### Video Transcript

Use Eulerβs formula two drive a
formula for cos two π and sin two π in terms of sin π and cos π.

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

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

Now, these two formulae are useful
in their own right. But what we could actually do is
replace π with π and we get the double angle formulae. In the first one, we get cos two π
equals cos squared π minus sin squared π. And with our second identity, we
get sin two π equals two cos π sin π. 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 π to the two ππ
and then writing that as π to the ππ squared. This time, when we apply Eulerβs
formula, on the left-hand side, we get cos two π plus π sin two π. And on the right-hand side, we get
cos π plus π sin π all squared.

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

Now, youβve probably noticed there
isnβt 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. Itβs also useful to know that we
can incorporate the binomial theorem to derive multiple angle formulae in sine and
cosine.