can make it factorize 2 sign X and sign X.
Because multiplied together, these two will give Me 2 sine
squared one and one because multiplied together, these two
will give me one, but one of them needs to be minus. To make
this a minus sign here. So I think I'll have minus there and
plus there because two sign X times by minus one gives me.
Minus 2 sign X one times by sign X gives me sign X and if I
combine sign X with minus two sign XI get minus sign X.
I have two numbers multiplied together. This number 2 sign X
Plus One and this number sign X minus one. They multiply
together to give me 0, so one or both of them must be 0. Let's
write that down.
2 sign X Plus One is equal to 0 and sign X minus one is equal
to 0, so this tells me that sign of X is equal. To take one away
from both sides and divide by two. So sign X is minus 1/2 and
this one tells me that sign X is equal to 1.
I'm now in a position to solve these two separate equations.
So let me take this one first.
Now. We were working between North and 2π, so we'll have a
sketch between North and 2π.
Of the sine curve and we want sign X equals one. Well, there's
one and there's where it meets, and that's pie by two, so we can
see that X is equal to pie by
two. Sign X equals minus 1/2. Again, the range that we've been
given is between North and 2π. So let's sketch between Norton
2π There's 2π.
Three π by 2.
Pie pie by two 0 - 1/2, so that's coming along between
minus one and plus one that's going to come along there.
And meet the curve there and there. Now the one thing that we
do know is the angle who sign is
plus 1/2. Is π by 6, so we're looking at plus 1/2. It will be
there and it would be pie by 6.
So it's π by 6 in from there, so symmetry tells us that this
must be pie by 6 in from there, so we've got X is equal to π
+ π by 6, and symmetry tells us it's pie by 6 in. From there, 2π
- Π by 6.
There are six sixths in pie, so that's Seven π by 6. There is
1216, two Pi. We're taking one of them away, so it will be
11 Pi over 6.
So we've shown there how to solve some trig equations.
The important thing is the sketch the graph. Find the
initial value and then workout where the others are
from the graphs. Remember, the graphs are all symmetric
and they're all periodic, so they repeat themselves every
2π or every 360 degrees.