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

Use Euler’s formula two drive a formula for cos 2πœƒ and sin 2πœƒ in terms of sin πœƒ and cos πœƒ.

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Video Transcript

Use Euler’s formula two drive a formula for cos two πœƒ and sin two πœƒ in terms of sin πœƒ and cos πœƒ.

There are actually two methods we could use to derive the formulae for cos two πœƒ and sin two πœƒ. The first is to consider this expression; it’s 𝑒 to the π‘–πœƒ plus πœ™. We know that this must be the same as 𝑒 to the π‘–πœƒ times 𝑒 to the π‘–πœ™. We’re going to apply Euler’s formula to both parts of this equation. On the left-hand side, we can see that 𝑒 to the π‘–πœƒ plus πœ™ is equal to cos πœƒ plus πœ™ plus 𝑖 sin of πœƒ plus πœ™. And on the right, we have cos πœƒ plus 𝑖 sin πœƒ times cos πœ™ plus 𝑖 sin πœ™. We’re going to distribute the parentheses on the right-hand side. And when we do, we get the given expression. Remember though 𝑖 squared is equal to negative one. And we can simplify and we get cos πœƒ cos πœ™ minus sin πœƒ sin πœ™ plus 𝑖 cos πœƒ sin πœ™ plus 𝑖 cos πœ™ sin πœƒ.

Our next step is to equate the real and imaginary parts of the equation. On the left-hand side, the real part is cos πœƒ plus πœ™ and on the right is cos πœƒ cos πœ™ minus sin πœƒ sin πœ™. And so, we see that cos πœƒ plus πœ™ is equal to cos πœƒ cos πœ™ minus sin πœƒ sin πœ™. Next, we equate the imaginary parts. On the left-hand side, we have sin πœƒ plus πœ™. And on the right, we have cos πœƒ sin πœ™ plus cos πœ™ sin πœƒ. And we can see then that sin πœƒ plus πœ™ is equal to cos πœƒ sin πœ™ plus cos πœ™ sin πœƒ.

Now, these two formulae are useful in their own right. But what we could actually do is replace πœ™ with πœƒ and we get the double angle formulae. In the first one, we get cos two πœƒ equals cos squared πœƒ minus sin squared πœƒ. And with our second identity, we get sin two πœƒ equals two cos πœƒ sin πœƒ. And there is an alternative approach we could have used. This time, we could have gone straight to the double angle formulae by choosing the expression 𝑒 to the two π‘–πœƒ and then writing that as 𝑒 to the π‘–πœƒ squared. This time, when we apply Euler’s formula, on the left-hand side, we get cos two πœƒ plus 𝑖 sin two πœƒ. And on the right-hand side, we get cos πœƒ plus 𝑖 sin πœƒ all squared.

Then, distributing these parentheses, we see that the right-hand side becomes cos squared πœƒ plus two 𝑖 cos πœƒ sin πœƒ plus 𝑖 squared sin squared πœƒ. And once again, 𝑖 squared is equal to negative one. So we can rewrite the right-hand side as cos squared πœƒ minus sin squared πœƒ plus two 𝑖 cos πœƒ sin πœƒ. This time when we equate the real parts, we see that cos two πœƒ equals cos squared πœƒ minus sin squared πœƒ. And when we equate the imaginary parts, we see that sin two πœƒ is equal to two cos πœƒ sin πœƒ.

Now, you’ve probably noticed there isn’t a huge amount of difference in these two methods. The latter is slightly more succinct. However, the former has the benefit of deriving those extra identities for cosine and sine. It’s also useful to know that we can incorporate the binomial theorem to derive multiple angle formulae in sine and cosine.

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