# Video: Using De Moivre’s Theorem to Derive Trigonometric Identities

Use de Moivre’s theorem to express tan 5𝜃 in terms of power of tan 𝜃.

06:04

### Video Transcript

Use de Moivre’s theorem to express tan of five 𝜃 in powers of tan 𝜃.

We’ll begin by recalling de Moivre’s theorem. It says quite simply that 𝑒 to the power of 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃. Now, of course, we’re going to be working eventually with tan of five 𝜃. So we’ll use the trigonometric identity tan 𝜃 equals sin 𝜃 over cos 𝜃. But that will be a little bit later on down the line. Now, in fact, we’re working with tan five 𝜃. And we know that tan five 𝜃 is equal to sin five 𝜃 over cos five 𝜃. So we’re going to need to find a way to express de Moivre’s theorem in terms of five 𝜃. Well, we’re going to begin by raising both sides of de Moivre’s theorem as a power of five.

When we do on the left-hand side, we get 𝑒 to the five 𝑖𝜃. And on the right, we have cos 𝜃 plus 𝑖 sin 𝜃 to the fifth power. De Moivre’s theorem says though that we can express 𝑒 to the power of five 𝑖𝜃 as cos five 𝜃 plus 𝑖 sin 𝜃. But on the right-hand side, we’re going to have to use the binomial theorem to distribute our parentheses. This says that 𝑎 plus 𝑏 to the 𝑛th power is equal to the sum from 𝑘 equal zero to 𝑛 of 𝑛 choose 𝑘 times 𝑎 to the power of 𝑛 minus 𝑘 times 𝑏 to the 𝑘th power. When 𝑛 is equal to five then, 𝑎 plus 𝑏 to the fifth power is 𝑎 to the fifth power plus five choose one times 𝑎 to the fourth power 𝑏 plus five choose two 𝑎 cubed 𝑏 squared and so one.

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

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

We write this as 𝑖 to the fifth power times sin 𝜃 to the fifth power. And then 𝑖 to the fifth power is equal to 𝑖 to the fourth power times 𝑖. Well, that’s one times 𝑖, which is simply 𝑖. And so our sixth and final term is 𝑖 sin 𝜃 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 𝜃 plus 𝑖 sin of five 𝜃 equals cos 𝜃 to the fifth power minus 10 cos cubed 𝜃 sin squared 𝜃 plus five cos 𝜃 sin 𝜃 to the fourth power plus 𝑖 times five cos 𝜃 to the fourth power sin 𝜃 minus 10 cos squared 𝜃 sin cubed 𝜃 plus sin 𝜃 to the fifth power. And we’re now ready to equate the real and imaginary parts.

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

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

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

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

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