In this video, we're going to be looking at basic trig identities
and how to use them to solve trigonometric equations. Trig
equation is an equation that involves a trig function or
functions. When we solve it, what we do is find a value for
the trig function and then find the angle that corresponds to
that particular trig function. But what we want to start is
with the idea of a right angle triangle and go back.
To the well known theorem of
Pythagoras. So let's begin with our right angle triangle. There
the right angle in and let's label the sides and the angles.
So will have this.
As a side of length a, this is a side of length
B and the hypotenuse. The side that is opposite the
right angle will call C.
And our label, this angle here the angle a.
Now Pythagoras theorem tells us that if we take the square of
this side. And the square of
this side. Add them together. We'll get the square of this
side, so Pythagoras theorem tells us that A squared plus B
squared is equal to C squared.
Now. Let's divide throughout by C squared, divided every term in
this equation by C squared.
So we have a squared over C squared plus B squared over C
squared is equal to and this side we would have C squared
over C squared, but that of course is just one.
Now we can rewrite A squared over C squared as a oversee all
squared. And we can do the same here with B squared over C
squared. We can rewrite that as B. Oversee all squared.
And it still equal to 1.
Let's go back to this triangle again.
What does a oversee represent?
Well, a over C is the side opposite to the angle a.
Divided by the hypotenuse.
And so opposite divided by hypotenuses sign. So this, a
oversee is sign of a sign of the angle A and we have to
square it. Now we could write it like that, sign a squared,
and for the moment I will plus and let's have a look at B over
C or B is the adjacent side to the angle A and C is the
hypotenuse, so B oversee is adjacent over hypotenuse and
that's cosine. So we can replace B oversee by cause A and we need
to square it still.
Equal to 1.
Now this notation sign.
Squared, I just said sine squared, so rather than writing
it to sign a squared, which might be confused with squaring
the a, let's put the square on the sign and so the notation for
the sign of a times by the sign of a is sine squared a written
like that plus, and we use exactly the same technique cost
squared A. Equals 1.
Now that is an identity because it is true for all angles
a like this in a right
angle triangle. However.
I could have done this for the definitions of sine and cosine
that come from a unit circle.
In which case?
This identity would be true for all angles a no matter what
their size, and that's the case. This is a basic trig identity
that sine squared of an angle plus cost squared of an angle
equals 1. It's true for
all angles. What developed this
identity now? To give us two more basic
identity's. So let's
begin. With sine squared A plus cost
squared A equals 1.
Our basic fundamental identity, one that you really must learn
and know and come to recognize every time that you see it.
What I want to do is divide everything by this term
here cause squared A.
So sine squared a divided by Cos squared A.
Plus cost squared a divided by Cos squared a is
one over cause squared A.
Now, because I've divided everything by Cos squared, this
is still a true equation. Still in fact an identity.
So sine squared over cost
squared. Will sign over. Cause is tan and so this is in
fact TAN squared A.
Plus cost squared divided by Cos squared is just one.
Now, one over cause is sick, and so we can rewrite this as sex
squared A. And so we have
another identity. And normally we would write this as sex
squared a is 1 + 10 squared a. So there's our second
basic fundamental identity that's derived directly from
sine squared plus cost squared
is one. Well, if we can divide this equation by Cos
squared, surely we can do the same thing but with sine
squared. So we can divide the whole of this equation by sine
squared. We start again by writing down our basic
fundamental identity. Sine squared plus cost squared is
one. And as we said, instead of dividing everything
by Cos squared, we're going to divide everything by sine
squared. So that we have everything in the
identity divided by sine squared. So it's still
true for all angles. A sine squared divided
by sine squared. That's
just one. House.
Now we've. 'cause squared divided by sine squared. So
we've caused divided by sign all squared and cons divided by sign
is just caught.
So that is cot squared.
A. Equals and here with one over sine squared.
One over sine is kosach and so one over sine
squared is cosec squared.
And so there we have our third fundamental identity, one
plus Scott squared is cosec
squared. So we've now got three basic
fundamental identity's. I just write them down here in
this corner sign square day.
Cost Square day.
Is one.
1. Post and squared a is sex
squared A and one plus cot? Squared a? Is cosec
squared A? Now the use that we're going to make of these is
to help us solve particular kinds of trigonometric
equations. So first of all, let's look at this one 2.
10 squared X.
Is equal to sex squared X.
What we need to do is looking at this equation related to
one of these three identity's and it's fairly obvious that
this is the one.
So what we need to do then is get everything in terms of
either tans. Or sex.
Well, our identity says that's X squared is equal to 1 + 10
squared. So let's replace the sex squared here by 1 + 10
squared. So we have two 10 squared X is equal to 1 plus.
10 squared X.
And now we can take 10 squared away from each side, which will
leave us with one 10 squared this side and equals 1 there.
So now we can take the square root of both sides, so Tan X is
equal to 1. And let's not forget when we take a square root.
We've got 2 answers, plus or minus one in this case.
Now, the one thing that we didn't specify at the beginning
of this question was what was the range of values that we were
going to be working with 4X.
Well, since we didn't specify at the beginning, I think we're
entitled to put in any range of values that we want. So for the
moment, let's say that we're going to look at this for X.
Between equal to 0 but less than 2π.
Sotan axes one or minus one. Let's sketch the graph.
Of tanks between North and 2π, so it looks like that.
Asymptotes. Like that asymptotes and like that.
This is 2π this one we know is π by 2.
This where it crosses the X axis is π, and this one here is 3 Pi
by two. 10 X equals
1. That's one of those special angles.
45 degrees if we were working in degrees or pie by 4 radians. If
we're working in radians. So for the one bit we want Thai by 4,
but where else do we want to be? Here's one, and if we go across
the tan graph we can see we meet
it here. That's the pie by 4 and we meet it here.
And that is going to be in there halfway between pie and three Pi
by two, and so that is going to be at five π by 4. So there's
our second answer.
Coming from one and then we've got the minus one, so let's go
across at minus one.
Till we meet the graph.
There and there.
This is half way between pie by two and Π, and so that's going
to be three π by 4.
And this one is halfway between three Pi by two an 2π and so
that's going to be 7 Pi by 4, and so there are.
Four solutions to this question.
Let's take another example and this time I'm going to take one
that will make use of one more of these particular basic
identity's. So two sine squared
X. Close
call sex. Equals 1.
Now this has got sine squared's
in it. And the cause?
Well, fairly obviously, I think we ought to be using sine
squared plus cost. Squared is
one. But what do we replace? Do we try and replace the cause or
do we try and replace the sign? We've got a choice. Well, the
identity says sign squared plus cost squared equals 1.
So if it's sine squared that's in the identity, then perhaps
it's the sine squared that we ought to replace. So let's
make that replacement instead of sine squared.
Be'cause sine squared plus cost squared is one sign. Squared
must be 1 minus Cos squared. So in there will write 1
minus Cos squared X plus cause X equals 1.
Multiply out the brackets 2 - 2 cost
squared X plus cause X equals 1.
Well, if we simplify this, what we're going to end up with is a
quadratic equation where the variable is going to be cause X.
So let's move this term minus two cost squared X over to this
side of the equation. By adding two cost squared X to each side
so that we get it positive at this side. And So what I want to
end up with is an equation that's equal to 0, so.
Equals 0 at this to both sides. To cost squared X, take
this away from both sides, because that's plus cause X, so
minus cause X.
Take the two away from both sides, so that's one. Take away
these two is minus one.
As we said before, this is now just a quadratic.
So let's see if we can
factorize it. Remembering that the variable is cause X, well we
want two numbers that will multiply together is to give us
2 cost squared X, which suggests perhaps two Cos X and cause X.
We want two numbers that will multiply together to give us
minus one. Well, let's put ones in for the moment and worry
about the sign now.
If I take 2 cause X times by one here, I will get
just to cause X.
If I take 1 by cause X here I will get just cause X and I want
to end up with minus Cos X, which means I've really got to
take away the result of doing
this multiplication. So the minus sign there and a plus side
there. So what does this tell us? If this expression is
equal to 0, then either 2 calls X plus one equals 0
or cause X.
Minus one equals 0. This gives us a nice little equation that
says cause X is equal to. I'll take one away from both sides,
so that's minus one and divide both sides by two, so cause X is
equal to minus 1/2 or.
Cause X is equal to 1.
And so in order to solve this equation at the top, I've now
reduced it to solving these two much simpler equations at the
bottom. So I'll turn over the page now and take these two with
me. So cause X is equal to minus
1/2. All. Kohl's X is equal to wall.
Now again, when we start at solving this, we did not have a
range of values for Cos X. So let's say that again we're going
to work with this range of values. X is greater than or
equal to 0, but less than two Pi. What we need to do first is
sketch the graph of Cos X in that range. So the graph of Cos
X in that range looks like that.
This is π by 2.
This is pie.
Three Pi by two
and 2π. This is one on the Y axis and minus one on
the X axis. So what are our values? Well, let's take this
equation first cause X equals one. We go across at one.
We've got this value here.
X equals 0.
And this value here X equals 2π,
but. This here says X is strictly less than 2π, so
if I included it.
Be right across it out because it's not within the range of
values. Let's now have a look at this cause X equals minus
1/2. Well minus 1/2 is there. So let's go across and see
where this meets the graph there and there.
Right, this again, is connected with one of those very special
angles. If concept X had been equal to 1/2, then
X would be equal to 60 degrees or pie by three.
That's about there.
These curves are symmetric, so
this one. Instead of being pie by three from there is π
by three back from pie.
And so this tells us that X is equal to 2π by 3.
Or it's pie by three on from there, which gives us four Pi by
three. So there we've got our answers. Two of them there, and
one of them there.
So. Let's take another example.
Three cop squared X is equal to Cosec
X. Minus one.
So the identity that we want is the one that talks to
us about cot squared and Cosec squared. But which term should
we replace now? Let's recall the identity is one plus cot
squared. X is cosec squared X.
Well, as we saw in the last example, we want to arrive at a
quadratic that we can factorise it therefore makes no sense to
try and substitute for the Cosec 'cause to do that we have to get
square roots in it.
But if we substitute for the cot squared, we can do so much
better. Because we will just have a direct substitution that
will involve cosec squared and hopefully get a quadratic. So
instead of caught squared will replace it. By changing this
around, that tells us that caught squared X is equal to
cosine X squared X minus one, so that will be 3.
Cosec squared X minus one is equal to
cosec X minus one.
Multiply out the brackets.
3. Cosec squared X minus three. Don't forget when
you multiply out brackets, you must multiply everything inside
the bracket by what's outside, so we've got to have the three
times by the minus one there equals cosec X minus one.
Let's get everything on one side of the equation so it says
equals 0. Keep the square term to be the positive term.
That makes factorization easier, so free KOs X
squared X takeaway. This code set from each side.
And now I've minus three here, and I've minus one here. I want
to take the minus one over to that site, so I have to add 1 to
each side, so minus three plus One is minus 2.
Equals 0.
Now. Factorise
this equation. The variable is kosach, so we're
going to have three cosec in
here. And cosec X in there because that kinds by that will
give us that term there three cosec squared X. And now I've
got the minus two to deal with.
Well to itself is 2 times by one. If I put the two in here
then I'm going to multiply the two by the three, and that's
going to give me 6, which is a very big number in Association
with the cosack. I only want minus one cosec, so let's put
the two in there on the one in there. Now I've got to balance
the signs I want, minus two
here. So one of these has got to be negative and I see that
what's going to make the decision for me. Is this minus
cosec X? So I need the bigger bit in size to be negative.
Which seems to me that the three cosec X has got to go with us
minus one, so three cosec times by minus one is minus three
kosek. And then I've got +2 kocek there gives me the minus
the single cosec that I want.
Solving a quadratic two brackets multiplied together equal 0.
That means one of these brackets or the other one.
Has got to be
equal to. 0.
So. But this one I can take two away from each side
and divide by three. So we have cosec X is equal to minus two
over three or minus 2/3.
And here kosek.
X is equal to 1, so again we've reduced.
An equation like this to solving two much smaller, much
simpler equations. So taking these two over the
page cosec X is minus 2/3 or?
Cosec X is one.
Now what is cosec? Cosack is one over sine X.
So let's write
that down. If you just look at this one over sine X
equals 1. While that can only mean cynex itself is equal to 1,
and if we can turn this upside down, we can do the same this
side. So turning that back upside down, sign X is equal to
minus three over 2.
Now when we began this equation and we began to solve it, we
didn't state a range of values of X, so let's use the range
again that we've been using and that is X greater than or equal
to 0, but less than 2π.
Let's just sketch the graph of cynex over that range of values.
Graph of Cynex will look like that going from North.
Through pie by two.
Pie.
Three π by 2.
And 2π and it will range between plus one and minus one. So
again, if we look at this solution here, we can see
straightaway. We've got the one solution here at X equals π by
2. Let's have a look at this one. Sign X equals minus three
over two. Well, that's here.
And of course there's a problem.
This doesn't mean the graph anywhere. There are no values of
X that will produce minus three
over 2. Be'cause sign is contained between plus one
and minus one. That doesn't mean that we've done it
wrong. All it means is that there are no values of X
that come from this equation, and so the only
solutions are those that come from this equation
here, so this.
Is our answer and our only answer.
Will take one more.
Example, this one is cost squared X minus sign squared X
is equal to 0.
Now we've got an identity that says cost squared plus
sign squared is one, so I could choose to replace
either the cost squared all the sine squared.
But The reason I've chosen this example is that you can
do it another way.
So let's have a look at the
other way. This is cost squared minus sign squared. So
in algebra terms it's the difference of two squares. It's
A squared minus B squared and that has a standard
factorization of A-B A+B.
So this factorizes as cause X
minus sign X.
And cause X plus sign X equals 0.
So one of these two brackets, one or the other, is equal to
0, so cause X minus sign X equals 0 or.
Cause X plus sign X equals 0.
Let's develop this one first, cause X minus sign X equals 0
means that they must be the same cause X and sign X are the same.
Divide both sides by Cos X and we get sign over calls which is
tan. And so Tan X is equal to dividing both sides by Cos X Cos
divided by cause is just one.
Or Look at this one. Take kozaks away from each
side and we have sine X is equal to minus Cos X. Divide both
sides by cause X sign X over cause exusia gain tanks and
minus cause X divided by Cos X is minus one.
We've broken that down into two separate equations.
Let's have a look at how we solve them. Again, let's assume
that the range of values of X is not to pie.
And let's sketch the graph of tan in that range.
Sketches don't have to be accurate, just enough to give us
a picture of the symmetry of the curve to help us solve the
equation. Tan X is one we know that this is one of those
special angles that its 45 degrees in degrees. But since
we're working in radians, it's X equals π by 4. In other words,
were across here.
One there is π by 4, halfway between North and pie by
two. So again, this is.
Pie and the one we want is there that will be halfway between pie
and three Pi by two. So this is going to give us the one that is
halfway between pie and three Pi by two, five π by 4.
With this one, where minus one.
So we're down here, meet, sit there half way between pie by
two and Π, and so X there is going to be three π by 4.
And meets the curve again here halfway between three Pi by two
an 2π. So that's going to be 7 Pi by 4.
So by spotting that we could factorise this equation, we
didn't need to use the identity and we came up with
these solutions.
If you want, you can use the
identity. Notice what we get here are the all the pie by
force if you like.
Only old pie by falls pie by 4 five. PI43 Pi 4 Seven π by 4.
So let's have a look at this equation again, Cos squared X.
Minus sign squared X equals 0.
And we're going to use our basic trig identity to solve it, so we
know that sine squared X plus cost squared X is equal to 1.
I'm going to replace the sine squared here, so let's have a
look what is sine squared according to our identity sign
squared X? If we take away Cos squared from each side is 1
minus Cos squared X. So I'm going to take that, put it in
there. Cos squared X minus, then a bracket 1 minus
Cos squared X.
And I use the bracket because I'm taking away all of this
expression, not just a little bit of it, but all of it. So the
bracket show that now I need to remove the brackets.
Cos squared X minus one and minus minus gives me a plus.
Cos squared X equals 0.
So now I've cost squared plus cost squared. That's two of
them. Two cost squared X equals.
Wall by adding this one to both sides. Now let me divide by two.
Cost squared X is one over 2.
Now at this point I could say one over 2 and half. That's not
.5 and get my Calculator out because I'm going to have to
take a square root.
But I don't want to do that, why not? Well, half is a nice
number and I happen to know, for instance, that sign 30 is 1/2
cost, 60 is 1/2. I also know that sign of 45 and cause of 45
are both one over the square root of 2, so there are enough
indications here to suggest to me that there is a nice
relationship. Between the angle and the cosine that I'm going to
get when I take the square root, so I don't want to spoil that
relationship by messing it up with a lot of decimals through a
Calculator. So let's take that square root cause of X is one
over the square root of 2.
But I've taken a square root so that means not only must I have
plus, but I must have minus.
Now we didn't say at the beginning what was the range
of values of X, so let's take the range that we've been
working with, namely between North and 2π.
So a sketch of the graph just to help us see where we are.
Here. Pie by two.
Here pie.
Three Pi by two there and 2π there and our cosine function
ranges between one and minus
one. We know that the cosine of X is one over Route 2. We know
that's 45 degrees or radians π by 4, so we know that we're
here. Arc PY by 4 and of course right across there
and again halfway between these two, so this bit is telling us
X is π by 4 or its partner is here halfway between
three. Pi by two and 2π Seven Π by 4.
And then 4 - 1 over Route 2, we're going to be about there.
Go across to the graph.
Up to the X axis and again this by the symmetry of the curve
must be half way between pivi 2 and Π, and so that gives us
three π by 4.
And here again, halfway between pie and three Pi by two. So
again, that gives us five π by 4. So in solving this equation a
different way we've got the same set of answers. And again we can
recognize them, because these are the odd pie by force pie by
4, three π by 4, five π by 4, and Seven π by 4.
So whether we do this solution of the equation by.
Using this method.
Using the identity or the method that we had before where we
factorized it doesn't matter. And that's true in solving any
of these trig equations. The method that you use shouldn't
matter. It should always give the same set of answers.
But let's just recap where we started from these three basic
and fundamental identity's sign squared X plus cost squared X
is equal to 1.
1
plus. Hot squared X is equal
tool cosec squared X.
And 1 + 10 squared X is equal to sex squared X.
Those are our three fundamental trigonometric
identities, and they must be learned. They must be known,
and you must be able to recognize them whenever you
see them.