Video: Applications of de Moivre’s Theorem

Use De Moivre’s theorem to express cos 3πœƒ and sin 3πœƒ in terms of cos πœƒ and sin πœƒ respectively.

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Video Transcript

Use De Moivre’s theorem to express cos three πœƒ and sin three πœƒ in terms of cos πœƒ and sin πœƒ, respectively.

First, let us recall De Moivre’s theorem. This states that cos πœƒ plus 𝑖 sin πœƒ raised to the power of 𝑛 is equal to cos of π‘›πœƒ plus 𝑖 sin of π‘›πœƒ. Looking at the right-hand side of this equation, we see that the two terms are very similar to the two terms that we must express in terms of cos πœƒ and sin πœƒ, respectively, specifically in the case where 𝑛 is equal to three.

Let us rewrite De Moivre’s theorem in the case where 𝑛 equals three to see this more clearly. Now that we have written out this equation, we can see that the right-hand side has a real part, cos three πœƒ, and an imaginary part, sin three πœƒ. We can express both of these parts in terms of cos πœƒ and sin πœƒ by isolating the real and imaginary parts of the left-hand side of the equation.

In order to do this, we will need to multiply out our parentheses and separate the terms. Looking at the left-hand side of the equation, we can see that we have a binomial to the power of three. A general form for this would be π‘Ž plus 𝑏 cubed. In order to expand our parentheses, we could multiply out each of our terms by hand or use the binomial theorem. Instead, we’re going to use a slightly different shortcut.

Looking at our parentheses, it is fairly easy to convince ourselves that each term will have some power of π‘Ž and some power of 𝑏. By simple inspection, we can see that the highest possible power of π‘Ž that can be reached by multiplying these parentheses is π‘Ž cubed. By similar reasoning, the highest possible power of 𝑏 that can be reached by multiplying out these parentheses is 𝑏 cubed.

Our parentheses will also give us the terms in between these two, which will have decreasing powers of π‘Ž and increasing powers of 𝑏. There are multiple ways to reach these in between terms, giving the π‘Žs and 𝑏s within our parentheses. We will hence have a multiple of them. In other words, they will have a coefficient.

In order to find these coefficients, we can take the numbers from the corresponding tier of Pascal’s triangle. Since we are dealing with a binomial to the power of three, we will use the third tier. The coefficients for our four terms are therefore one π‘Ž cubed plus three π‘Ž squared 𝑏 plus three π‘Žπ‘ squared plus one 𝑏 cubed.

Now that we have this expansion, let’s apply it to our equation. We can substitute in our terms to this expansion by taking π‘Ž to be cos πœƒ and 𝑏 to be 𝑖 sin πœƒ. We first have π‘Ž cubed, which becomes cos πœƒ cubed. We next have three π‘Ž squared 𝑏, which becomes three times cos πœƒ squared times 𝑖 sin πœƒ. We then have three π‘Žπ‘ squared, which becomes three times cos πœƒ times 𝑖 sin πœƒ squared. And finally, we have 𝑏 cubed, which is 𝑖 sin πœƒ cubed.

As our next step, we’re going to multiply out the squares and the cubes of our parentheses and move all of our 𝑖s to the front of the term. Let us now simplify by recalling the definition of 𝑖. We know that 𝑖 denotes an imaginary number. And it’s the square root of negative one. 𝑖 squared is then the square root of negative one times the square root of negative one, in other words, negative one.

To make the simplification easier, we can write 𝑖 cubed as 𝑖 squared times 𝑖. Since we now know that 𝑖 squared is negative one, we can use this in our definition. And we can say that 𝑖 cubed is equal to negative 𝑖. Let us use these identities to replace the 𝑖 squared and the 𝑖 cubed within our equation.

Here we have simplified our equation. We can now see that, of our four terms, we have two that do not have a factor of 𝑖 or real components. And we have two terms that do have a factor of 𝑖 or imaginary components. Let us now group together the real and imaginary terms.

We’re now in a position to look back on our original equation derived from De Moivre’s theorem. Let’s review. We have expanded out the left-hand side of our equation and separated out the real and imaginary terms. To proceed, we can equate the real and imaginary parts of our expansion with the real and imaginary parts of the right-hand side of our equation. Let us show this now.

We now have an equation for cos three πœƒ and an equation for sin three πœƒ. However, we have one final simplification to perform. Our question requires that we express cos three πœƒ in terms of cos πœƒ. But we still have a sin in this equation. Likewise, the question requires that we express sin three πœƒ in terms of sin πœƒ. But we still have a cos in this equation.

We can substitute these terms by recalling one of the Pythagorean identities. This states that cos squared πœƒ plus sin squared πœƒ is equal to one. By subtracting cos squared πœƒ from both sides of this equation, we get that sin squared πœƒ is equal to one minus cos squared πœƒ.

We can do exactly the same to find cos squared πœƒ by subtracting sin squared πœƒ from both sides of our original identity. Let us work on our equation for cos three πœƒ using the first rearrangement of the identity. We substitute sin squared πœƒ for one minus cos squared πœƒ. We then multiply out the parentheses. Finally, we collect our cos cubed πœƒ terms.

We can now perform the exact same method on the equation for sin three πœƒ. We replace cos squared πœƒ with one minus sin squared πœƒ. We then multiply out the parentheses and collect our sin cubed πœƒ terms.

Now that we have done this, we have completed our question. And we have expressed cos three πœƒ in terms of cos πœƒ and we have expressed sin three πœƒ in terms of sin πœƒ.

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