### Video Transcript

In this video, we’ll explore the
derivation of a number of trigonometric identities using Euler’s formula. There’s a very good chance you will
have already worked with some of these identities extensively but perhaps aren’t
quite sure where they come from. So in this lesson, we’ll see how
Euler’s formula links to the double angle formulae, additional multiple angle
formulae, and the product to sum formulae.

We begin by recalling Euler’s
formula, sometimes called Euler’s relation. This states that for a real number
𝜃, 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃. Here, 𝑖 is the imaginary unit
denoted as the solution to the equation 𝑥 squared equals negative one and 𝜃 must
be a real number given in radians. As we can see, the formula provides
a powerful connection between complex analysis and trigonometry. But it also has many applications
in physics, engineering, and quantum mechanics. In our first example, we’ll see how
to derive a well-used trigonometric identity by considering the properties of the
exponential function and Euler’s formula.

1) Use Euler’s formula to express
𝑒 to the negative 𝑖𝜃 in terms of sine and cosine. 2) Given that 𝑒 to the 𝑖𝜃 times
𝑒 to the negative 𝑖𝜃 equals one, what trigonometric identity can be derived by
expanding the exponential in terms of trigonometric functions?

For part one, we’ll begin by
rewriting 𝑒 to the negative 𝑖𝜃. It’s the same as 𝑒 to the 𝑖 times
negative 𝜃. We can now apply Euler’s
formula. Since Euler’s formula says that 𝑒
to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can see that 𝑒 to the 𝑖 negative
𝜃 is equal to cos of negative 𝜃 plus 𝑖 sin of negative 𝜃. And then, we recall the properties
of the cosine and sine functions. Cos is an even function. So cos of negative 𝜃 is equal to
cos of 𝜃. Sin, however, is an odd function. So sin of negative 𝜃 is the same
as negative sin 𝜃. And we can therefore rewrite our
expression. And we see that 𝑒 to the negative
𝑖𝜃 is the same as cos 𝜃 minus 𝑖 sin 𝜃.

Now, let’s consider part two of
this question. We’re going to use the answer we
got from part one. When we do, we can see that 𝑒 to
the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 plus 𝑖 sin 𝜃 times
cos 𝜃 minus 𝑖 sin 𝜃. Let’s distribute these parentheses, perhaps noticing that this is an expression factored using the difference of two squares. cos 𝜃 times cos 𝜃 is cos squared 𝜃. cos 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 cos 𝜃 sin 𝜃. We then get plus 𝑖 sin 𝜃 cos 𝜃
and 𝑖 sin 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 squared sin squared 𝜃. We see then negative 𝑖 cos 𝜃 sin
𝜃 plus 𝑖 cos 𝜃 sin 𝜃 is zero. And of course, we know that 𝑖
squared is equal to negative one.

So we can simplify this
somewhat. And we see that 𝑒 to the 𝑖𝜃
times 𝑒 to the negative 𝑖𝜃 is cos squared 𝜃 plus sin squared 𝜃. We were told however that 𝑒 to the
𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 was equal to one. So you can see that we’ve derived
the formula sin squared 𝜃 plus cos squared 𝜃 equals one. This is a fairly succinct
derivation of the trigonometric identity sin squared 𝜃 plus cos squared 𝜃 equals
one.

We can perform a similar process to
help us derive the double angle formulae. Let’s see what that might look
like.

Use Euler’s formula two drive a
formula for cos two 𝜃 and sin two 𝜃 in terms of sin 𝜃 and cos 𝜃.

There are actually two methods we
could use to derive the formulae for cos two 𝜃 and sin two 𝜃. The first is to consider this
expression; it’s 𝑒 to the 𝑖𝜃 plus 𝜙. We know that this must be the same
as 𝑒 to the 𝑖𝜃 times 𝑒 to the 𝑖𝜙. We’re going to apply Euler’s
formula to both parts of this equation. On the left-hand side, we can see
that 𝑒 to the 𝑖𝜃 plus 𝜙 is equal to cos 𝜃 plus 𝜙 plus 𝑖 sin of 𝜃 plus
𝜙. And on the right, we have cos 𝜃
plus 𝑖 sin 𝜃 times cos 𝜙 plus 𝑖 sin 𝜙. We’re going to distribute the
parentheses on the right-hand side. And when we do, we get the given
expression. Remember though 𝑖 squared is equal
to negative one. And we can simplify and we get cos
𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙 plus 𝑖 cos 𝜃 sin 𝜙 plus 𝑖 cos 𝜙 sin 𝜃.

Our next step is to equate the real
and imaginary parts of the equation. On the left-hand side, the real
part is cos 𝜃 plus 𝜙 and on the right is cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙. And so, we see that cos 𝜃 plus 𝜙
is equal to cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙. Next, we equate the imaginary
parts. On the left-hand side, we have sin
𝜃 plus 𝜙. And on the right, we have cos 𝜃
sin 𝜙 plus cos 𝜙 sin 𝜃. And we can see then that sin 𝜃
plus 𝜙 is equal to cos 𝜃 sin 𝜙 plus cos 𝜙 sin 𝜃.

Now, these two formulae are useful
in their own right. But what we could actually do is
replace 𝜙 with 𝜃 and we get the double angle formulae. In the first one, we get cos two 𝜃
equals cos squared 𝜃 minus sin squared 𝜃. And with our second identity, we
get sin two 𝜃 equals two cos 𝜃 sin 𝜃. And there is an alternative
approach we could have used. This time, we could have gone
straight to the double angle formulae by choosing the expression 𝑒 to the two 𝑖𝜃
and then writing that as 𝑒 to the 𝑖𝜃 squared. This time, when we apply Euler’s
formula, on the left-hand side, we get cos two 𝜃 plus 𝑖 sin two 𝜃. And on the right-hand side, we get
cos 𝜃 plus 𝑖 sin 𝜃 all squared.

Then, distributing these
parentheses, we see that the right-hand side becomes cos squared 𝜃 plus two 𝑖 cos
𝜃 sin 𝜃 plus 𝑖 squared sin squared 𝜃. And once again, 𝑖 squared is equal
to negative one. So we can rewrite the right-hand
side as cos squared 𝜃 minus sin squared 𝜃 plus two 𝑖 cos 𝜃 sin 𝜃. This time when we equate the real
parts, we see that cos two 𝜃 equals cos squared 𝜃 minus sin squared 𝜃. And when we equate the imaginary
parts, we see that sin two 𝜃 is equal to two cos 𝜃 sin 𝜃.

Now, you’ve probably noticed there
isn’t a huge amount of difference in these two methods. The latter is slightly more
succinct. However, the former has the benefit
of deriving those extra identities for cosine and sine. It’s also useful to know that we
can incorporate the binomial theorem to derive multiple angle formulae in sine and
cosine.

The binomial theorem says that for
integer values of 𝑛, we can write 𝑎 plus 𝑏 to the power of 𝑛 as 𝑎 to the power
of 𝑛 plus 𝑛 choose one 𝑎 to the power of 𝑛 minus one 𝑏. And we continue this pattern with
descending powers of 𝑎 and ascending powers of 𝑏 all the way through to 𝑏 to the
power of 𝑛. Our next example is going to use
the binomial theorem to help us evaluate multiple angles in terms of powers of
trigonometric functions.

1) Use Euler’s formula to derive a
formula for cos of four 𝜃 in terms of cos 𝜃. 2) Use Euler’s formula to drive a
formula for sin of four 𝜃 in terms of cos 𝜃 and sin 𝜃.

For part one, we’ll use the
properties of the exponential function. And we’ll write 𝑒 to the four 𝑖𝜃
as 𝑒 to the 𝑖𝜃 to the power of four. And now, we can use Euler’s
formula. And we write the left-hand side as
cos four 𝜃 plus 𝑖 sin four 𝜃. And on the right-hand side, we can
say that this is equal to cos 𝜃 plus 𝑖 sin 𝜃 all to the power of four. Now, we’re going to apply the
binomial theorem to distribute cos 𝜃 plus 𝑖 sin 𝜃 to the power of four.

In our equation, 𝑎 is equal to cos
of 𝜃, 𝑏 is equal to 𝑖 sin of 𝜃, and 𝑛 is the power; it’s four. And this means we can say that cos
𝜃 plus 𝑖 sin 𝜃 to the power of four is the same as cos 𝜃 to the power of four
plus four choose one cos cubed 𝜃 times 𝑖 sin 𝜃 and so on. We know that four choose one is
four, four choose two is six, and four choose three is also four. We also know that 𝑖 squared is
negative one, 𝑖 cubed is negative 𝑖, and 𝑖 to the power of four is one. And we can further rewrite our
equation as shown.

Now, we’re going to equate the real
parts of this equation. And that will give us a formula for
cos of four 𝜃 in terms of cos 𝜃 and sin 𝜃. Let’s clear some space. The real part on the left-hand side
is cos four 𝜃. And then on the right-hand side, we
have cos 𝜃 to the power of four. We’ve got negative cos squared 𝜃
sin squared 𝜃. And we’ve got sin 𝜃 to the power
of four. So we equate these. But we’re not quite finished. We were asked to derive a formula
for cos four 𝜃 in terms of cos 𝜃 only.

So here, we use the identity cos
squared 𝜃 plus sin squared 𝜃 is equal to one. And we rearrange this. And we say that well, that means
that sin squared 𝜃 must be equal to one minus cos squared 𝜃. And we can rewrite this as cos 𝜃
to the power of four plus six cos squared 𝜃 times one minus cos squared 𝜃 plus one
minus cos squared 𝜃 squared. We distribute the parentheses. And our final step is to collect like terms. And we see that we’ve derived the formula for cos of four 𝜃 in terms of cos 𝜃. cos four 𝜃 is equal to eight cos 𝜃 to the power of four minus eight cos squared 𝜃 plus one.

For part two, we can repeat this
process equating the imaginary parts. They are sin of four 𝜃 on the
left. And then on the right, we have four
cos cubed 𝜃 sin 𝜃, negative four cos 𝜃 sin cubed 𝜃. And we see this sin four 𝜃 must be
equal to four cos cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃. And we could — if we so wish —
factor four cos 𝜃 sin 𝜃. And we’ll be left with four cos 𝜃
sin 𝜃 times cos squared 𝜃 minus sin squared 𝜃. And last, we’ve been asked to
derive a formula for sin four 𝜃 in terms of cos 𝜃 and sin 𝜃. You might now see a link between
sin four 𝜃 and the double angle formulae.

Now, an interesting application of
Euler’s formula is that we can use it to derive an expression for sin 𝜃 and cos 𝜃
in terms of 𝑒 to the 𝑖𝜃. We’ve already seen that we can
write 𝑒 to the negative 𝑖𝜃 as cos 𝜃 minus 𝑖 sin 𝜃. And since 𝑒 to the 𝑖𝜃 is equal
to cos 𝜃 plus 𝑖 sin 𝜃, we can say that the sum of 𝑒 to the 𝑖𝜃 and 𝑒 to the
negative 𝑖𝜃 is cos 𝜃 plus 𝑖 sin 𝜃 plus cos 𝜃 minus 𝑖 sin 𝜃. Well, this expression on the
right-hand side simplifies to two cos 𝜃. And we can make cos 𝜃 the subject
by dividing through by two. And we see we have an expression
for cos 𝜃 in terms of powers of 𝑒 to the 𝑖𝜃. It’s a half 𝑒 to the 𝑖𝜃 plus 𝑒
to the negative 𝑖𝜃.

Similarly, we can find the
difference. And we get cos 𝜃 plus 𝑖 sin 𝜃
minus cos 𝜃 minus 𝑖 sin 𝜃. This simplifies to two 𝑖 sin
𝜃. This time we divide through by two
𝑖. And we can see that sin 𝜃 is equal
to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. At these two formulae have plenty
of applications in their own right. But for the purposes of this video,
we’ll look at one final example. And we’ll see how they can be used
to derive further trigonometric identities.

Use Euler’s formula to express sin
cubed 𝜃 cos squared 𝜃 in the form 𝑎 sin 𝜃 plus 𝑏 sin three 𝜃 plus 𝑐 sin five
𝜃, where 𝑎, 𝑏, and 𝑐 are constants to be found. Hence, find the solutions of sin
five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal
to zero and less than 𝜋. Give your answer in exact form.

We begin by recalling the fact that
sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative
𝑖𝜃. And cos 𝜃 is equal to a half times
𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃. This means we can find the product
of sin cubed 𝜃 and cos squared 𝜃. We can write it as one over two 𝑖
times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃 cubed times a half times 𝑒 to
the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. One over two 𝑖 cubed is negative
one over eight 𝑖. And a half squared is
one-quarter. So we can rewrite our expression a
little bit further. We find the product of negative one
over eight 𝑖 and a quarter. And we get negative one over
32𝑖. And we can rewrite the rest of our
expression as shown.

We’re now going to use the binomial
theorem to expand each of the sets of parentheses. The first part becomes 𝑒 to three
𝑖𝜃 plus three choose one 𝑒 to the two 𝑖𝜃 times negative 𝑒 to the negative 𝑖𝜃
and so on. And this simplifies to 𝑒 to the
three 𝑖𝜃 minus three 𝑒 to the 𝑖𝜃 plus three 𝑒 to the negative 𝑖𝜃 minus 𝑒 to
the negative three 𝑖𝜃. Let’s repeat this process for 𝑒 to
the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. When we do, we get 𝑒 to the two
𝑖𝜃 plus two 𝑒 to the zero which is just two plus 𝑒 to the negative two 𝑖𝜃.

We’re going to need to find the
product of these two expressions. We’ll need to do that really
carefully. We’ll need to ensure that each term
in the first expression is multiplied by each term in the second expression. And we can write sin cubed 𝜃 cos
squared 𝜃 as shown. Now, there’s quite a lot going on
here. So you might wish to pause the
video and double-check your answer against mine. We’re going to gather the
corresponding powers of 𝑒 together.

We’ll gather 𝑒 to the five 𝑖𝜃
and 𝑒 to the negative five 𝑖𝜃. We’ll collect 𝑒 to the plus and
minus three 𝑖𝜃. And we’ll gather 𝑒 the 𝑖𝜃 and 𝑒
to the negative 𝑖𝜃. Let’s neaten things up
somewhat. We end up with negative one over
32𝑖 times 𝑒 to the five 𝑖𝜃 minus 𝑒 to the negative five 𝑖𝜃 minus 𝑒 to the
three 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃 minus two times 𝑒 to the 𝑖𝜃 minus
𝑒 to the negative 𝑖𝜃. And now, you might be able to spot
why we chose to do this. We can now go back to the given
formulae. Let’s clear some space for the next
step.

We kind of unfactorize a
little. And we can rewrite sin cubed 𝜃 cos
squared 𝜃 as shown. And we can therefore replace 𝑒 to
the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 with sin 𝜃 and so on. And we can see that sin cubed 𝜃
cos squared 𝜃 is equal to a 16th times two sin 𝜃 plus sin three 𝜃 minus sin five
𝜃. Since 𝑎, 𝑏, and 𝑐 are constants
to be found, we can say that 𝑎, the coefficient of sin 𝜃, is one-eighth. 𝑏, the coefficient of sin three
𝜃, is a 16th. And 𝑐, the coefficient sin five
𝜃, is negative one 16th.

Let’s now consider part two of this
question. We begin by using our answer to
part one and multiplying both sides by 16. We then subtract two sin 𝜃 from
both sides and multiply through by negative one. And we can now see that we’ve got
an equation in sin five 𝜃 minus sin three 𝜃. We’re told that sin five 𝜃 minus
sin three 𝜃 is equal to zero. So we let two sin 𝜃 minus 16 sin
cubed 𝜃 cos squared 𝜃 be equal to zero. And then, we factor by two sin
𝜃. Since the product of these two
terms is equal to zero, this means that either of these terms must be equal to
zero. So either two sin 𝜃 is equal to
zero and dividing by two, we could see that sin 𝜃 is equal to zero or one minus
eight sin squared 𝜃 cos squared 𝜃 is equal to zero. Given the interval 𝜃 is greater
than or equal to zero and less than 𝜋, we can see that one of our solutions is when
𝜃 is equal to zero.

We’re going to rewrite our other
equations somewhat. We know that sin two 𝜃 is equal to
two sin 𝜃 cos 𝜃. Squaring this, we get sin squared
two 𝜃 equals four sin squared 𝜃 cos squared 𝜃. And that’s in turn means that our
equation is one minus two sin squared two 𝜃 equals zero. Rearranging to make sin two 𝜃 the
subject, we see that sin two 𝜃 is equal to plus or minus one over root two. Starting with the positive square
root for 𝜃 in the interval given, we know that sin two 𝜃 is equal to one over root
two when 𝜃 is equal to 𝜋 by eight or three 𝜋 by eight. Similarly, we can solve for the
negative square root. And we get five 𝜋 by eight and
seven 𝜋 by eight. And there are therefore five
solutions to the equation sin five 𝜃 minus sin three 𝜃 equals zero in the interval
𝜃 is greater than or equal to zero and less than 𝜋. They are zero 𝜋 by eight, three 𝜋
by eight, five 𝜋 by eight, and seven 𝜋 by eight.

In this video, we’ve seen that we
can use Euler’s formula in conjunction with the properties of the exponential
functions. And we can derive many
trigonometric identities such as the Pythagorean identity and multiple angle
formulae. We also saw the we can use the
theorem to express sine and cosine in terms of the complex exponential function as
shown. We’ve also seen that we can use the
identities derived from Euler’s formula to help us simplify expressions. And these in turn can help us to
solve trigonometric equations.