Video: Multiple-Angle Formulas from Euler’s Formula

Use Euler’s formula to derive a formula for cos 5πœƒ and sin 5πœƒ in terms of sin πœƒ and cos πœƒ.

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Video Transcript

Use Euler’s formula to derive a formula for cos five πœƒ and sin five πœƒ in terms of sin πœƒ and cos πœƒ.

We recall that Euler’s formula says that 𝑒 to the power of π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. So how do we apply this to derive a formula for cos of five πœƒ and sin of five πœƒ? Well, we’re going to begin by raising both sides of this formula to the fifth power. Now, we can say that 𝑒 to the π‘–πœƒ all to the fifth power is equal to 𝑒 to the five π‘–πœƒ. But then, of course, we could use Euler’s formula to rewrite that as cos five πœƒ plus 𝑖 sin of five πœƒ. And so we have the equation cos of five πœƒ plus 𝑖 sin of five πœƒ equals cos of πœƒ plus 𝑖 sin πœƒ all to the fifth power. And now, we can use the binomial theorem to distribute these parentheses.

This says that π‘Ž plus 𝑏 to 𝑛th power is 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 five, we have π‘Ž plus 𝑏 to the fifth power being equal to π‘Ž to the fifth power plus five choose one π‘Ž to fourth power 𝑏 plus five choose two times π‘Ž cubed times 𝑏 squared and so on. Now in fact, five choose one and five choose four are five, whilst five choose two and five choose three are equal to 10. So we have the following formula to help us distribute the parentheses cos πœƒ plus 𝑖 sin πœƒ to the fifth power. The first term is simply cos πœƒ to the fifth power, whilst the second term is five cos πœƒ to the fourth power times 𝑖 sin πœƒ.

But in fact, let’s move the 𝑖 to the front and write this as five 𝑖 cos πœƒ to the fourth power sin πœƒ. Then the third term is 10 cos cubed πœƒ times 𝑖 sin πœƒ squared, which we can write as 10 cos cubed πœƒ times 𝑖 squared times sin squared πœƒ. But we know that 𝑖 squared is equal to negative one. So we can rewrite this further as negative 10 cos cubed πœƒ sin squared πœƒ. Our fourth term is then 10 cos squared πœƒ times 𝑖 sin πœƒ cubed. And if we consider 𝑖 cubed as equal to 𝑖 times 𝑖 squared, we see we can rewrite this whole expression as negative 10 𝑖 times cos squared πœƒ times sin cubed πœƒ. We then have five cos πœƒ times 𝑖 sin πœƒ to the fourth power. And since 𝑖 to the fourth power is equal to 𝑖 squared squared, that’s negative one squared, which is simply one. And this term becomes five cos πœƒ sin πœƒ to the fourth power.

Our final term is 𝑖 sin πœƒ to the fifth power. 𝑖 to the fifth power is 𝑖 to the fourth power times 𝑖. So we have simply 𝑖 sin πœƒ to the fifth power. And so our equation is now cos five πœƒ plus 𝑖 sin five πœƒ equals cos πœƒ to the fifth power plus five 𝑖 cos πœƒ to the fourth power times sin πœƒ minus 10 cos cubed πœƒ sin squared πœƒ and so on. And we’re now ready to derive a formula for cos of five πœƒ. We do this by equating or comparing the real parts on each side of our equation. on the left-hand side, that’s just cos five πœƒ, whereas on the right-hand side, we have cos πœƒ to the fifth power minus 10 cos cubed πœƒ sin squared πœƒ plus five cos πœƒ sin πœƒ to the fourth power.

Since we know that the real components on each side of our equation must be equal, we create the following equation. And we could leave it like that. But we could also recall that sin squared πœƒ plus cos squared πœƒ equals one. And then by writing sin squared πœƒ is one minus cos squared πœƒ, we find cos five πœƒ is equal to cos πœƒ to the fifth power minus 10 cos cubed πœƒ times one minus cos squared πœƒ plus five cos πœƒ times one minus cos squared πœƒ squared.

Finally, we distribute the parentheses. And we find that the right-hand side of this equation becomes 16 cos πœƒ to the fifth power minus 20 cos cubed πœƒ plus five cos πœƒ. And so we’ve derived our formula for cos of five πœƒ. Now in fact, we repeat this process for sin of five πœƒ. This time, though, we’re going to compare the imaginary parts. On the left-hand side, we have sin five πœƒ, whereas on the right-hand side, we have five cos πœƒ to the fourth power sin πœƒ minus 10 cos squared πœƒ sin cubed πœƒ plus sin πœƒ to the fifth power. And so our equation for sin five πœƒ becomes sin five πœƒ equals five cos πœƒ to the fourth power sin πœƒ minus 10 cos squared πœƒ sin cubed πœƒ plus sin πœƒ to the fifth power. Notice when equating the imaginary parts, we’d no longer need the 𝑖, essentially, when comparing the coefficient of 𝑖.

Now, we could leave it like that or we can create an equation simply in terms of sin. This time, we replace cos squared πœƒ with one minus sin squared πœƒ. So we have five times one minus sin squared πœƒ squared times sin πœƒ minus 10 times one minus sin squared πœƒ times sin cubed πœƒ plus sin πœƒ to the fifth power. Distributing our parentheses, and we find that this is equal to 16 sin πœƒ to the fifth power minus 20 sin cubed πœƒ plus five sin πœƒ. And so we’ve used Euler’s formula to derive formulae for cos of five πœƒ and sin of five πœƒ. Cos of five πœƒ is equal to 16 times cos πœƒ to the fifth power minus 20 cos cubed πœƒ plus five cos πœƒ, whereas sin five πœƒ is equal to 16 times sin πœƒ to the fifth power minus 20 sin cubed πœƒ plus five sin πœƒ.

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