Video: Triple-Angle Formulas from Euler’s Formula

Using Euler’s formula, derive a formula for cos (3πœƒ) and sin (3πœƒ) in terms of sin πœƒ and cos πœƒ.

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Video Transcript

Using Euler’s formula, derive a formula for cos of three πœƒ and sin of three πœƒ in terms of sin πœƒ and cos πœƒ.

We recall that Euler’s formula says that 𝑒 to the π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. So, how do we use this to derive a formula for cos of three πœƒ and sin of three πœƒ? 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 π‘–πœƒ all to the power of three is equal to 𝑒 to the power of three π‘–πœƒ. But of course, using Euler’s formula, we can write this as cos of three πœƒ plus 𝑖 sin of three πœƒ. So, cos of three πœƒ plus 𝑖 sin of three πœƒ is equal to cos of πœƒ plus 𝑖 sin of πœƒ all cubed.

Now, on this right-hand side, we can also use the binomial theorem to distribute our parentheses. This says that π‘Ž plus 𝑏 to the 𝑛th power is equal to the sum from π‘˜ equals zero to 𝑛 of 𝑛 choose π‘˜ times π‘Ž to the power of 𝑛 minus π‘˜ times 𝑏 to the π‘˜th power. When 𝑛 is equal to three, this is π‘Ž cubed plus three choose one π‘Ž squared 𝑏 plus three choose two π‘Žπ‘ squared plus 𝑏 cubed.

And actually, three choose one and three choose two are equal to three. So, we have π‘Ž cubed plus three π‘Ž squared 𝑏 plus three π‘Žπ‘ squared plus 𝑏 cubed. And so, we find that the first term in the expansion of cos πœƒ plus 𝑖 sin πœƒ cubed is cos cubed πœƒ. Our second term is three cos squared πœƒ 𝑖 sin πœƒ, which we’ll choose to rewrite as three 𝑖 cos squared πœƒ sin πœƒ. Our third term is three cos πœƒ times 𝑖 sin πœƒ squared.

We’ll distribute the exponent over this set of parentheses and we get 𝑖 squared sin squared πœƒ. But of course, 𝑖 squared is equal to negative one. So, this third term becomes negative three cos πœƒ sin squared πœƒ. Then, the fourth and final term is 𝑖 sin πœƒ all cubed. This is 𝑖 cubed times sin cubed πœƒ. 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 πœƒ. So, we see that cos three πœƒ plus 𝑖 sin three πœƒ is equal to cos cubed πœƒ plus three 𝑖 cos squared sin πœƒ minus three cos πœƒ sin squared πœƒ minus 𝑖 sin cubed πœƒ. We’re now going to compare real and imaginary parts.

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

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

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

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