Video Transcript
Using Euler’s formula, derive a formula for cos of three 𝜃 and sin of three 𝜃 in terms of sin 𝜃 and cos 𝜃.
We recall that Euler’s formula says that 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃. So, how do we use this to derive a formula for cos of three 𝜃 and sin of three 𝜃? Well, let’s begin by cubing both sides of our equation. When we do, on this left-hand side, we see that 𝑒 to the 𝑖𝜃 all to the power of three is equal to 𝑒 to the power of three 𝑖𝜃. But of course, using Euler’s formula, we can write this as cos of three 𝜃 plus 𝑖 sin of three 𝜃. So, cos of three 𝜃 plus 𝑖 sin of three 𝜃 is equal to cos of 𝜃 plus 𝑖 sin of 𝜃 all cubed.
Now, on this right-hand side, we can also use the binomial theorem to distribute our parentheses. This says that 𝑎 plus 𝑏 to the 𝑛th power is equal to the sum from 𝑘 equals zero to 𝑛 of 𝑛 choose 𝑘 times 𝑎 to the power of 𝑛 minus 𝑘 times 𝑏 to the 𝑘th power. When 𝑛 is equal to three, this is 𝑎 cubed plus three choose one 𝑎 squared 𝑏 plus three choose two 𝑎𝑏 squared plus 𝑏 cubed.
And actually, three choose one and three choose two are equal to three. So, we have 𝑎 cubed plus three 𝑎 squared 𝑏 plus three 𝑎𝑏 squared plus 𝑏 cubed. And so, we find that the first term in the expansion of cos 𝜃 plus 𝑖 sin 𝜃 cubed is cos cubed 𝜃. Our second term is three cos squared 𝜃 𝑖 sin 𝜃, which we’ll choose to rewrite as three 𝑖 cos squared 𝜃 sin 𝜃. Our third term is three cos 𝜃 times 𝑖 sin 𝜃 squared.
We’ll distribute the exponent over this set of parentheses and we get 𝑖 squared sin squared 𝜃. But of course, 𝑖 squared is equal to negative one. So, this third term becomes negative three cos 𝜃 sin squared 𝜃. Then, the fourth and final term is 𝑖 sin 𝜃 all cubed. This is 𝑖 cubed times sin cubed 𝜃. And we can write 𝑖 cubed as 𝑖 times 𝑖 squared, which is of course negative one 𝑖 or negative 𝑖. And our fourth term becomes negative 𝑖 times sin cubed 𝜃. So, we see that cos three 𝜃 plus 𝑖 sin three 𝜃 is equal to cos cubed 𝜃 plus three 𝑖 cos squared sin 𝜃 minus three cos 𝜃 sin squared 𝜃 minus 𝑖 sin cubed 𝜃. We’re now going to compare real and imaginary parts.
On the left-hand side, the real part is cos three 𝜃, whereas the imaginary part is the coefficient of 𝑖. It’s sin three 𝜃. On the right-hand side, the real parts are cos cubed 𝜃 minus three cos 𝜃 sin squared 𝜃. And on the right-hand side — and remember, we’re interested in the coefficient of 𝑖 — we have three cos squared 𝜃 sin 𝜃 minus sin cubed 𝜃. So, comparing the real parts, we see that cos three 𝜃 must be equal to cos cubed 𝜃 minus three cos 𝜃 sin squared 𝜃. And comparing the imaginary parts, we find that sin three 𝜃 equals three cos squared 𝜃 sin 𝜃 minus sin cubed 𝜃.
Now, we could leave these like this, but it’s ideal to express cos three 𝜃 purely in terms of cos and sin three 𝜃 purely in terms of sin. So, we use the well-known trigonometric identity cos squared 𝜃 plus sin squared 𝜃 equals one. And if we rearrange, we find that sin squared 𝜃 is equal to one minus cos squared 𝜃. Distributing these parentheses, and we find that this is equal to cos cubed 𝜃 minus three cos 𝜃 plus three cos cubed 𝜃. And finally, we collect like terms. And we find that cos of three 𝜃 is equal to four cos cubed 𝜃 minus three cos 𝜃.
Let’s repeat this process for sin three 𝜃. This time, we replace cos squared 𝜃 with one minus sin squared 𝜃. Distributing the parentheses, and this becomes three sin 𝜃 minus three sin cubed 𝜃 minus sin cubed 𝜃, which simplifies to three sin 𝜃 minus four sin cubed 𝜃. And so, we’ve used Euler’s formula to derive a formula for cos three 𝜃 and sin three 𝜃 in terms of sin 𝜃 and cos 𝜃. Cos three 𝜃 is equal to four cos cubed 𝜃 minus three cos 𝜃, whereas sin three 𝜃 is equal to three sin 𝜃 minus four sin cubed 𝜃.
