Video: Applications of De Moivre’s Theorem to Trigonometric Identities

In this video, we will learn how to apply De Moivre’s theorem to discover trignometric identities.

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Video Transcript

In this lesson, we’ll learn how to use De Moivre’s theorem to prove trigonometric identities. There’s a very good chance you will have been using some of these identities for a significant period of time without really realising where they come from. And this lesson will give you some insight into this. We’ll begin by recapping how to apply De Moivre’s theorem and the binomial theorem for distributing parentheses before deriving a number of identities and looking at how to use them to solve equations involving trigonometric functions.

Remember De Moivre’s theorem says that for integer values of 𝑛, a complex number written in polar form 𝑟 cos 𝜃 plus 𝑖 sin 𝜃 to the power of 𝑛 is equal to 𝑟 to the power of 𝑛 times cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃. Similarly, the binomial theorem shows us how to evaluate 𝑎 plus 𝑏 all to the power of 𝑛. You might also know the binomial series as the expansion of one plus 𝑥 to the power of 𝑛 though we won’t be applying this method during this video. Let’s see how we can use these two concepts to derive the multiple angle formula.

1) Use De Moivre’s theorem to express sin five 𝜃 in terms of powers of sin 𝜃. 2) By considering the solutions of sin five 𝜃 equals zero, find an exact representation for sin squared of 𝜋 by five.

To answer part 1) of this question, we’ll actually use the inverse of De Moivre’s theorem to evaluate cos of five 𝜃 plus 𝑖 sin of five 𝜃. According to De Moivre’s theorem, this is equal to cos of 𝜃 plus 𝑖 sin 𝜃 all to the power of five. So let’s use the binomial theorem to distribute these parentheses. We’ll compare cos 𝜃 plus 𝑖 sin 𝜃 to the power of five to the binomial theorem. We see that 𝑎 is equal to cos 𝜃, 𝑏 is equal to 𝑖 sin 𝜃, and 𝑛 is equal to five.

The first term in the expansion of cos 𝜃 plus 𝑖 sin 𝜃 to the power of five is therefore cos 𝜃 to the power of five. The second term is five choose one times cos 𝜃 to the power of four times 𝑖 sin 𝜃. The third term is five choose two cos cubed 𝜃 plus 𝑖 sin 𝜃 squared. And we can evaluate the remaining terms as shown. Then, we recall that five choose one is five, five choose two is 10, five choose three is 10 again, and five choose four is also five. We also know that 𝑖 squared is going to be negative one, 𝑖 cubed is going to be negative 𝑖, 𝑖 to the power of four is going to be one, and 𝑖 to the power of five is going to be 𝑖. And our expressions simplifies as shown.

And since we initially said that cos five 𝜃 plus 𝑖 sin five 𝜃 was equal to cos 𝜃 plus 𝑖 sin 𝜃 to the power of five, we can equate this entire expression to cos five 𝜃 plus 𝑖 sin five 𝜃. Now, remember we’re trying to find an expression for sin five 𝜃. So we’re going to equate the imaginary parts on each side of our equation. On the left-hand side, that’s simply sin five 𝜃. And on the right, it’s five cos 𝜃 to the power of four times sin 𝜃 minus 10 cos squared 𝜃 sin cubed 𝜃 plus sin 𝜃 to the power of five.

We are going to need to clear a little bit of space here. And at this stage, we’re going to recall the fact that sin squared 𝜃 plus cos squared 𝜃 is always equal to one. Rearranging this, we see that cos squared 𝜃 is equal to one minus sin squared 𝜃. And we’ve done this because it will allow us to replace cos squared 𝜃 and cos 𝜃 to the power of four in our expression because we’re trying to write it in terms of powers of sine.

And when we do, we see that sin five 𝜃 is equal to five times one minus sin squared 𝜃 squared times sin 𝜃 minus 10 times one minus sin squared 𝜃 sin cubed 𝜃 plus sin 𝜃 to the power of five. And distributing these parentheses and simplifying, we see that sin five 𝜃 is equal to 16 sin 𝜃 to the power of five minus 20 sin cubed 𝜃 plus five sin 𝜃.

Now that we have sin five 𝜃 expressed in terms of powers of sin 𝜃, we can answer part 2) of this question. We need to consider the solutions of sin five 𝜃 equals zero. We know that sin 𝜃 is equal to zero has solutions at integer multiples of 𝜋. So this must mean that sin of five 𝜃 equals zero has solutions when 𝜃 is equal to 𝑛𝜋 over five or integer multiples of 𝜋 over five. Now comparing this to our equation in part one, we see that 16 sin 𝜃 to the power of five minus 20 sin cubed 𝜃 plus five sin 𝜃 equals zero must also have solutions when 𝜃 is equal to 𝑛𝜋 over five.

Let’s perform some manipulation to the expression on the left-hand side of our equation, remembering that our aim is to find the exact value of sin squared 𝜋 by five. We’ll begin by factoring sin 𝜃. And we see that sin 𝜃 times 16 sin 𝜃 to the power of four minus 20 sin squared 𝜃 plus five must be equal to zero. And let’s assume our value for 𝜃 is 𝜋 by five. In other words, 𝑛 is equal to one. Sin of 𝜋 by five is not equal to zero. So for the product of these two parentheses to be equal to zero, this must mean that 16 sin 𝜋 by five to the power of four minus 20 sin squared 𝜋 by five plus five is equal to zero. And notice this looks a little bit like a quadratic equation. We’ll set 𝑥 as sin squared 𝜋 by five.

And our quadratic equation is now 16𝑥 squared minus 20𝑥 plus five equals zero. And we can solve this equation using any method we like for solving quadratic such as completing the square or the quadratic formula. And when we do, we see that 𝑥 is equal to five plus or minus root five all divided by eight. And of course, we said that 𝑥 is equal to sin squared of 𝜋 by five. So our solutions for sin squared of 𝜋 by five are five plus or minus root five over eight. Now in fact, if we check these on our calculator, we see that sin squared 𝜋 by five is five minus root five divided by eight. And we’ve answered question 2). The exact representation for sin squared of 𝜋 by five is five minus root five over eight.

We’ve just seen how to use De Moivre’s theorem to evaluate sin of multiples of 𝜃. But we can also use the theorem to derive identities for sin 𝜃 to the power of 𝑛 and cos 𝜃 to the power of 𝑛. Let’s say we have a complex number 𝑧 which is simply cos 𝜃 plus 𝑖 sin 𝜃. We know that the reciprocal of 𝑧 is 𝑧 to the power of negative one. And we can use De Moivre’s theorem to evaluate this. It’s cos of negative 𝜃 plus 𝑖 sin of negative 𝜃.

Now, remember cos is an even function. So cos of negative 𝜃 is simply the same as cos of 𝜃. Sin however is an odd function. So sin of negative 𝜃 is the same as the negative sin of 𝜃. And we can therefore see that the reciprocal of 𝑧 can be written as cos 𝜃 minus 𝑖 sin 𝜃. So why is this useful? Well, it allows us to evaluate 𝑧 plus the reciprocal of 𝑧. 𝑧 plus the reciprocal of 𝑧 is simply two cos 𝜃. Similarly, it allows us to evaluate their difference. 𝑧 minus the reciprocal of 𝑧 is two 𝑖 sin 𝜃. And we can actually generalize this for higher powers of 𝑧.

Using De Moivre’s theorem and applying the odd and even identities for sine and cosine, we get these two equations. Rearranging these by dividing the first equation by two and the second by two 𝑖 for a complex number 𝑧 in exponential form, we see that cos of 𝑛𝜃 is equal to one-half 𝑒 to the 𝑖𝑛𝜃 plus 𝑒 to the negative 𝑖𝑛𝜃. And sin 𝑛𝜃 is equal to one over two 𝑖 multiplied by 𝑒 to the 𝑖𝑛𝜃 plus 𝑒 to the negative 𝑖𝑛𝜃. These formulae are incredibly powerful for deriving a number of trigonometric identities. And they should be committed to memory for easy recall. Let’s have a look at some examples for where these might be helpful.

Using De Moivre’s theorem, find the exact value of the integral of sin 𝜃 to the power of seven with respect to 𝜃 between the limits 𝜋 by two and zero.

We’ll begin by expressing sin 𝜃 to the power of seven in terms of multiple angles since those are quite straightforward to integrate. We’ll also use the fact that 𝑧 minus the reciprocal of 𝑧 is equal to two 𝑖 sin 𝜃. And since we’re interested in sin 𝜃 to the power of seven, we’re going to raise this entire equation to the power of seven. It’s 𝑧 minus one over 𝑧 all to the power of seven and that’s equal to two 𝑖 sin 𝜃 to the power of seven.

Two to the power of seven is 128 and 𝑖 to the power of seven is negative 𝑖. We divide both sides of this equation by negative 128𝑖. We’re going to evaluate negative one over 128𝑖 by multiplying both the numerator and the denominator by 𝑖. When we do, we get negative 𝑖 over 128𝑖 squared. But since 𝑖 squared is equal to negative one, this simplifies to 𝑖 over 128.

Our next step is to apply the binomial theorem to 𝑧 minus one over 𝑧 all to the power of seven. When we use the binomial theorem, we see that sin 𝜃 to the power of seven is as shown. We then recall the fact that 𝑧 to the power of 𝑛 minus one over 𝑧 to the power of 𝑛 is two 𝑖 sin 𝑛𝜃. And we can see that we can write this expression in terms of multiple angles by gathering powers of 𝑧.

This means that sin 𝜃 to the power of seven is equal to 𝑖 over 28 [128] times two 𝑖 sin seven 𝜃 minus seven times two 𝑖 sin five 𝜃 plus 21 times two 𝑖 sin three 𝜃 minus 35 times two 𝑖 sin 𝜃. And then fully simplified, we see that this is equal to one over 64 times 35 sin 𝜃 minus 21 sin three 𝜃 plus seven sin five 𝜃 minus sin seven 𝜃.

Let’s clear some space and replace sin 𝜃 to the power of seven in our integral with this expression. We then recall the fact that the integral of sin 𝑛𝜃 with respect to 𝜃 is negative one over 𝑛 times cos 𝑛𝜃 plus obviously that constant of integration. And this means our integral is negative 35 cos 𝜃 plus seven cos three 𝜃 minus seven-fifths times cos of five 𝜃 plus one-seventh of cos of seven 𝜃. And since we’ll be evaluating this between the limits of 𝜋 by two and zero, we don’t need the constant of integration. This becomes one sixty-fourth times 35 minus seven plus seven-fifths minus a seventh which is sixteen thirty-fifths.

Notice how this process actually made what was quite a tricky integral really quite simple to evaluate. And this process isn’t limited to just powers of sine or cosine. We can also use it to find expressions for the products of powers of these functions. But they also work with the tangent function. By recalling the fact that tan 𝜃 is equal to sin 𝜃 divided by cos 𝜃, we can express tan of some integer multiple of 𝜃 in terms of powers of tan. Let’s see what this looks like.

Express tan six 𝜃 in terms of powers of tan 𝜃.

We begin by recalling the fact that tan 𝜃 is equal to sin 𝜃 divided by cos 𝜃. And this in turn means that tan six 𝜃 equals sin six 𝜃 divided by cos six 𝜃. So we’ll evaluate sin six 𝜃 and cos six 𝜃 in terms of powers of sine and cosine. De Moivre’s theorem says that cos six 𝜃 plus 𝑖 sin six 𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃 all to the power of six. We distribute this parenthesis by using the binomial theorem and simplify by evaluating powers of 𝑖. And we see that cos 𝜃 plus 𝑖 sin 𝜃 to the power of six is as shown. And of course, we said that cos six 𝜃 plus 𝑖 sin six 𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃 to the power of six. So we can equate this expansion to cos six 𝜃 plus 𝑖 sin six 𝜃.

And now we see that we can equate the real and the imaginary parts of the equation. The real part of the left-hand side is cos six 𝜃. On the right-hand side, we got cos 𝜃 to the power of six, negative 15 cos 𝜃 to the power of four sin squared 𝜃, 15 cos squared 𝜃 times sin 𝜃 to the power of four, and negative sin 𝜃 to the power of six. And then equating the imaginary parts, on the left-hand side, we have sin six 𝜃. On the right-hand side, we have six cos 𝜃 to the power of five times sin 𝜃, negative 20 cos cubed 𝜃 sin cubed 𝜃, and six cos 𝜃 times sin 𝜃 to the power of five.

And we’re now ready to evaluate tan of six 𝜃. It’s six cos 𝜃 to the power of five times sin 𝜃 minus 20 cos cubed 𝜃 sin cubed 𝜃 plus six cos 𝜃 times sin 𝜃 to the power of five all over cos 𝜃 to the power of six minus 15 cos 𝜃 to the power of four times sin squared 𝜃 plus 15 cos squared 𝜃 sin 𝜃 to the power of four minus sin 𝜃 to the power of six. To express this in terms of tan, we’re going to divide everything by cos 𝜃 to the power of six.

On the numerator, six cos 𝜃 to the power of five times sin 𝜃 divided by cos 𝜃 to the power of six is simply six tan 𝜃. We have negative 20 tan cubed 𝜃 and six times tan 𝜃 to the power of five. On the denominator, we have one minus 15 tan squared 𝜃 plus 15 tan 𝜃 to the power of four minus tan 𝜃 to the power of six. And we’ve successfully expressed tan of six 𝜃 in terms of powers of tan 𝜃.

In this video, we’ve seen that we can use De Moivre’s theorem and the binomial theorem to derive multiple angle formulae for different sine, cosine, and tangent functions. We’ve also seen that for a complex number 𝑧 equals cos 𝜃 plus 𝑖 sin 𝜃, 𝑧 plus the reciprocal of 𝑧 is two cos 𝜃 and 𝑧 minus the reciprocal of 𝑧 is two 𝑖 sin 𝜃. And we’ve also extended this idea into 𝑧 to the power of 𝑛.

We’ve used these equations to find expressions for the powers of sine and cosine. And we said that we can even find it for their products. We’ve even seen that we can use these techniques for deriving trigonometric identities to simplify more complicated integrals.

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