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Question Video: Identities for Powers of Cosine Mathematics

Express cos⁢ πœƒ in terms of cos 6πœƒ, cos 5πœƒ, cos 4πœƒ, cos 3πœƒ, cos 2πœƒ, cos πœƒ, and any constant terms.

06:55

Video Transcript

Express cos to the sixth power πœƒ in terms of cos six πœƒ, cos five πœƒ, cos four πœƒ, cos three πœƒ, cos two πœƒ, cos πœƒ, and any constant terms.

In this question, we want to find an expression for the cos to the sixth power of πœƒ. And this expression needs to consist of terms which only involve cos of integer multiple values of πœƒ and constant terms. In some questions like this, it’s possible to find expressions for each of these multiple angle formula for cosine and then get our answer to be equal to what we want, in this case cos to the sixth power πœƒ. However, to do this, we would need to do this for each of the multiple angle formulas in the question. So although this method might be possible, it would be very time consuming, and it would take a lot of trial and error.

Instead, we’re going to need to recall we can find expressions like this by using De Moivre’s theorem. So to answer this question, we’ll start by recalling De Moivre’s theorem. One version of this says, for any integer value of 𝑛 and real number πœƒ, the cos of πœƒ plus 𝑖 sin of πœƒ all raised to the 𝑛th power is equal to the cos of π‘›πœƒ plus 𝑖 sin of π‘›πœƒ. However, if we were to try and directly use this expression to answer our question, we might struggle. Since we want to find an expression for cos to the sixth power of πœƒ, we’d need to set 𝑛 equal to six to get cos to the sixth power to appear in our expression. However, when we distribute the 𝑛th power over our parentheses, we would end up with fifth power cos terms, fourth power cos terms, all the way down to zeroth power cos terms.

And this seems like it would be very difficult to get rid of all of these terms. So instead, we’re going to need to recall a very useful result. If we call the complex number inside our parentheses 𝑧, then De Moivre’s theorem is telling us, 𝑧 to the 𝑛th power is equal to the cos of π‘›πœƒ plus 𝑖 sin of π‘›πœƒ. And we can then use De Moivre’s theorem to find two useful results, 𝑧 to the 𝑛th power plus one over 𝑧 to the 𝑛th power is equal to two cos of π‘›πœƒ and 𝑧 to the 𝑛th power minus one over 𝑧 to the 𝑛th power is equal to two 𝑖 sin of π‘›πœƒ. And to prove these results, we directly use De Moivre’s theorem, which we remember is true for any integer value of 𝑛. And we know one over 𝑧 to the 𝑛th power is the same as 𝑧 to the power of negative 𝑛.

These results are very useful and they’re worth committing to memory. We’re going to focus on the top result, which we’re going to use to find an expression for cos to the sixth power of πœƒ. We start off by setting our value of 𝑛 equal to one in this expression. This gives us two cos of πœƒ is equal to 𝑧 plus one over 𝑧. And remember, the question wants us to find an expression for the cos to the sixth power πœƒ, so we’re going to raise both sides of this equation to the sixth power. On the left-hand side of our equation, we can simplify by taking the sixth power of both of our two factors. This gives us two to the sixth power times cos to the sixth power of πœƒ. And we can simplify this since we know two to the sixth power is equal to 64.

So the left-hand side of our equation simplifies to give us 64 cos to the sixth power of πœƒ. And on the right-hand side of our equation, we have the sum of two numbers raised to an exponent. We can do this by using the binomial formula. We recall, this tells us for any positive integer π‘š, π‘Ž plus 𝑏 all raised to the π‘šth power is equal to the sum from π‘Ÿ equals zero to π‘š of π‘š choose π‘Ÿ times π‘Ž to the power of π‘Ÿ multiplied by 𝑏 to the power of π‘š minus π‘Ÿ. So we can use this to distribute the exponent of six over the right-hand side of our equation. We get that six choose zero 𝑧 to the sixth power plus six choose one times 𝑧 to the fifth power times one over 𝑧 plus six choose two times 𝑧 to the fourth power multiplied by one over 𝑧 all squared. And then we keep adding terms of this form all the way up to six choose six multiplied by one over 𝑧 all raised to the sixth power.

Remember, the question wants us to find an expression for the cos to the sixth power of πœƒ. But it wants this in terms of cos of multiple angles of πœƒ. We can rearrange this to get the cos to the sixth power of πœƒ. However, the right-hand side of this expression is in terms of 𝑧; it’s not yet in terms of cos of angles of πœƒ. This means we’re going to need to simplify the right-hand side of our equation. We’ll do this term by term. In our first term, six choose zero is equal to one, so the first term is just 𝑧 to the sixth power. In our second term, six choose one is equal to six and 𝑧 to the fifth power multiplied by one over 𝑧 can be simplified by using our laws of exponents. It’s equal to 𝑧 to the fourth power. So our second term is just six 𝑧 to the fourth power.

In our third term, six choose two is 15 and 𝑧 to the fourth power multiplied by one over 𝑧 squared is 𝑧 to the fourth power over 𝑧 squared, which is just equal to 𝑧 squared. So our third term is 15𝑧 squared. And we can simplify the rest of the terms in this expansion in the same way. We get 20, 15 over 𝑧 squared, six over 𝑧 to the fourth power, and one over 𝑧 to the sixth power. And now we can notice something interesting. We can simplify this expression by using De Moivre’s theorem. For example, we have the first term in this expansion plus the last term in this expansion is 𝑧 to the sixth power plus one over 𝑧 to the sixth power. By setting 𝑛 equal to six in our useful result, we know this is equal to two times the cos of six πœƒ. And this isn’t the only time this result appears in our expansion. We can use this two more times.

So we’ll pair up the terms in our expansion to give us 𝑧 to the sixth power plus one over 𝑧 to the sixth power plus six 𝑧 to the fourth power plus six over 𝑧 to the fourth power plus 15𝑧 squared plus 15 over 𝑧 squared plus 20. And before we apply De Moivre’s theorem, we can notice something interesting. In our first set of parentheses, we can take out the shared factor of six. And similarly, in our third set of parentheses, we can take out the shared factor of 15. This gives us the following expression. And now we can simplify all three of our sets of parentheses by using De Moivre’s theorem. By setting 𝑛 equal to six, we know 𝑧 to the sixth power plus one over 𝑧 to the sixth power is two cos of six πœƒ. With 𝑛 as four, 𝑧 to the fourth power plus one over 𝑧 to the fourth power is two cos four πœƒ. And with 𝑛 as two, 𝑧 squared plus one over 𝑧 squared is two cos of two πœƒ.

So by substituting these expressions in and remembering we need to multiply by our coefficients and we still need to add 20 at the end, we get 64 cos to the sixth power of πœƒ is equal to two cos of six πœƒ plus 12 cos of four πœƒ plus 30 cos of two πœƒ plus 20. And now this is almost in exactly the form we want. All we need now is to rearrange to make cos to the sixth power of πœƒ the subject. And to do this, we’re going to need to divide both sides of our equation through by 64. So to do this, let’s start by clearing some space. We have cos to the sixth power of πœƒ is equal to the following expression. And to evaluate this, we’re going to divide every term in this expression by 64.

Doing this, we get two cos six πœƒ all over 64 plus 12 cos four πœƒ all over 64 plus 30 cos two πœƒ all over 64 plus 20 divided by 64. All that’s left to do now is cancel all of the shared factors of two in both our numerator and denominator for all four of our terms. We get the following expression. And this gives us our final answer. Therefore, by using De Moivre’s theorem, we were able to express cos to the sixth power of πœƒ in terms of cosines of integer multiple angles of πœƒ. We got cos to the sixth power of πœƒ is equal to one over 32 times cos six πœƒ plus three over 16 times cos four πœƒ plus 15 over 32 times cos of two πœƒ plus five over 16.

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