# Question Video: Finding a Mutliple Angle Formula for Cosine Mathematics

Express cos 6π in terms of powers of sin π and cos π.

04:12

### Video Transcript

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 Moivreβs theorem.

We recall one part of de Moivreβs 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 πth power is equal to the cos of ππ plus π times the sin of ππ. 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 weβll set our value of π equal to six. So by setting our value of π equal to six in this version of de Moivreβs 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 itβs not in terms of powers of sin π and cos π. So weβre 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 πth power is equal to the sum from π equals zero to π of π choose π times π to the πth 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 weβre 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 weβll 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 Moivreβs theorem, we know this is equal to cos six π plus π sin six π. But weβre 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.