www.mathcentre.ac.uk/.../5.6Trigonometric%20Identities.mp4

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.