### Video Transcript

Use de Moivreβs theorem to express
tan of five π in powers of tan π.

Weβll begin by recalling de
Moivreβs theorem. It says quite simply that π to the
power of ππ is equal to cos π plus π sin π. Now, of course, weβre going to be
working eventually with tan of five π. So weβll use the trigonometric
identity tan π equals sin π over cos π. But that will be a little bit later
on down the line. Now, in fact, weβre working with
tan five π. And we know that tan five π is
equal to sin five π over cos five π. So weβre going to need to find a
way to express de Moivreβs theorem in terms of five π. Well, weβre going to begin by
raising both sides of de Moivreβs theorem as a power of five.

When we do on the left-hand side,
we get π to the five ππ. And on the right, we have cos π
plus π sin π to the fifth power. De Moivreβs theorem says though
that we can express π to the power of five ππ as cos five π plus π sin π. But on the right-hand side, weβre
going to have to use the binomial theorem to distribute our parentheses. This says that π plus π to the
πth power is equal to the sum from π equal zero to π of π choose π times π to
the power of π minus π times π to the πth power. When π is equal to five then, π
plus π to the fifth power is π to the fifth power plus five choose one times π to
the fourth power π plus five choose two π cubed π squared and so one.

We replace π with cos π and π
with π sin π. And we find the first term in our
binomial expansion is cos π to the fifth power. Five choose one is simply five. So our second term becomes five cos
π to the fourth power times π sin π. And actually, convention dictates
that we move the π to the front of this term. Our third term is five choose two
which is 10, so 10 cos cubed π times π sin π squared. We distribute the two over our
parentheses, and this becomes π squared sin squared π. But of course, we know that π
squared is equal to negative one. So our third term is in fact
negative 10 cos cubed π sin squared π. Five choose three is also 10. So our fourth term is 10 cos
squared π times π sin π cubed.

We write π sin π cubed as π
cubed sin cubed π. And then if we consider π cubed to
be equal to π squared times π, we see that itβs also equal to negative π. And so instead, we write our fourth
term as shown. We then have five cos π times π
sin π to the fourth power. And then this time, when we
distribute our parentheses, we get π to the fourth power time sin π to the fourth
power. π to the fourth power is π
squared times π squared, which is negative one times negative one which is simply
one. And so our fifth term becomes five
cos π sin π to the fourth power. And our final term becomes π sin
π to the fifth power.

We write this as π to the fifth
power times sin π to the fifth power. And then π to the fifth power is
equal to π to the fourth power times π. Well, thatβs one times π, which is
simply π. And so our sixth and final term is
π sin π to the fifth power. We can neat this up a little bit by
gathering the real and imaginary parts. And we see that our equation
becomes cos of five π plus π sin of five π equals cos π to the fifth power minus
10 cos cubed π sin squared π plus five cos π sin π to the fourth power plus π
times five cos π to the fourth power sin π minus 10 cos squared π sin cubed π
plus sin π to the fifth power. And weβre now ready to equate the
real and imaginary parts.

On the left-hand side, the real
part of the expression is cos five π. And on the right-hand side, itβs
cos π to the fifth power minus 10 cos cubed π sin squared π plus five cos π sin
π to the fourth power. And now, we have an equation for
cos five π. Next, we equate the imaginary
parts. Remember, these are the coefficient
of π. So on the left-hand side, itβs sin
five π. And on the right, itβs five cos π
to the fourth power sin π minus 10 cos squared π sin cubed π plus sin π to the
fifth power. And so we have an equation for sin
five π.

Weβre going to clear some space for
the next step. At this stage, weβre almost
there. We now need to go back to our
earlier identity. That is, tan five π equals sin
five π over cos five π. Weβre going to divide the entire
expression for sin five π by the entire expression for cos five π. And when we do, we obtain an
expression for tan five π but in terms of cos π and sin π. Now, of course, in this question,
we want it in terms of tan π only. So what weβre going to do is divide
both the numerator and the denominator of this fraction by cos π to the fifth
power. And it will become evident why
weβre going to do that in a moment.

The first term in our numerator
five cos π to the fourth power sin π becomes five cos π to the fourth power sin
π divided by cos π to the fifth power. We can then divide both the
numerator and the denominator of this individual expression by cos π to the fourth
power. And weβll be left with five sin π
on the numerator and cos π on the denominator. Now, of course sin π over cos π
is tan π. So we end up with five tan π as
the first term on our numerator. We repeat this with the second term
in our numerator. We get 10 cos squared π sin cubed
π over cos π to the fifth power.

We then divide through both the
numerator and denominator of this single term by cos squared π. We end up with 10 sin cubed π over
cos cubed π. But of course, sin cubed π over
cos cubed π will be tan cubed π. So the second term in the numerator
is negative 10 tan cubed π. Our third term in our numerator
becomes sin π to the fifth power over cos π to the fifth power, which is tan π to
the fifth power. Weβll repeat this for the
denominator. Cos π to fifth power divided by
cos π to the fifth power, thatβs this first term here, is one. Our second term is negative 10 cos
cubed π sin squared π over cos π to the fifth power.

This time we divide both the
numerator and denominator by cos cubed. So weβre left with negative 10 sin
squared π over cos squared π, which is negative 10 tan squared π. Our third term on the denominator
is five cos π sin π to the fourth power all over cos π to the fifth power. This time we divide through simply
by cos π. And we can write sin π to the
fourth power over cos π to the fourth power as tan π to the fourth Power. And so we have five tan π minus 10
tan cubed π plus tan π to the fifth power over one minus 10 tan squared π plus
five tan π to the fourth power.