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Use Eulerβs formula two drive a formula for cos two π and sin two π in terms of sin π and cos π.
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There are actually two methods we could use to derive the formulae for cos two π and sin two π.
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The first is to consider this expression; itβs π to the ππ plus π.
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We know that this must be the same as π to the ππ times π to the ππ.
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Weβre going to apply Eulerβs formula to both parts of this equation.
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On the left-hand side, we can see that π to the ππ plus π is equal to cos π plus π plus π sin of π plus π.
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And on the right, we have cos π plus π sin π times cos π plus π sin π.
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Weβre going to distribute the parentheses on the right-hand side.
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And when we do, we get the given expression.
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Remember though π squared is equal to negative one.
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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.
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Our next step is to equate the real and imaginary parts of the equation.
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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.
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And so, we see that cos π plus π is equal to cos π cos π minus sin π sin π.
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Next, we equate the imaginary parts.
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On the left-hand side, we have sin π plus π.
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And on the right, we have cos π sin π plus cos π sin π.
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And we can see then that sin π plus π is equal to cos π sin π plus cos π sin π.
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Now, these two formulae are useful in their own right.
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But what we could actually do is replace π with π and we get the double angle formulae.
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In the first one, we get cos two π equals cos squared π minus sin squared π.
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And with our second identity, we get sin two π equals two cos π sin π.
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And there is an alternative approach we could have used.
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This time, we could have gone straight to the double angle formulae by choosing the expression π to the two ππ and then writing that as π to the ππ squared.
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This time, when we apply Eulerβs formula, on the left-hand side, we get cos two π plus π sin two π.
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And on the right-hand side, we get cos π plus π sin π all squared.
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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.
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And once again, π squared is equal to negative one.
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So we can rewrite the right-hand side as cos squared π minus sin squared π plus two π cos π sin π.
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This time when we equate the real parts, we see that cos two π equals cos squared π minus sin squared π.
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And when we equate the imaginary parts, we see that sin two π is equal to two cos π sin π.
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Now, youβve probably noticed there isnβt a huge amount of difference in these two methods.
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The latter is slightly more succinct.
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However, the former has the benefit of deriving those extra identities for cosine and sine.
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Itβs also useful to know that we can incorporate the binomial theorem to derive multiple angle formulae in sine and cosine.