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Question Video: Finding a Specific Multiple Angle Formula for Cosine Mathematics • Third Year of Secondary School

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Use de Moivre’s theorem to express cos 5𝜃 in terms of powers of cos 𝜃.

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Video Transcript

Use de Moivre’s theorem to express cos of five 𝜃 in terms of powers of cos of 𝜃.

In this question, we want to find an expression for the cos of five 𝜃 in terms of powers of cos of 𝜃. We’re told to do this by using de Moivre’s theorem. So let’s start by recalling this. De Moivre’s theorem tells us for any integer 𝑛 and real value 𝜃, the cos of 𝜃 plus 𝑖 sin of 𝜃 all raised to the 𝑛th power is equal to the cos of 𝑛𝜃 plus 𝑖 sin of 𝑛𝜃. If we want to use this to find an expression for cos of five 𝜃, then we’re going to want cos of five 𝜃 to appear in this expression. So we’re going to want to set our value of the integer 𝑛 equal to five.

So by using this statement of de Moivre’s theorem and setting 𝑛 equal to five, we get cos of five 𝜃 plus 𝑖 sin of five 𝜃 must be equal to cos 𝜃 plus 𝑖 sin 𝜃 all raised to the fifth power. But this is not yet an expression for cos five 𝜃. And it’s definitely not in terms of just powers of cos 𝜃. So we’re going to need to manipulate this. We can see on the right-hand side of our equation, we can try distributing the fifth power over our parentheses. And to do this, we need to notice inside our parentheses we have two terms. We have a binomial. So one way of distributing the exponent over our parentheses will be to use the binomial formula.

This tells us for positive integer 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 𝑟th power multiplied by 𝑏 to the power of 𝑚 minus 𝑟. So by applying the binomial theorem, we can expand the right-hand side of our equation. We get five choose zero times cos to the fifth power of 𝜃 plus five choose one multiplied by cos to the fourth power of 𝜃 multiplied by 𝑖 sin 𝜃 plus five choose two times cos cubed of 𝜃 multiplied by 𝑖 sin 𝜃 all squared plus five choose three times cos squared 𝜃 multiplied by 𝑖 sin 𝜃 all cubed plus five choose four times cos 𝜃 multiplied by 𝑖 sin 𝜃 all raised to the fourth power plus five choose five times 𝑖 sin 𝜃 all raised to the fifth power.

And now we can see we’re making some progress. On the right-hand side of our equation, our expressions are now involving powers of cos 𝜃. However, we also have powers of sin 𝜃. So we’re going to need to simplify further. We’ll start by simplifying each term individually. In our first term, five choose zero is equal to one. So our first term is just cos to the fifth power of 𝜃. In our second term, five choose one is equal to five. And remember, we have a factor of 𝑖, so our second term simplifies to give us five 𝑖 multiplied by cos to the fourth power of 𝜃 times sin 𝜃. In our third term, five choose two is equal to 10. And we could distribute the square over our parentheses. We get 𝑖 squared times sin squared 𝜃.

But remember, 𝑖 is the square root of negative one. So 𝑖 squared is just equal to negative one. So our third term simplifies to give us negative 10 cos cubed 𝜃 times sin squared 𝜃. And we can evaluate the last three terms in this expansion in exactly the same way. The last three terms in this expansion are negative 10𝑖 cos squared 𝜃 sin cubed 𝜃 plus five times cos 𝜃 sin to the fourth power 𝜃 plus 𝑖 times sin to the fifth power of 𝜃. And don’t forget by de Moivre’s theorem, we know this is equal to cos of five 𝜃 plus 𝑖 sin of five 𝜃.

We want to use this to find an expression for the cos of five 𝜃. And to do this, we need to notice something interesting. The left-hand side of this equation is just a complex number, and we can see cos of five 𝜃 is the real part of this complex number. So we can find an expression for the cos of five 𝜃 by just taking the real part of the right-hand side of this equation. That’s just going to be all of the terms which don’t have a factor of 𝑖. Doing this, we get cos of five 𝜃 is equal to cos to the fifth power of 𝜃 minus 10 cos cubed of 𝜃 times sin squared of 𝜃 plus five multiplied by cos of 𝜃 times sin to the fourth power of 𝜃. So we’ve now successfully found an expression for cos of five 𝜃.

However, remember, the question wants us to give our answer only in powers of cos 𝜃. So to do this, we’re going to need to find expressions for sin squared of 𝜃 and sin to the fourth power of 𝜃 in terms of cos 𝜃. And we can do this by using the following trigonometric identity. We know for any value of 𝜃, cos squared 𝜃 plus sin squared 𝜃 is equal to one. Subtracting cos squared from both sides, we get sin squared 𝜃 is equivalent to one minus cos squared 𝜃. And because we also need an expression for sin to the fourth power of 𝜃, we can square both sides. We get sin to the fourth power of 𝜃 is equivalent to one minus cos squared 𝜃 all squared. We can use these to rewrite sin squared 𝜃 and sin to the fourth power of 𝜃 in terms of powers of cos 𝜃.

This gives us cos of five 𝜃 is equal to cos to the fifth power of 𝜃 minus 10 cos cubed of 𝜃 multiplied by one minus cos squared of 𝜃 plus five cos of 𝜃 times one minus cos squared of 𝜃 all squared. And now this expression only consists of powers of cos of 𝜃, so we could stop here. However, we can simplify this expression by distributing over our parentheses. Distributing over our first set of parentheses, we get negative 10 cos cubed of 𝜃 plus 10 cos to the fifth power of 𝜃. Distributing the square over our second set of parentheses, we get one minus two cos squared of 𝜃 plus cos to the fourth power of 𝜃. And we’re going to need to multiply each of these three terms by five cos of 𝜃. And by doing this, we get five cos of 𝜃 minus 10 cos cubed of 𝜃 plus five times the cos to the fifth power of 𝜃.

Now all that’s left to do is collect like terms. Doing this, we get cos of five 𝜃 is equal to 16 cos to the fifth power of 𝜃 minus 20 cos cubed of 𝜃 plus five cos of 𝜃, which is our final answer. In this question, we were able to use de Moivre’s theorem to find an expression for the cos of five 𝜃 entirely in terms of powers of the cos of 𝜃. To do this, we did need to expand by using the binomial theorem and notice we need to take the real parts of both sides of our equation and then finally simplify our expression by using our trigonometric identities. However, we were able to show the cos of five 𝜃 is equal to 16 cos to the fifth power of 𝜃 minus 20 cos cubed of 𝜃 plus five cos of 𝜃.

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