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Express cos six π in terms of powers of sin π and cos π.

In this question, we need to find an expression for the cos of six π in terms of powers of the sin of π and the cos of π. We have a few options for doing this. For example, we could write cos six π as cos three π plus three π and then use our angle addition formula for cosine to find an expression for cos six π in terms of angles of three π. We could then repeat this process to eventually get an expression in terms of sin π and cos π. And this would work, and it would give us the correct answer. However, there is a simpler method involving de Moivreβs theorem.

We recall one part of de Moivreβs theorem tells us for any integer value of π and real number π, the cos of π plus π sin of π all raised to the πth power is equal to the cos of ππ plus π times the sin of ππ. And we can use this to find expressions for our trigonometric functions evaluated at integer multiples of π. For example, in the question, we want to find an expression for cos six π. So weβll set our value of π equal to six. So by setting our value of π equal to six in this version of de Moivreβs theorem and switching the sides of our equation around, we get cos of six π plus π sin of six π is equal to cos π plus π sin π all raised to the sixth power.

However, this is not an expression for cos six π yet, and itβs not in terms of powers of sin π and cos π. So weβre going to need to manipulate this expression even further. And to do this, we need to notice on the right-hand side of our equation, we have the sum of two values raised to an exponent. This is a binomial expression, so we can distribute the exponent over our parentheses by using the binomial formula. And we recall this tells us for any positive integer π, π plus π all raised to the πth power is equal to the sum from π equals zero to π of π choose π times π to the πth power multiplied by π to the power of π minus π.

So we can distribute our exponent of six over the right-hand side of our equation by using binomial expansion. We get that this is equal to six choose zero times cos to the sixth power of π plus six choose one times cos to the fifth power of π multiplied by π sin π plus six choose two times cos to the fourth power of π multiplied by π sin π all squared plus six choose three times cos cubed π multiplied by π sin π all cubed. And we keep adding terms of this form all the way up to six choose six times π sin π all raised to the sixth power. And we can see weβre starting to get closer. We now have an expression in terms of powers of sin π and cos π. However, this is not an expression for cos six π, and we can simplify the right-hand side of our equation.

So weβll simplify the right-hand side of our equation term by term. In our first term, six choose zero is equal to one. So our first term is cos to the sixth power of π. In our second term, six choose one is equal to six. And remember, we have a factor of π, so we get the second term is six π cos to the fifth power of π multiplied by sin π. In our third term, we have six choose two is equal to 15, and we can distribute the square over our parentheses to get π squared times sin squared π. However, remember, π is the square root of negative one, so π squared is going to be equal to negative one. So our third term can be simplified to give us negative 15 cos to the fourth power of π multiplied by sin squared π.

And we can follow the exact same process to simplify the final four terms of our expansion. We get negative 20π cos cubed π times sin cubed π plus 15 cos squared π times sin to the fourth power of π plus six π cos π times sin to the fifth power of π minus sin to the sixth power of π. And remember, by de Moivreβs theorem, we know this is equal to cos six π plus π sin six π. But weβre not yet done. Remember, the question wants us to find an expression only for the cos of six π. And on the left-hand side of our equation, we can see we have a complex number with real part cos six π. So we can find an expression for the cos of six π by taking the real parts of both sides of our equation. These have to be equal.

And on the right-hand side of our equation, the real parts will be the terms without a factor of π. So equating the real parts of both sides of our equation, we get our final answer. cos of six π is equal to cos to the sixth power of π minus 15 cos to the fourth power of π times sin squared π plus 15 cos squared π multiplied by sin to the fourth power of π minus sin to the sixth power of π.