### Video Transcript

In this video, we’ll explore the derivation of a number of trigonometric identities using Euler’s formula. There’s a very good chance you will have already worked with some of these identities extensively but perhaps aren’t quite sure where they come from. So in this lesson, we’ll see how Euler’s formula links to the double angle formulae, additional multiple angle formulae, and the product to sum formulae.

We begin by recalling Euler’s formula, sometimes called Euler’s relation. This states that for a real number 𝜃, 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃. Here, 𝑖 is the imaginary unit denoted as the solution to the equation 𝑥 squared equals negative one and 𝜃 must be a real number given in radians. As we can see, the formula provides a powerful connection between complex analysis and trigonometry. But it also has many applications in physics, engineering, and quantum mechanics. In our first example, we’ll see how to derive a well-used trigonometric identity by considering the properties of the exponential function and Euler’s formula.

1) Use Euler’s formula to express 𝑒 to the negative 𝑖𝜃 in terms of sine and cosine. 2) Given that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 equals one, what trigonometric identity can be derived by expanding the exponential in terms of trigonometric functions?

For part one, we’ll begin by rewriting 𝑒 to the negative 𝑖𝜃. It’s the same as 𝑒 to the 𝑖 times negative 𝜃. We can now apply Euler’s formula. Since Euler’s formula says that 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can see that 𝑒 to the 𝑖 negative 𝜃 is equal to cos of negative 𝜃 plus 𝑖 sin of negative 𝜃. And then, we recall the properties of the cosine and sine functions. cos is an even function. So cos of negative 𝜃 is equal to cos of 𝜃. sin, however, is an odd function. So sin of negative 𝜃 is the same as negative sin 𝜃. And we can therefore rewrite our expression. And we see that 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 minus 𝑖 sin 𝜃.

Now, let’s consider part two of this question. We’re going to use the answer we got from part one. When we do, we can see that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 plus 𝑖 sin 𝜃 times cos 𝜃 minus 𝑖 sin 𝜃. Let’s distribute these parentheses, perhaps noticing that this is an expression factored using the difference of two squares. cos 𝜃 times cos 𝜃 is cos squared 𝜃. cos 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 cos 𝜃 sin 𝜃. We then get plus 𝑖 sin 𝜃 cos 𝜃 and 𝑖 sin 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 squared sin squared 𝜃. We see then negative 𝑖 cos 𝜃 sin 𝜃 plus 𝑖 cos 𝜃 sin 𝜃 is zero. And of course, we know that 𝑖 squared is equal to negative one.

So we can simplify this somewhat. And we see that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is cos squared 𝜃 plus sin squared 𝜃. We were told however that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 was equal to one. So you can see that we’ve derived the formula sin squared 𝜃 plus cos squared 𝜃 equals one. This is a fairly succinct derivation of the trigonometric identity sin squared 𝜃 plus cos squared 𝜃 equals one. We can perform a similar process to help us derive the double angle formulae. Let’s see what that might look like.

Use Euler’s formula two drive a formula for cos two 𝜃 and sin two 𝜃 in terms of sin 𝜃 and cos 𝜃.

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

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

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

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

Now, you’ve probably noticed there isn’t 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. It’s also useful to know that we can incorporate the binomial theorem to derive multiple angle formulae in sine and cosine.

The binomial theorem says that for integer values of 𝑛, we can write 𝑎 plus 𝑏 to the power of 𝑛 as 𝑎 to the power of 𝑛 plus 𝑛 choose one 𝑎 to the power of 𝑛 minus one 𝑏. And we continue this pattern with descending powers of 𝑎 and ascending powers of 𝑏 all the way through to 𝑏 to the power of 𝑛. Our next example is going to use the binomial theorem to help us evaluate multiple angles in terms of powers of trigonometric functions.

1) Use Euler’s formula to derive a formula for cos of four 𝜃 in terms of cos 𝜃. 2) Use Euler’s formula to drive a formula for sin of four 𝜃 in terms of cos 𝜃 and sin 𝜃.

For part one, we’ll use the properties of the exponential function. And we’ll write 𝑒 to the four 𝑖𝜃 as 𝑒 to the 𝑖𝜃 to the power of four. And now, we can use Euler’s formula. And we write the left-hand side as cos four 𝜃 plus 𝑖 sin four 𝜃. And on the right-hand side, we can say that this is equal to cos 𝜃 plus 𝑖 sin 𝜃 all to the power of four. Now, we’re going to apply the binomial theorem to distribute cos 𝜃 plus 𝑖 sin 𝜃 to the power of four.

In our equation, 𝑎 is equal to cos of 𝜃, 𝑏 is equal to 𝑖 sin of 𝜃, and 𝑛 is the power; it’s four. And this means we can say that cos 𝜃 plus 𝑖 sin 𝜃 to the power of four is the same as cos 𝜃 to the power of four plus four choose one cos cubed 𝜃 times 𝑖 sin 𝜃 and so on. We know that four choose one is four, four choose two is six, and four choose three is also four. We also know that 𝑖 squared is negative one, 𝑖 cubed is negative 𝑖, and 𝑖 to the power of four is one. And we can further rewrite our equation as shown.

Now, we’re going to equate the real parts of this equation. And that will give us a formula for cos of four 𝜃 in terms of cos 𝜃 and sin 𝜃. Let’s clear some space. The real part on the left-hand side is cos four 𝜃. And then on the right-hand side, we have cos 𝜃 to the power of four. We’ve got negative cos squared 𝜃 sin squared 𝜃. And we’ve got sin 𝜃 to the power of four. So we equate these. But we’re not quite finished. We were asked to derive a formula for cos four 𝜃 in terms of cos 𝜃 only.

So here, we use the identity cos squared 𝜃 plus sin squared 𝜃 is equal to one. And we rearrange this. And we say that well, that means that sin squared 𝜃 must be equal to one minus cos squared 𝜃. And we can rewrite this as cos 𝜃 to the power of four plus six cos squared 𝜃 times one minus cos squared 𝜃 plus one minus cos squared 𝜃 squared. We distribute the parentheses. And our final step is to collect like terms. And we see that we’ve derived the formula for cos of four 𝜃 in terms of cos 𝜃. cos four 𝜃 is equal to eight cos 𝜃 to the power of four minus eight cos squared 𝜃 plus one.

For part two, we can repeat this process equating the imaginary parts. They are sin of four 𝜃 on the left. And then on the right, we have four cos cubed 𝜃 sin 𝜃, negative four cos 𝜃 sin cubed 𝜃. And we see this sin four 𝜃 must be equal to four cos cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃. And we could — if we so wish — factor four cos 𝜃 sin 𝜃. And we’ll be left with four cos 𝜃 sin 𝜃 times cos squared 𝜃 minus sin squared 𝜃. And last, we’ve been asked to derive a formula for sin four 𝜃 in terms of cos 𝜃 and sin 𝜃. You might now see a link between sin four 𝜃 and the double angle formulae.

Now, an interesting application of Euler’s formula is that we can use it to derive an expression for sin 𝜃 and cos 𝜃 in terms of 𝑒 to the 𝑖𝜃. We’ve already seen that we can write 𝑒 to the negative 𝑖𝜃 as cos 𝜃 minus 𝑖 sin 𝜃. And since 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can say that the sum of 𝑒 to the 𝑖𝜃 and 𝑒 to the negative 𝑖𝜃 is cos 𝜃 plus 𝑖 sin 𝜃 plus cos 𝜃 minus 𝑖 sin 𝜃. Well, this expression on the right-hand side simplifies to two cos 𝜃. And we can make cos 𝜃 the subject by dividing through by two. And we see we have an expression for cos 𝜃 in terms of powers of 𝑒 to the 𝑖𝜃. It’s a half 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃.

Similarly, we can find the difference. And we get cos 𝜃 plus 𝑖 sin 𝜃 minus cos 𝜃 minus 𝑖 sin 𝜃. This simplifies to two 𝑖 sin 𝜃. This time we divide through by two 𝑖. And we can see that sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. At these two formulae have plenty of applications in their own right. But for the purposes of this video, we’ll look at one final example. And we’ll see how they can be used to derive further trigonometric identities.

Use Euler’s formula to express sin cubed 𝜃 cos squared 𝜃 in the form 𝑎 sin 𝜃 plus 𝑏 sin three 𝜃 plus 𝑐 sin five 𝜃, where 𝑎, 𝑏, and 𝑐 are constants to be found. Hence, find the solutions of sin five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal to zero and less than 𝜋. Give your answer in exact form.

We begin by recalling the fact that sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. And cos 𝜃 is equal to a half times 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃. This means we can find the product of sin cubed 𝜃 and cos squared 𝜃. We can write it as one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃 cubed times a half times 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. One over two 𝑖 cubed is negative one over eight 𝑖. And a half squared is one-quarter. So we can rewrite our expression a little bit further. We find the product of negative one over eight 𝑖 and a quarter. And we get negative one over 32𝑖. And we can rewrite the rest of our expression as shown.

We’re now going to use the binomial theorem to expand each of the sets of parentheses. The first part becomes 𝑒 to three 𝑖𝜃 plus three choose one 𝑒 to the two 𝑖𝜃 times negative 𝑒 to the negative 𝑖𝜃 and so on. And this simplifies to 𝑒 to the three 𝑖𝜃 minus three 𝑒 to the 𝑖𝜃 plus three 𝑒 to the negative 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃. Let’s repeat this process for 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. When we do, we get 𝑒 to the two 𝑖𝜃 plus two 𝑒 to the zero which is just two plus 𝑒 to the negative two 𝑖𝜃.

We’re going to need to find the product of these two expressions. We’ll need to do that really carefully. We’ll need to ensure that each term in the first expression is multiplied by each term in the second expression. And we can write sin cubed 𝜃 cos squared 𝜃 as shown. Now, there’s quite a lot going on here. So you might wish to pause the video and double-check your answer against mine. We’re going to gather the corresponding powers of 𝑒 together.

We’ll gather 𝑒 to the five 𝑖𝜃 and 𝑒 to the negative five 𝑖𝜃. We’ll collect 𝑒 to the plus and minus three 𝑖𝜃. And we’ll gather 𝑒 the 𝑖𝜃 and 𝑒 to the negative 𝑖𝜃. Let’s neaten things up somewhat. We end up with negative one over 32𝑖 times 𝑒 to the five 𝑖𝜃 minus 𝑒 to the negative five 𝑖𝜃 minus 𝑒 to the three 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃 minus two times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. And now, you might be able to spot why we chose to do this. We can now go back to the given formulae. Let’s clear some space for the next step.

We kind of unfactorize a little. And we can rewrite sin cubed 𝜃 cos squared 𝜃 as shown. And we can therefore replace 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 with sin 𝜃 and so on. And we can see that sin cubed 𝜃 cos squared 𝜃 is equal to a 16th times two sin 𝜃 plus sin three 𝜃 minus sin five 𝜃. Since 𝑎, 𝑏, and 𝑐 are constants to be found, we can say that 𝑎, the coefficient of sin 𝜃, is one-eighth. 𝑏, the coefficient of sin three 𝜃, is a 16th. And 𝑐, the coefficient sin five 𝜃, is negative one 16th.

Let’s now consider part two of this question. We begin by using our answer to part one and multiplying both sides by 16. We then subtract two sin 𝜃 from both sides and multiply through by negative one. And we can now see that we’ve got an equation in sin five 𝜃 minus sin three 𝜃. We’re told that sin five 𝜃 minus sin three 𝜃 is equal to zero. So we let two sin 𝜃 minus 16 sin cubed 𝜃 cos squared 𝜃 be equal to zero. And then, we factor by two sin 𝜃. Since the product of these two terms is equal to zero, this means that either of these terms must be equal to zero. So either two sin 𝜃 is equal to zero and dividing by two, we could see that sin 𝜃 is equal to zero or one minus eight sin squared 𝜃 cos squared 𝜃 is equal to zero. Given the interval 𝜃 is greater than or equal to zero and less than 𝜋, we can see that one of our solutions is when 𝜃 is equal to zero.

We’re going to rewrite our other equations somewhat. We know that sin two 𝜃 is equal to two sin 𝜃 cos 𝜃. Squaring this, we get sin squared two 𝜃 equals four sin squared 𝜃 cos squared 𝜃. And that’s in turn means that our equation is one minus two sin squared two 𝜃 equals zero. Rearranging to make sin two 𝜃 the subject, we see that sin two 𝜃 is equal to plus or minus one over root two. Starting with the positive square root for 𝜃 in the interval given, we know that sin two 𝜃 is equal to one over root two when 𝜃 is equal to 𝜋 by eight or three 𝜋 by eight. Similarly, we can solve for the negative square root. And we get five 𝜋 by eight and seven 𝜋 by eight. And there are therefore five solutions to the equation sin five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal to zero and less than 𝜋. They are zero 𝜋 by eight, three 𝜋 by eight, five 𝜋 by eight, and seven 𝜋 by eight.

In this video, we’ve seen that we can use Euler’s formula in conjunction with the properties of the exponential functions. And we can derive many trigonometric identities such as the Pythagorean identity and multiple angle formulae. We also saw the we can use the theorem to express sine and cosine in terms of the complex exponential function as shown. We’ve also seen that we can use the identities derived from Euler’s formula to help us simplify expressions. And these in turn can help us to solve trigonometric equations.