Video: Using Euler’s Formula to Derive Formulas for Trigonometric Functions

1. Use Euler’s formula to derive a formula for cos 4πœƒ in terms of cos πœƒ. 2. Use Euler’s formula to drive a formula for sin 4πœƒ in terms of cos πœƒ and sin πœƒ.

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Video Transcript

1) Use Euler’s formula to derive a formula for cos of four πœƒ in terms of cos πœƒ. 2) Use Euler’s formula to drive a formula for sin of four πœƒ in terms of cos πœƒ and sin πœƒ.

For part one, we’ll use the properties of the exponential function. And we’ll write 𝑒 to the four π‘–πœƒ as 𝑒 to the π‘–πœƒ to the power of four. And now, we can use Euler’s formula. And we write the left-hand side as cos four πœƒ plus 𝑖 sin four πœƒ. And on the right-hand side, we can say that this is equal to cos πœƒ plus 𝑖 sin πœƒ all to the power of four. Now, we’re going to apply the binomial theorem to distribute cos πœƒ plus 𝑖 sin πœƒ to the power of four.

In our equation, π‘Ž is equal to cos of πœƒ, 𝑏 is equal to 𝑖 sin of πœƒ, and 𝑛 is the power; it’s four. And this means we can say that cos πœƒ plus 𝑖 sin πœƒ to the power of four is the same as cos πœƒ to the power of four plus four choose one cos cubed πœƒ times 𝑖 sin πœƒ 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 𝑖 squared is negative one, 𝑖 cubed is negative 𝑖, and 𝑖 to the power of four is one. And we can further rewrite our equation as shown.

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

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

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

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