# Video: Using Eulerβs Formula to Derive Formulas for Trigonometric Functions

1. Use Eulerβs formula to derive a formula for cos 4π in terms of cos π. 2. Use Eulerβs formula to drive a formula for sin 4π in terms of cos π and sin π.

03:05

### Video Transcript

1) Use Eulerβs formula to derive a formula for cos of four π in terms of cos π. 2) Use Eulerβs formula to drive a formula for sin of four π in terms of cos π and sin π.

For part one, weβll use the properties of the exponential function. And weβll write π to the four ππ as π to the ππ to the power of four. And now, we can use Eulerβs formula. And we write the left-hand side as cos four π plus π sin four π. And on the right-hand side, we can say that this is equal to cos π plus π sin π all to the power of four. Now, weβre going to apply the binomial theorem to distribute cos π plus π sin π to the power of four.

In our equation, π is equal to cos of π, π is equal to π sin of π, and π is the power; itβs four. And this means we can say that cos π plus π sin π to the power of four is the same as cos π to the power of four plus four choose one cos cubed π times π sin π and so on. We know that four choose one is four, four choose two is six, and four choose three is also four. We also know that π squared is negative one, π cubed is negative π, and π to the power of four is one. And we can further rewrite our equation as shown.

Now, weβre going to equate the real parts of this equation. And that will give us a formula for cos of four π in terms of cos π and sin π. Letβs clear some space. The real part on the left-hand side is cos four π. And then on the right-hand side, we have cos π to the power of four. Weβve got negative cos squared π sin squared π. And weβve got sin π to the power of four. So we equate these. But weβre not quite finished. We were asked to derive a formula for cos four π in terms of cos π only.

So here, we use the identity cos squared π plus sin squared π is equal to one. And we rearrange this. And we say that well, that means that sin squared π must be equal to one minus cos squared π. And we can rewrite this as cos π to the power of four plus six cos squared π times one minus cos squared π plus one minus cos squared π squared. We distribute the parentheses. And our final step is to collect like terms. And we see that weβve derived the formula for cos of four π in terms of cos π. cos four π is equal to eight cos π to the power of four minus eight cos squared π plus one.

For part two, we can repeat this process equating the imaginary parts. They are sin of four π on the left. And then on the right, we have four cos cubed π sin π, negative four cos π sin cubed π. And we see this sin four π must be equal to four cos cubed π sin π minus four cos π sin cubed π. And we could β if we so wish β factor four cos π sin π. And weβll be left with four cos π sin π times cos squared π minus sin squared π. And last, weβve been asked to derive a formula for sin four π in terms of cos π and sin π. You might now see a link between sin four π and the double angle formulae.