Video: De Moivre’s Theorem to Express Sin 4π‘₯ in Terms of Powers of Sin π‘₯ and Cos π‘₯

Use de Moivre’s theorem to express sin 4πœƒ in terms of power of sin πœƒ and cos πœƒ.

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Video Transcript

Use de Moivre’s theorem to express sin of four πœƒ in terms of powers of sin πœƒ and cos πœƒ.

We begin by recalling what de Moivre’s theorem tells us. It says that 𝑒 to the power of π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. Now, of course, we’re looking to express sin of four πœƒ in terms of powers of sin πœƒ and cos πœƒ. So, how do we achieve this? Well, by simply multiplying πœƒ by four, we see that we can rewrite de Moivre’s theorem as 𝑒 to the power of four π‘–πœƒ equals cos of four πœƒ plus 𝑖 sin of four πœƒ. It’s a good start, but we still have 𝑒 to the power of four π‘–πœƒ.

So, we’re going to go back to the original expression, 𝑒 to the π‘–πœƒ equals cos πœƒ plus 𝑖 sin πœƒ, and we’re going to raise both sides as a power of four. And we can see that 𝑒 to the π‘–πœƒ all to the power of four is actually the same as 𝑒 to the power of four π‘–πœƒ. And this in turn means that cos of four πœƒ plus 𝑖 sin of four πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ to the fourth power. And that’s great because we can now use the binomial theorem to distribute the parentheses on the right-hand side. 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 power of π‘˜.

When 𝑛 is equal to four, we have π‘Ž plus 𝑏 to the fourth power equals π‘Ž to the fourth power plus four choose one π‘Ž cubed 𝑏 plus four choose two π‘Ž squared 𝑏 squared plus four choose three π‘Žπ‘ plus 𝑏 to the fourth power. Now, in fact, four choose one and four choose three are four, and four choose two is equal to six. And so, this is the formula we’re going to use to distribute our parentheses. The first term is cos πœƒ to the fourth power. The second is four cos cubed πœƒ times 𝑖 sin πœƒ, although convention dictates that we write this as four 𝑖 cos cubed πœƒ sin πœƒ.

We then have six cos squared πœƒ 𝑖 sin πœƒ squared. And since we can write 𝑖 sin πœƒ squared as 𝑖 squared sin squared πœƒ and 𝑖 squared equals to negative one, this whole term becomes negative six cos squared πœƒ sin squared πœƒ. Our fourth term is four cos πœƒ 𝑖 sin πœƒ cubed. And then, we write 𝑖 sin πœƒ cubed as 𝑖 cubed sin cubed πœƒ. We then write 𝑖 cubed as 𝑖 times 𝑖 squared. We said that 𝑖 squared was equal to negative one, so this is simply negative 𝑖. And then, this time is simply negative four 𝑖 cos πœƒ sin cubed πœƒ.

Our final term is 𝑖 sin πœƒ to the fourth power. We distribute the parentheses and write this as 𝑖 to the fourth power times sin πœƒ to the fourth power. 𝑖 to the fourth power, though, is 𝑖 squared times 𝑖 which is negative one times negative one, which is simply one. This means the final term in the distribution of the parentheses cos πœƒ plus 𝑖 sin πœƒ to the fourth power is sin πœƒ to the fourth power. Now, going back to the question, we want to express sin four πœƒ in terms of powers of sin πœƒ and cos πœƒ. And if we notice on the left-hand side that sin πœƒ is the coefficient of 𝑖, that gives us a pretty good hint of what we’re going to do next. We’re going to equate the imaginary parts on both sides of our equation.

We just saw that the imaginary part on the left-hand side, the coefficient of 𝑖, is sin four πœƒ. On the right-hand side, we have four cos cubed πœƒ sin πœƒ minus four cos πœƒ sin cubed πœƒ. And so, we can say that the coefficients of 𝑖 on the left-hand side must be equal to the coefficient of 𝑖 on the right. That is, sin four πœƒ equals four cos cubed πœƒ sin πœƒ minus four cos πœƒ sin cubed πœƒ. And so, we’re done. We’ve used de Moivre’s theorem to express sin four πœƒ in terms of powers of sin and cos. It’s sin four πœƒ equals four cos cubed πœƒ sin πœƒ minus four cos πœƒ sin cubed πœƒ.

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