### Video Transcript

Using Eulerβs formula, derive a formula for cos of three π and sin of three π in terms of sin π and cos π.

We recall that Eulerβs formula says that π to the ππ is equal to cos π plus π sin π. So, how do we use this to derive a formula for cos of three π and sin of three π? Well, letβs begin by cubing both sides of our equation. When we do, on this left-hand side, we see that π to the ππ all to the power of three is equal to π to the power of three ππ. But of course, using Eulerβs formula, we can write this as cos of three π plus π sin of three π. So, cos of three π plus π sin of three π is equal to cos of π plus π sin of π all cubed.

Now, on this right-hand side, we can also use the binomial theorem to distribute our parentheses. This says that π plus π to the πth power is equal to the sum from π equals zero to π of π choose π times π to the power of π minus π times π to the πth power. When π is equal to three, this is π cubed plus three choose one π squared π plus three choose two ππ squared plus π cubed.

And actually, three choose one and three choose two are equal to three. So, we have π cubed plus three π squared π plus three ππ squared plus π cubed. And so, we find that the first term in the expansion of cos π plus π sin π cubed is cos cubed π. Our second term is three cos squared π π sin π, which weβll choose to rewrite as three π cos squared π sin π. Our third term is three cos π times π sin π squared.

Weβll distribute the exponent over this set of parentheses and we get π squared sin squared π. But of course, π squared is equal to negative one. So, this third term becomes negative three cos π sin squared π. Then, the fourth and final term is π sin π all cubed. This is π cubed times sin cubed π. And we can write π cubed as π times π squared, which is of course negative one π or negative π. And our fourth term becomes negative π times sin cubed π. So, we see that cos three π plus π sin three π is equal to cos cubed π plus three π cos squared sin π minus three cos π sin squared π minus π sin cubed π. Weβre now going to compare real and imaginary parts.

On the left-hand side, the real part is cos three π, whereas the imaginary part is the coefficient of π. Itβs sin three π. On the right-hand side, the real parts are cos cubed π minus three cos π sin squared π. And on the right-hand side β and remember, weβre interested in the coefficient of π β we have three cos squared π sin π minus sin cubed π. So, comparing the real parts, we see that cos three π must be equal to cos cubed π minus three cos π sin squared π. And comparing the imaginary parts, we find that sin three π equals three cos squared π sin π minus sin cubed π.

Now, we could leave these like this, but itβs ideal to express cos three π purely in terms of cos and sin three π purely in terms of sin. So, we use the well-known trigonometric identity cos squared π plus sin squared π equals one. And if we rearrange, we find that sin squared π is equal to one minus cos squared π. Distributing these parentheses, and we find that this is equal to cos cubed π minus three cos π plus three cos cubed π. And finally, we collect like terms. And we find that cos of three π is equal to four cos cubed π minus three cos π.

Letβs repeat this process for sin three π. This time, we replace cos squared π with one minus sin squared π. Distributing the parentheses, and this becomes three sin π minus three sin cubed π minus sin cubed π, which simplifies to three sin π minus four sin cubed π. And so, weβve used Eulerβs formula to derive a formula for cos three π and sin three π in terms of sin π and cos π. Cos three π is equal to four cos cubed π minus three cos π, whereas sin three π is equal to three sin π minus four sin cubed π.