Video Transcript
Express cos cubed 𝜃 in terms of the cosines of multiples of 𝜃.
In this question, we want to find an expression for the cos cubed of 𝜃 in terms of cosines of multiples of our angle 𝜃. The easiest way to answer this question will be to use De Moivre’s theorem. One version of De Moivre’s theorem tells us for any integer value of 𝑛 and a real number 𝜃, cos 𝜃 plus 𝑖 sin 𝜃 all raised to the 𝑛th power will be equal to cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃. If we wanted to use this to answer our question, we would need to choose our value of 𝑛 equal to three, so cos cubed 𝜃 appears in our expression. However, if we did this, when we distribute over our parentheses, we would end up with terms involving sin 𝜃. And the question only wants us to use terms involving cos 𝜃, so we’re going to need to use an extra result.
If we set 𝑧 to be the expression inside of our parentheses in De Moivre’s theorem, then De Moivre’s theorem tells us 𝑧 to the 𝑛th power is equal to cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃. We can then use this to get two very useful results. 𝑧 to the 𝑛th power plus one over 𝑧 to the 𝑛th power is equal to two cos of 𝑛𝜃, and 𝑧 to the 𝑛th power minus one over 𝑧 to the 𝑛th power is two 𝑖 sin of 𝑛𝜃. And we can prove these results by remembering De Moivre’s theorem works for any integer value of 𝑛 and one over 𝑧 to the 𝑛th power is equal 𝑧 to the power of negative 𝑛. These two results are really useful and are worth committing to memory.
We want to use these to find an expression for cos cubed of 𝜃, so we’re going to use the expression involving cosine. And we’ll start by setting our value of 𝑛 equal to one because we want to find an expression for cos cubed of just 𝜃. So by setting 𝑛 equal to one, we get two cos of 𝜃 will be equal to 𝑧 plus one over 𝑧. Next, we want an expression for cos cubed of 𝜃, so we’re going to need to cube both sides of our equation, and we could simplify both sides separately. On the left-hand side of our equation, we need to cube each of the factors separately. This gives us two cubed, which is eight multiplied by cos cubed of 𝜃. And on the right-hand side of this expression, we have the sum of two terms raised to the power of three. This is a binomial.
So we can expand the right-hand side of this expression by using the binomial theorem. Remember, this tells us for any positive integer value of 𝑚, 𝑎 plus 𝑏 all raised to the 𝑚th power is equal to the sum from 𝑟 equals zero to 𝑚 of 𝑚 choose 𝑟 times 𝑎 to the power of 𝑟 multiplied by 𝑏 to the power of 𝑚 minus 𝑟. So, by using the binomial formula, we can expand the right-hand side of our equation. We get three choose zero times 𝑧 cubed plus three choose one multiplied by 𝑧 squared times one over 𝑧 plus three choose two times 𝑧 multiplied by one over 𝑧 all squared plus three choose three times one over 𝑧 all cubed.
And we can simplify the right-hand side of this expression. We’ll do this term by term. In our first term, three choose zero is equal to one. So our first term simplifies to give us 𝑧 cubed. In our second term, three choose one is equal to three and 𝑧 squared times one over 𝑧 is 𝑧 squared over 𝑧, which simplifies to give us 𝑧. So our second term simplifies to give us three 𝑧. In our third term, three choose two is equal to three and 𝑧 times one over 𝑧 all squared is 𝑧 divided by 𝑧 squared, which simplifies to give us one over 𝑧. So our third term simplifies to give us three over 𝑧. And in our fourth and final term, three choose three is equal to one and one over 𝑧 all cubed is one over 𝑧 cubed. So our fourth term is just one over 𝑧 cubed.
So, so far, we’ve shown that eight cos cubed of 𝜃 is equal to 𝑧 cubed plus three 𝑧 plus three over 𝑧 plus one over 𝑧 cubed. But this isn’t enough because, remember, the question wants us to find an expression for cos cubed of 𝜃 only in terms of cosines of multiples of 𝜃. And to do this, we’re going to need to notice something. On the right-hand side of our expression, we can actually simplify. For example, on the right-hand side of our expression, we have 𝑧 cubed plus one over 𝑧, which we can simplify by using De Moivre’s theorem. 𝑧 to the 𝑛th power plus one over 𝑧 to the 𝑛th power is two cos of 𝑛𝜃. So 𝑧 cubed plus one over 𝑧 cubed is two cos of three 𝜃.
So we’ll pair these two terms up on the right-hand side of our expression. And we will be able to see we can also do this with three 𝑧 plus three over 𝑧. First, we need to take out the shared factor of three from both of these terms. This leaves us with 𝑧 plus one over 𝑧 inside of our parentheses. And 𝑧 plus one over 𝑧 is this exact same result with 𝑛 set to be one. By De Moivre’s theorem, 𝑧 plus one over 𝑧 is two cos of 𝜃 and 𝑧 cubed plus one over 𝑧 is two cos three 𝜃. So we can substitute both of these into our expression. And remember, we need to multiply two cos of 𝜃 by three. This gives us two cos of three 𝜃 plus six cos 𝜃. And remember, this is all equal to eight cos cubed of 𝜃.
Now, we can find an expression for cos cubed of 𝜃 in terms of cosines of multiples of 𝜃 by dividing both sides of our equation through by eight. We’ll divide each term separately on the right-hand side by eight. This gives us cos cubed of 𝜃 is equal to two cos of three 𝜃 all over eight plus six cos of 𝜃 all over eight. And we can simplify the right-hand side. We can cancel the shared factor of two in the numerator and denominator of both terms, giving us cos of three 𝜃 all over four plus three cos of 𝜃 all over four. And we’ll simplify this by taking out the shared factor of one-quarter. This gives us one-quarter multiplied by cos of three 𝜃 plus three cos of 𝜃.
And the last piece of simplification we’ll do is reorder the two terms inside of our parentheses. And this gives us our final answer of one-quarter multiplied by three cos 𝜃 plus cos of three 𝜃. Therefore, by using De Moivre’s theorem and several results we get from De Moivre’s theorem, we were able to find an expression for cos cubed of 𝜃 in terms of cosines of multiples of 𝜃. We were able to show cos cubed of 𝜃 is equal to one-quarter multiplied by three cos of 𝜃 plus cos of three 𝜃.
