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

# Question Video: Using Eulerβs Formula to Derive Formulas for Trigonometric Functions Mathematics

Use Eulerβs formula two drive a formula for cos 2π and sin 2π in terms of sin π and cos π.

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### Video Transcript

Use Eulerβs formula two drive a formula for cos two π and sin two π in terms of sin π and cos π.

There are actually two methods we could use to derive the formulae for cos two π and sin two π. The first is to consider this expression; itβs π to the ππ plus π. We know that this must be the same as π to the ππ times π to the ππ. Weβre going to apply Eulerβs formula to both parts of this equation. On the left-hand side, we can see that π to the ππ plus π is equal to cos π plus π plus π sin of π plus π. And on the right, we have cos π plus π sin π times cos π plus π sin π. Weβre going to distribute the parentheses on the right-hand side. And when we do, we get the given expression. Remember though π squared is equal to negative one. And we can simplify and we get cos π cos π minus sin π sin π plus π cos π sin π plus π cos π sin π.

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 plus π and on the right is cos π cos π minus sin π sin π. And so, we see that cos π plus π is equal to cos π cos π minus sin π sin π. Next, we equate the imaginary parts. On the left-hand side, we have sin π plus π. And on the right, we have cos π sin π plus cos π sin π. And we can see then that sin π plus π is equal to cos π sin π plus cos π sin π.

Now, these two formulae are useful in their own right. But what we could actually do is replace π with π and we get the double angle formulae. In the first one, we get cos two π equals cos squared π minus sin squared π. And with our second identity, we get sin two π equals two cos π sin π. 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 π to the two ππ and then writing that as π to the ππ squared. This time, when we apply Eulerβs formula, on the left-hand side, we get cos two π plus π sin two π. And on the right-hand side, we get cos π plus π sin π all squared.

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

Now, youβve probably noticed there isnβ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. Itβ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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