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

Use de Moivre’s theorem to express tan 5πœƒ in terms of power of tan πœƒ.

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Video Transcript

Use de Moivre’s theorem to express tan of five πœƒ in powers of tan πœƒ.

We’ll begin by recalling de Moivre’s theorem. It says quite simply that 𝑒 to the power of π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. Now, of course, we’re going to be working eventually with tan of five πœƒ. So we’ll use the trigonometric identity tan πœƒ equals sin πœƒ over cos πœƒ. But that will be a little bit later on down the line. Now, in fact, we’re working with tan five πœƒ. And we know that tan five πœƒ is equal to sin five πœƒ over cos five πœƒ. So we’re going to need to find a way to express de Moivre’s theorem in terms of five πœƒ. 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 π‘–πœƒ. And on the right, we have cos πœƒ plus 𝑖 sin πœƒ to the fifth power. De Moivre’s theorem says though that we can express 𝑒 to the power of five π‘–πœƒ as cos five πœƒ plus 𝑖 sin πœƒ. But on the right-hand side, we’re going to have to use the binomial theorem to distribute our parentheses. This says that π‘Ž plus 𝑏 to the 𝑛th power is equal to the sum from π‘˜ equal zero to 𝑛 of 𝑛 choose π‘˜ times π‘Ž to the power of 𝑛 minus π‘˜ times 𝑏 to the π‘˜th power. When 𝑛 is equal to five then, π‘Ž plus 𝑏 to the fifth power is π‘Ž to the fifth power plus five choose one times π‘Ž to the fourth power 𝑏 plus five choose two π‘Ž cubed 𝑏 squared and so one.

We replace π‘Ž with cos πœƒ and 𝑏 with 𝑖 sin πœƒ. And we find the first term in our binomial expansion is cos πœƒ to the fifth power. Five choose one is simply five. So our second term becomes five cos πœƒ to the fourth power times 𝑖 sin πœƒ. 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 πœƒ times 𝑖 sin πœƒ squared. We distribute the two over our parentheses, and this becomes 𝑖 squared sin squared πœƒ. But of course, we know that 𝑖 squared is equal to negative one. So our third term is in fact negative 10 cos cubed πœƒ sin squared πœƒ. Five choose three is also 10. So our fourth term is 10 cos squared πœƒ times 𝑖 sin πœƒ cubed.

We write 𝑖 sin πœƒ cubed as 𝑖 cubed sin cubed πœƒ. 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 πœƒ times 𝑖 sin πœƒ to the fourth power. And then this time, when we distribute our parentheses, we get 𝑖 to the fourth power time sin πœƒ 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 πœƒ sin πœƒ to the fourth power. And our final term becomes 𝑖 sin πœƒ to the fifth power.

We write this as 𝑖 to the fifth power times sin πœƒ 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 πœƒ 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œƒ plus 𝑖 sin of five πœƒ equals cos πœƒ to the fifth power minus 10 cos cubed πœƒ sin squared πœƒ plus five cos πœƒ sin πœƒ to the fourth power plus 𝑖 times five cos πœƒ to the fourth power sin πœƒ minus 10 cos squared πœƒ sin cubed πœƒ plus sin πœƒ 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 πœƒ. And on the right-hand side, it’s cos πœƒ to the fifth power minus 10 cos cubed πœƒ sin squared πœƒ plus five cos πœƒ sin πœƒ to the fourth power. And now, we have an equation for cos five πœƒ. Next, we equate the imaginary parts. Remember, these are the coefficient of 𝑖. So on the left-hand side, it’s sin five πœƒ. And on the right, it’s five cos πœƒ to the fourth power sin πœƒ minus 10 cos squared πœƒ sin cubed πœƒ plus sin πœƒ to the fifth power. And so we have an equation for sin five πœƒ.

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 πœƒ equals sin five πœƒ over cos five πœƒ. We’re going to divide the entire expression for sin five πœƒ by the entire expression for cos five πœƒ. And when we do, we obtain an expression for tan five πœƒ but in terms of cos πœƒ and sin πœƒ. Now, of course, in this question, we want it in terms of tan πœƒ only. So what we’re going to do is divide both the numerator and the denominator of this fraction by cos πœƒ 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 πœƒ to the fourth power sin πœƒ becomes five cos πœƒ to the fourth power sin πœƒ divided by cos πœƒ to the fifth power. We can then divide both the numerator and the denominator of this individual expression by cos πœƒ to the fourth power. And we’ll be left with five sin πœƒ on the numerator and cos πœƒ on the denominator. Now, of course sin πœƒ over cos πœƒ is tan πœƒ. So we end up with five tan πœƒ as the first term on our numerator. We repeat this with the second term in our numerator. We get 10 cos squared πœƒ sin cubed πœƒ over cos πœƒ to the fifth power.

We then divide through both the numerator and denominator of this single term by cos squared πœƒ. We end up with 10 sin cubed πœƒ over cos cubed πœƒ. But of course, sin cubed πœƒ over cos cubed πœƒ will be tan cubed πœƒ. So the second term in the numerator is negative 10 tan cubed πœƒ. Our third term in our numerator becomes sin πœƒ to the fifth power over cos πœƒ to the fifth power, which is tan πœƒ to the fifth power. We’ll repeat this for the denominator. Cos πœƒ to fifth power divided by cos πœƒ to the fifth power, that’s this first term here, is one. Our second term is negative 10 cos cubed πœƒ sin squared πœƒ over cos πœƒ 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 πœƒ over cos squared πœƒ, which is negative 10 tan squared πœƒ. Our third term on the denominator is five cos πœƒ sin πœƒ to the fourth power all over cos πœƒ to the fifth power. This time we divide through simply by cos πœƒ. And we can write sin πœƒ to the fourth power over cos πœƒ to the fourth power as tan πœƒ to the fourth Power. And so we have five tan πœƒ minus 10 tan cubed πœƒ plus tan πœƒ to the fifth power over one minus 10 tan squared πœƒ plus five tan πœƒ to the fourth power.

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