Question Video: De Moivre’s Theorem to Express Sin 4𝑥 in Terms of Powers of Sin 𝑥 and Cos 𝑥 | Nagwa Question Video: De Moivre’s Theorem to Express Sin 4𝑥 in Terms of Powers of Sin 𝑥 and Cos 𝑥 | Nagwa

# Question Video: De Moivreβs Theorem to Express Sin 4π₯ in Terms of Powers of Sin π₯ and Cos π₯ Mathematics • Third Year of Secondary School

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Use de Moivreβs theorem to express sin 4π in terms of power of sin π and cos π.

03:18

### Video Transcript

Use de Moivreβs theorem to express sin of four π in terms of powers of sin π and cos π.

We begin by recalling what de Moivreβs theorem tells us. It says that π to the power of ππ is equal to cos π plus π sin π. Now, of course, weβre looking to express sin of four π in terms of powers of sin π and cos π. So, how do we achieve this? Well, by simply multiplying π by four, we see that we can rewrite de Moivreβs theorem as π to the power of four ππ equals cos of four π plus π sin of four π. Itβs a good start, but we still have π to the power of four ππ.

So, weβre going to go back to the original expression, π to the ππ equals cos π plus π sin π, and weβre going to raise both sides as a power of four. And we can see that π to the ππ all to the power of four is actually the same as π to the power of four ππ. And this in turn means that cos of four π plus π sin of four π is equal to cos π plus π sin π to the fourth power. And thatβs great because we can now use the binomial theorem to distribute the parentheses on the right-hand side. This says that π plus π to the πth power is equal to the sum from π equals zero to π of π choose π times π to the power of π minus π times π to the power of π.

When π is equal to four, we have π plus π to the fourth power equals π to the fourth power plus four choose one π cubed π plus four choose two π squared π squared plus four choose three ππ plus π 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 weβre going to use to distribute our parentheses. The first term is cos π to the fourth power. The second is four cos cubed π times π sin π, although convention dictates that we write this as four π cos cubed π sin π.

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

Our final term is π sin π to the fourth power. We distribute the parentheses and write this as π to the fourth power times sin π to the fourth power. π to the fourth power, though, is π squared times π which is negative one times negative one, which is simply one. This means the final term in the distribution of the parentheses cos π plus π sin π to the fourth power is sin π to the fourth power. Now, going back to the question, we want to express sin four π in terms of powers of sin π and cos π. And if we notice on the left-hand side that sin π is the coefficient of π, that gives us a pretty good hint of what weβre going to do next. Weβ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 π, is sin four π. On the right-hand side, we have four cos cubed π sin π minus four cos π sin cubed π. And so, we can say that the coefficients of π on the left-hand side must be equal to the coefficient of π on the right. That is, sin four π equals four cos cubed π sin π minus four cos π sin cubed π. And so, weβre done. Weβve used de Moivreβs theorem to express sin four π in terms of powers of sin and cos. Itβs sin four π equals four cos cubed π sin π minus four cos π sin cubed π.

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