Video Transcript
Which of the following is equivalent to cos three 𝜃? Option (A) cos cubed 𝜃 minus three cos 𝜃 times sin squared 𝜃. Option (B) one minus sin cubed 𝜃. Option (C) cos cubed of 𝜃 plus three cos 𝜃 times the sin squared of 𝜃. Option (D) cos cubed of 𝜃 minus three times the cos squared of 𝜃 times the sin of 𝜃. Or option (E) cos cubed of 𝜃 plus three times the cos squared of 𝜃 multiplied by sin 𝜃.
In this question, we need to find an equivalent expression to the cos of three 𝜃. And we have a few options for doing this. For example, we could write this as the cos of two 𝜃 plus 𝜃 and then use our angle addition formula for cos. Then we could also use the double-angle formula for cos to find an expression for cos of three 𝜃. And in fact, we could do this and it would give us the correct answer. So this is one method we could use. However, there is a second method. And the reason we use the second method is it’s a lot easier to use when we’re dealing with higher coefficients of 𝜃.
To use this method, we’re first going to need to recall de Moivre’s theorem. One version of de Moivre’s theorem tells us for any integer value of 𝑛 and real value 𝜃, the cos of 𝜃 plus 𝑖 sin of 𝜃 all raised to the power of 𝑛 is equal to the cos of 𝑛𝜃 plus 𝑖 times the sin of 𝑛𝜃. And this is particularly useful for helping us find angle identities involving cosine and sine. For example, in this question, we want to find an expression for the cos of three 𝜃. On the right-hand side of our expression, we have the cos of 𝑛𝜃. So we can set 𝑛 equal to three.
And we can see that the cos of 𝑛𝜃 on the right-hand side of our expression is the real part of this expression and the sin of 𝑛𝜃 is the imaginary part of this expression. So we can use this to find an expression for the cos of three 𝜃. So we’ll swap the two sides of this equation and set our value of 𝑛 equal to three. We get cos of three 𝜃 plus 𝑖 sin of three 𝜃 is equal to the cos of 𝜃 plus 𝑖 sin of 𝜃 all cubed. Now we want to simplify the right-hand side of our expression. And we can see that we have the sum of two values all raised to an exponent. This is a binomial. So to distribute the cube over our parentheses, we’re going to want to use the binomial formula.
We recall this tells us for a positive integer value of 𝑚, 𝑎 plus 𝑏 all raised to the to the 𝑚th power is equal to the sum from 𝑘 equals zero to 𝑚 of 𝑚 choose 𝑘 times 𝑎 to the 𝑘th power multiplied by 𝑏 to the power of 𝑚 minus 𝑘. Applying this to the cos of 𝜃 plus 𝑖 sin of 𝜃 all cubed, we get three choose zero times cos cubed of 𝜃 plus three choose one times cos squared of 𝜃 multiplied by 𝑖 sin 𝜃 plus three choose two multiplied by cos 𝜃 times 𝑖 sin 𝜃 all squared plus three choose three multiplied by 𝑖 sin 𝜃 all cubed.
Now we can simplify each term in this expression separately. First, three choose zero is equal to one. So the first term is just cos cubed 𝜃. In our next term, three choose one is equal to three. And remember, we have a factor of 𝑖, so we can rearrange this to get three 𝑖 cos squared 𝜃 times sin 𝜃. In our third term, three choose two is equal to three. And we can distribute the square over our parentheses to get 𝑖 squared times sin squared of 𝜃. But remember, 𝑖 is the square root of negative one. So 𝑖 squared is just equal to negative one.
So we can simplify our third term to be negative three cos of 𝜃 multiplied by the sin squared of 𝜃. And we can do something similar for our fourth and final term. Three choose three is equal to one and we can distribute the cube over our parentheses and simplify to get negative 𝑖 sin cubed of 𝜃. Now remember, by de Moivre’s theorem, this is all equal to the cos of three 𝜃 plus 𝑖 sin of three 𝜃. And we want to find an expression for the cos of three 𝜃, which we can see is the real part of the left-hand side of this equation.
So we can find an expression for the cos of three 𝜃 by just taking the real part of the right-hand side of this equation. That’s going to be all of our terms which don’t have a factor of 𝑖. This gives us cos of three 𝜃 is equal to the cos cubed of 𝜃 minus three cos 𝜃 times the sin squared of 𝜃 which we can see is option (A). Therefore, in this question, we were able to use de Moivre’s theorem to find an equivalent statement to the cos of three 𝜃. We were able to show it was equivalent to cos cubed 𝜃 minus three cos 𝜃 times sin squared 𝜃, which we can see is option (A).
