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Question Video: Identities for Powers of Cosine Mathematics • Third Year of Secondary School

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Express cosΒ³ π in terms of the cosines of multiples of π.

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Express cos cubed π in terms of the cosines of multiples of π.

In this question, we want to find an expression for the cos cubed of π in terms of cosines of multiples of our angle π. The easiest way to answer this question will be to use De Moivreβs theorem. One version of De Moivreβs theorem tells us for any integer value of π and a real number π, cos π plus π sin π all raised to the πth power will be equal to cos ππ plus π sin ππ. If we wanted to use this to answer our question, we would need to choose our value of π equal to three, so cos cubed π appears in our expression. However, if we did this, when we distribute over our parentheses, we would end up with terms involving sin π. And the question only wants us to use terms involving cos π, so weβre going to need to use an extra result.

If we set π§ to be the expression inside of our parentheses in De Moivreβs theorem, then De Moivreβs theorem tells us π§ to the πth power is equal to cos ππ plus π sin ππ. We can then use this to get two very useful results. π§ to the πth power plus one over π§ to the πth power is equal to two cos of ππ, and π§ to the πth power minus one over π§ to the πth power is two π sin of ππ. And we can prove these results by remembering De Moivreβs theorem works for any integer value of π and one over π§ to the πth power is equal π§ to the power of negative π. 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 π, so weβre going to use the expression involving cosine. And weβll start by setting our value of π equal to one because we want to find an expression for cos cubed of just π. So by setting π equal to one, we get two cos of π will be equal to π§ plus one over π§. Next, we want an expression for cos cubed of π, so weβ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 π. 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 π, π plus π all raised to the πth power is equal to the sum from π equals zero to π of π choose π times π to the power of π multiplied by π to the power of π minus π. So, by using the binomial formula, we can expand the right-hand side of our equation. We get three choose zero times π§ cubed plus three choose one multiplied by π§ squared times one over π§ plus three choose two times π§ multiplied by one over π§ all squared plus three choose three times one over π§ all cubed.

And we can simplify the right-hand side of this expression. Weβ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 π§ cubed. In our second term, three choose one is equal to three and π§ squared times one over π§ is π§ squared over π§, which simplifies to give us π§. So our second term simplifies to give us three π§. In our third term, three choose two is equal to three and π§ times one over π§ all squared is π§ divided by π§ squared, which simplifies to give us one over π§. So our third term simplifies to give us three over π§. And in our fourth and final term, three choose three is equal to one and one over π§ all cubed is one over π§ cubed. So our fourth term is just one over π§ cubed.

So, so far, weβve shown that eight cos cubed of π is equal to π§ cubed plus three π§ plus three over π§ plus one over π§ cubed. But this isnβt enough because, remember, the question wants us to find an expression for cos cubed of π only in terms of cosines of multiples of π. And to do this, weβ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 π§ cubed plus one over π§, which we can simplify by using De Moivreβs theorem. π§ to the πth power plus one over π§ to the πth power is two cos of ππ. So π§ cubed plus one over π§ cubed is two cos of three π.

So weβ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 π§ plus three over π§. First, we need to take out the shared factor of three from both of these terms. This leaves us with π§ plus one over π§ inside of our parentheses. And π§ plus one over π§ is this exact same result with π set to be one. By De Moivreβs theorem, π§ plus one over π§ is two cos of π and π§ cubed plus one over π§ is two cos three π. So we can substitute both of these into our expression. And remember, we need to multiply two cos of π by three. This gives us two cos of three π plus six cos π. And remember, this is all equal to eight cos cubed of π.

Now, we can find an expression for cos cubed of π in terms of cosines of multiples of π by dividing both sides of our equation through by eight. Weβll divide each term separately on the right-hand side by eight. This gives us cos cubed of π is equal to two cos of three π all over eight plus six cos of π 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 π all over four plus three cos of π all over four. And weβll simplify this by taking out the shared factor of one-quarter. This gives us one-quarter multiplied by cos of three π plus three cos of π.

And the last piece of simplification weβ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 π plus cos of three π. Therefore, by using De Moivreβs theorem and several results we get from De Moivreβs theorem, we were able to find an expression for cos cubed of π in terms of cosines of multiples of π. We were able to show cos cubed of π is equal to one-quarter multiplied by three cos of π plus cos of three π.

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