Express sin cubed 𝜃 in terms of the sines of multiples of 𝜃.
In this question, we need to find an expression for the sin cubed of 𝜃 in terms of sines of multiples of our angle 𝜃. We might be tempted to try using things like our double-angle formula for sine to try and rewrite sine cubed in terms of this. And this might work; however, it’s difficult and will require lots of experimentation. And in fact, there’s an easier method.
And to do this, we’re going to first need to recall de Moivre’s theorem. This 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 𝑛th power is equal to the cos of 𝑛𝜃 plus 𝑖 sin of 𝑛𝜃. And this is just one version of the theorem. There are several different equivalent statements. And there are some very useful direct results of this theorem that will help us answer this question. And these are worth committing to memory.
If we call our complex number cos 𝜃 plus 𝑖 sin 𝜃 equal to 𝑍, then by de Moivre’s theorem 𝑍 to the 𝑛th of power is the cos of 𝑛𝜃 plus 𝑖 times the sin of 𝑛𝜃. And we can prove the following two useful results for any integer value of 𝑛. 𝑍 to the 𝑛th power plus one over 𝑍 to the 𝑛th power will be equal to two times the cos of 𝑛𝜃 and 𝑍 to the 𝑛th power minus one over 𝑍 to the 𝑛th power will be two 𝑖 sin of 𝑛𝜃.
To prove these two results, we just remember that one over 𝑍 to the 𝑛th power is equal to 𝑍 to the power of negative 𝑛. And we can apply de Moivre’s theorem for any integer value of 𝑛. The result we’re interested in here is the bottom result. And the reason for this is it’s going to help us find an expression for the sin cubed of 𝜃. Let’s start by setting our value of 𝑛 equal to one. By setting our value of 𝑛 equal to one in our bottom result, we get two 𝑖 sin of 𝜃 will be equal to 𝑍 minus one over 𝑍.
Of course, in the question, we’re looking for an expression for the sin cubed of 𝜃. So, we’re going to have to cube both sides of this expression. We can then rearrange to find an expression for the sin cubed of 𝜃. Let’s start by simplifying the left-hand side of this expression. We’re going to want to cube each of our factors separately. This gives us eight times 𝑖 cubed multiplied by sin cubed of 𝜃. And on the right-hand side of our expression, we can see inside our parentheses we have two numbers, and we’re raising this to an exponent. This is a binomial expression.
So, we could write this out in full and multiply this by using the FOIL method or we could use the binomial theorem. Remember, this tells us 𝑎 plus 𝑏 all raised to the 𝑛th power will be 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, where 𝑚 is a positive integer. We’ll do this by using the binomial theorem. Expanded out, this gives us three choose zero times 𝑍 cubed plus three choose one times 𝑍 squared multiplied by negative one over 𝑍 plus three choose two times 𝑍 multiplied by negative one over 𝑍 all squared plus three choose three multiplied by negative one over 𝑍 all cubed.
And now, we can start simplifying. First, remember that 𝑖 is the square root of negative one. So, 𝑖 squared is equal to negative one. This means that 𝑖 cubed is just equal the negative 𝑖. So, we can simplify the left-hand side of our equation to give us negative eight 𝑖 times the sin cubed of 𝜃. And we can simplify each term on the right-hand side of our equation separately. First, three choose zero is just equal to one, so our first term is 𝑍 cubed.
To simplify our next term, recall that negative one over 𝑍 can be written as negative 𝑍 to the power of negative one. And remember, to multiply two terms of the same base together, we just add their exponents. And in our exponent, two minus one is equal to one. So, we get 𝑍 to the first power, which is just 𝑍. And the coefficient will be three choose one multiplied by negative one, which is negative three.
We can do exactly the same to evaluate our next term. However, this time, we have negative one over 𝑍 all squared. This is just going to simplify to give us 𝑍 to the power of negative two. So, 𝑍 times 𝑍 to the power of negative two is equal to 𝑍 to the power of negative one. And our coefficient is just three choose two, which we know is equal to three. So, our third term is three 𝑍 to the power of negative one.
Finally, in our last term, we can distribute the cube over our parentheses. We get negative one over 𝑍 cubed. And three choose three is just equal to one. So, our final term is negative one over 𝑍 cubed. So, our equation now reads negative eight 𝑖 sin cubed 𝜃 is equal to 𝑍 cubed minus three 𝑍 plus three 𝑍 to the power of negative one minus one over 𝑍 cubed. And we could notice something interesting. We have 𝑍 cubed minus one over 𝑍 cubed. And this is exactly one of our statements for de Moivre’s theorem with our value of 𝑛 set to be three. And in fact, this isn’t the only time this appears. Remember, we can rewrite 𝑍 to the power of negative one as one over 𝑍.
So, we can rewrite our third term as three divided by 𝑍. And we want to write this in the form 𝑍 to the 𝑛th power minus one over 𝑍 to the 𝑛th power. We want our value of 𝑛 to be one. And we can either take out a factor of three or a factor of negative three. We’ll take out the factor of negative three. Taking out our factor of negative three and rearranging our equation, we get negative eight 𝑖 sin cubed 𝜃 is equal to 𝑍 cubed minus one over 𝑍 cubed minus three times 𝑍 minus one over 𝑍.
We’re now ready to start writing this expression in terms of sines of multiples of 𝜃. And to do this, we’re going to once again need to use De Moivre’s theorem. By setting our value of 𝑛 equal to three, we know that 𝑍 cubed minus one over 𝑍 cubed is two 𝑖 times the sin of three 𝜃. And by setting our value of 𝑛 equal to one, we know 𝑍 minus one over 𝑍 is two 𝑖 sin 𝜃.
Substituting these expressions in and remembering we need to multiply two 𝑖 sin 𝜃 by negative three, we get the following equation. We get that negative eight 𝑖 sin cubed of 𝜃 is equal to two 𝑖 sin of three 𝜃 minus six 𝑖 sin 𝜃. Now, all we need to do is rearrange this equation for sin cubed 𝜃. To do this, we’re going to need to divide through both sides of our equation by negative eight 𝑖. To make this easier, to simplify, we’ll divide each term on the right-hand side of our equation separately by negative eight 𝑖. We get sin cubed 𝜃 is equal to two 𝑖 sin three 𝜃 all over negative eight 𝑖 plus negative six 𝑖 sin 𝜃 all divided by negative eight 𝑖.
And now, we can start simplifying. In both terms, we have a shared factor of 𝑖 in our numerator and denominator. Similarly, both the first and second term share a factor of two in both the numerator and our denominator. So, we can cancel this out. Finally, in our second term, we can cancel the shared factor of negative one in the numerator and denominator. This leaves us with sin three 𝜃 all over negative four plus three sin 𝜃 all over four. We could leave our answer like this. However, we’ll rearrange these two terms and we’ll take out the shared factor of one-quarter, giving us our final answer of one-quarter multiplied by three sin 𝜃 minus the sin of three 𝜃.
Therefore, in this question, we were able to use de Moivre’s theorem to express sin cubed 𝜃 in terms of the sines of multiples of 𝜃. We showed sin cubed 𝜃 was equal to one-quarter times three sin 𝜃 minus the sin of three 𝜃.