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In this video, we’ll explore the derivation of a number of trigonometric identities using Euler’s formula.
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There’s a very good chance you will have already worked with some of these identities extensively but perhaps aren’t quite sure where they come from.
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So in this lesson, we’ll see how Euler’s formula links to the double angle formulae, additional multiple angle formulae, and the product to sum formulae.
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We begin by recalling Euler’s formula, sometimes called Euler’s relation.
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This states that for a real number 𝜃, 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃.
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Here, 𝑖 is the imaginary unit denoted as the solution to the equation 𝑥 squared equals negative one and 𝜃 must be a real number given in radians.
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As we can see, the formula provides a powerful connection between complex analysis and trigonometry.
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But it also has many applications in physics, engineering, and quantum mechanics.
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In our first example, we’ll see how to derive a well-used trigonometric identity by considering the properties of the exponential function and Euler’s formula.
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1) Use Euler’s formula to express 𝑒 to the negative 𝑖𝜃 in terms of sine and cosine.
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2) Given that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 equals one, what trigonometric identity can be derived by expanding the exponential in terms of trigonometric functions?
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For part one, we’ll begin by rewriting 𝑒 to the negative 𝑖𝜃.
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It’s the same as 𝑒 to the 𝑖 times negative 𝜃.
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We can now apply Euler’s formula.
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Since Euler’s formula says that 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can see that 𝑒 to the 𝑖 negative 𝜃 is equal to cos of negative 𝜃 plus 𝑖 sin of negative 𝜃.
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And then, we recall the properties of the cosine and sine functions.
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Cos is an even function.
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So cos of negative 𝜃 is equal to cos of 𝜃.
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Sin, however, is an odd function.
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So sin of negative 𝜃 is the same as negative sin 𝜃.
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And we can therefore rewrite our expression.
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And we see that 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 minus 𝑖 sin 𝜃.
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Now, let’s consider part two of this question.
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We’re going to use the answer we got from part one.
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When we do, we can see that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 plus 𝑖 sin 𝜃 times cos 𝜃 minus 𝑖 sin 𝜃.
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Let’s distribute these parentheses, perhaps noticing that this is an expression factored using the difference of two squares.
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cos 𝜃 times cos 𝜃 is cos squared 𝜃.
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cos 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 cos 𝜃 sin 𝜃.
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We then get plus 𝑖 sin 𝜃 cos 𝜃 and 𝑖 sin 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 squared sin squared 𝜃.
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We see then negative 𝑖 cos 𝜃 sin 𝜃 plus 𝑖 cos 𝜃 sin 𝜃 is zero.
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And of course, we know that 𝑖 squared is equal to negative one.
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So we can simplify this somewhat.
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And we see that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is cos squared 𝜃 plus sin squared 𝜃.
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We were told however that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 was equal to one.
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So you can see that we’ve derived the formula sin squared 𝜃 plus cos squared 𝜃 equals one.
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This is a fairly succinct derivation of the trigonometric identity sin squared 𝜃 plus cos squared 𝜃 equals one.
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We can perform a similar process to help us derive the double angle formulae.
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Let’s see what that might look like.
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Use Euler’s formula two drive a formula for cos two 𝜃 and sin two 𝜃 in terms of sin 𝜃 and cos 𝜃.
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There are actually two methods we could use to derive the formulae for cos two 𝜃 and sin two 𝜃.
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The first is to consider this expression; it’s 𝑒 to the 𝑖𝜃 plus 𝜙.
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We know that this must be the same as 𝑒 to the 𝑖𝜃 times 𝑒 to the 𝑖𝜙.
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We’re going to apply Euler’s formula to both parts of this equation.
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On the left-hand side, we can see that 𝑒 to the 𝑖𝜃 plus 𝜙 is equal to cos 𝜃 plus 𝜙 plus 𝑖 sin of 𝜃 plus 𝜙.
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And on the right, we have cos 𝜃 plus 𝑖 sin 𝜃 times cos 𝜙 plus 𝑖 sin 𝜙.
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We’re 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 𝑖 squared is equal to negative one.
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And we can simplify and we get cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙 plus 𝑖 cos 𝜃 sin 𝜙 plus 𝑖 cos 𝜙 sin 𝜃.
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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 𝜃 plus 𝜙 and on the right is cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙.
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And so, we see that cos 𝜃 plus 𝜙 is equal to cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙.
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Next, we equate the imaginary parts.
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On the left-hand side, we have sin 𝜃 plus 𝜙.
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And on the right, we have cos 𝜃 sin 𝜙 plus cos 𝜙 sin 𝜃.
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And we can see then that sin 𝜃 plus 𝜙 is equal to cos 𝜃 sin 𝜙 plus cos 𝜙 sin 𝜃.
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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 𝜙 with 𝜃 and we get the double angle formulae.
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In the first one, we get cos two 𝜃 equals cos squared 𝜃 minus sin squared 𝜃.
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And with our second identity, we get sin two 𝜃 equals two cos 𝜃 sin 𝜃.
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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 𝑒 to the two 𝑖𝜃 and then writing that as 𝑒 to the 𝑖𝜃 squared.
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This time, when we apply Euler’s formula, on the left-hand side, we get cos two 𝜃 plus 𝑖 sin two 𝜃.
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And on the right-hand side, we get cos 𝜃 plus 𝑖 sin 𝜃 all squared.
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Then, distributing these parentheses, we see that the right-hand side becomes cos squared 𝜃 plus two 𝑖 cos 𝜃 sin 𝜃 plus 𝑖 squared sin squared 𝜃.
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And once again, 𝑖 squared is equal to negative one.
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So we can rewrite the right-hand side as cos squared 𝜃 minus sin squared 𝜃 plus two 𝑖 cos 𝜃 sin 𝜃.
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This time when we equate the real parts, we see that cos two 𝜃 equals cos squared 𝜃 minus sin squared 𝜃.
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And when we equate the imaginary parts, we see that sin two 𝜃 is equal to two cos 𝜃 sin 𝜃.
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Now, you’ve probably noticed there isn’t 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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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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The binomial theorem says that for integer values of 𝑛, we can write 𝑎 plus 𝑏 to the power of 𝑛 as 𝑎 to the power of 𝑛 plus 𝑛 choose one 𝑎 to the power of 𝑛 minus one 𝑏.
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And we continue this pattern with descending powers of 𝑎 and ascending powers of 𝑏 all the way through to 𝑏 to the power of 𝑛.
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Our next example is going to use the binomial theorem to help us evaluate multiple angles in terms of powers of trigonometric functions.
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1) Use Euler’s formula to derive a formula for cos of four 𝜃 in terms of cos 𝜃.
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2) Use Euler’s formula to drive a formula for sin of four 𝜃 in terms of cos 𝜃 and sin 𝜃.
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For part one, we’ll use the properties of the exponential function.
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And we’ll write 𝑒 to the four 𝑖𝜃 as 𝑒 to the 𝑖𝜃 to the power of four.
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And now, we can use Euler’s formula.
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And we write the left-hand side as cos four 𝜃 plus 𝑖 sin four 𝜃.
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And on the right-hand side, we can say that this is equal to cos 𝜃 plus 𝑖 sin 𝜃 all to the power of four.
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Now, we’re going to apply the binomial theorem to distribute cos 𝜃 plus 𝑖 sin 𝜃 to the power of four.
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In our equation, 𝑎 is equal to cos of 𝜃, 𝑏 is equal to 𝑖 sin of 𝜃, and 𝑛 is the power; it’s four.
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And this means we can say that cos 𝜃 plus 𝑖 sin 𝜃 to the power of four is the same as cos 𝜃 to the power of four plus four choose one cos cubed 𝜃 times 𝑖 sin 𝜃 and so on.
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We know that four choose one is four, four choose two is six, and four choose three is also four.
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We also know that 𝑖 squared is negative one, 𝑖 cubed is negative 𝑖, and 𝑖 to the power of four is one.
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And we can further rewrite our equation as shown.
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Now, we’re going to equate the real parts of this equation.
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And that will give us a formula for cos of four 𝜃 in terms of cos 𝜃 and sin 𝜃.
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Let’s clear some space.
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The real part on the left-hand side is cos four 𝜃.
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And then on the right-hand side, we have cos 𝜃 to the power of four.
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We’ve got negative cos squared 𝜃 sin squared 𝜃.
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And we’ve got sin 𝜃 to the power of four.
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So we equate these.
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But we’re not quite finished.
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We were asked to derive a formula for cos four 𝜃 in terms of cos 𝜃 only.
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So here, we use the identity cos squared 𝜃 plus sin squared 𝜃 is equal to one.
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And we rearrange this.
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And we say that well, that means that sin squared 𝜃 must be equal to one minus cos squared 𝜃.
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And we can rewrite this as cos 𝜃 to the power of four plus six cos squared 𝜃 times one minus cos squared 𝜃 plus one minus cos squared 𝜃 squared.
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We distribute the parentheses.
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And our final step is to collect like terms.
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And we see that we’ve derived the formula for cos of four 𝜃 in terms of cos 𝜃.
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cos four 𝜃 is equal to eight cos 𝜃 to the power of four minus eight cos squared 𝜃 plus one.
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For part two, we can repeat this process equating the imaginary parts.
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They are sin of four 𝜃 on the left.
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And then on the right, we have four cos cubed 𝜃 sin 𝜃, negative four cos 𝜃 sin cubed 𝜃.
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And we see this sin four 𝜃 must be equal to four cos cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃.
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And we could — if we so wish — factor four cos 𝜃 sin 𝜃.
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And we’ll be left with four cos 𝜃 sin 𝜃 times cos squared 𝜃 minus sin squared 𝜃.
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And last, we’ve been asked to derive a formula for sin four 𝜃 in terms of cos 𝜃 and sin 𝜃.
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You might now see a link between sin four 𝜃 and the double angle formulae.
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Now, an interesting application of Euler’s formula is that we can use it to derive an expression for sin 𝜃 and cos 𝜃 in terms of 𝑒 to the 𝑖𝜃.
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We’ve already seen that we can write 𝑒 to the negative 𝑖𝜃 as cos 𝜃 minus 𝑖 sin 𝜃.
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And since 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can say that the sum of 𝑒 to the 𝑖𝜃 and 𝑒 to the negative 𝑖𝜃 is cos 𝜃 plus 𝑖 sin 𝜃 plus cos 𝜃 minus 𝑖 sin 𝜃.
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Well, this expression on the right-hand side simplifies to two cos 𝜃.
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And we can make cos 𝜃 the subject by dividing through by two.
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And we see we have an expression for cos 𝜃 in terms of powers of 𝑒 to the 𝑖𝜃.
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It’s a half 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃.
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Similarly, we can find the difference.
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And we get cos 𝜃 plus 𝑖 sin 𝜃 minus cos 𝜃 minus 𝑖 sin 𝜃.
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This simplifies to two 𝑖 sin 𝜃.
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This time we divide through by two 𝑖.
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And we can see that sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃.
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At these two formulae have plenty of applications in their own right.
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But for the purposes of this video, we’ll look at one final example.
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And we’ll see how they can be used to derive further trigonometric identities.
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Use Euler’s formula to express sin cubed 𝜃 cos squared 𝜃 in the form 𝑎 sin 𝜃 plus 𝑏 sin three 𝜃 plus 𝑐 sin five 𝜃, where 𝑎, 𝑏, and 𝑐 are constants to be found.
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Hence, find the solutions of sin five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal to zero and less than 𝜋.
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Give your answer in exact form.
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We begin by recalling the fact that sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃.
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And cos 𝜃 is equal to a half times 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃.
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This means we can find the product of sin cubed 𝜃 and cos squared 𝜃.
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We can write it as one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃 cubed times a half times 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared.
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One over two 𝑖 cubed is negative one over eight 𝑖.
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And a half squared is one-quarter.
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So we can rewrite our expression a little bit further.
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We find the product of negative one over eight 𝑖 and a quarter.
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And we get negative one over 32𝑖.
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And we can rewrite the rest of our expression as shown.
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We’re now going to use the binomial theorem to expand each of the sets of parentheses.
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The first part becomes 𝑒 to three 𝑖𝜃 plus three choose one 𝑒 to the two 𝑖𝜃 times negative 𝑒 to the negative 𝑖𝜃 and so on.
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And this simplifies to 𝑒 to the three 𝑖𝜃 minus three 𝑒 to the 𝑖𝜃 plus three 𝑒 to the negative 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃.
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Let’s repeat this process for 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared.
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When we do, we get 𝑒 to the two 𝑖𝜃 plus two 𝑒 to the zero which is just two plus 𝑒 to the negative two 𝑖𝜃.
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We’re going to need to find the product of these two expressions.
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We’ll need to do that really carefully.
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We’ll need to ensure that each term in the first expression is multiplied by each term in the second expression.
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And we can write sin cubed 𝜃 cos squared 𝜃 as shown.
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Now, there’s quite a lot going on here.
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So you might wish to pause the video and double-check your answer against mine.
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We’re going to gather the corresponding powers of 𝑒 together.
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We’ll gather 𝑒 to the five 𝑖𝜃 and 𝑒 to the negative five 𝑖𝜃.
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We’ll collect 𝑒 to the plus and minus three 𝑖𝜃.
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And we’ll gather 𝑒 the 𝑖𝜃 and 𝑒 to the negative 𝑖𝜃.
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Let’s neaten things up somewhat.
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We end up with negative one over 32𝑖 times 𝑒 to the five 𝑖𝜃 minus 𝑒 to the negative five 𝑖𝜃 minus 𝑒 to the three 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃 minus two times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃.
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And now, you might be able to spot why we chose to do this.
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We can now go back to the given formulae.
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Let’s clear some space for the next step.
00:14:04.450 --> 00:14:06.710
We kind of unfactorize a little.
00:14:06.840 --> 00:14:10.240
And we can rewrite sin cubed 𝜃 cos squared 𝜃 as shown.
00:14:10.550 --> 00:14:16.380
And we can therefore replace 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 with sin 𝜃 and so on.
00:14:16.850 --> 00:14:24.550
And we can see that sin cubed 𝜃 cos squared 𝜃 is equal to a 16th times two sin 𝜃 plus sin three 𝜃 minus sin five 𝜃.
00:14:25.130 --> 00:14:30.440
Since 𝑎, 𝑏, and 𝑐 are constants to be found, we can say that 𝑎, the coefficient of sin 𝜃, is one-eighth.
00:14:30.630 --> 00:14:33.310
𝑏, the coefficient of sin three 𝜃, is a 16th.
00:14:33.630 --> 00:14:37.280
And 𝑐, the coefficient sin five 𝜃, is negative one 16th.
00:14:37.600 --> 00:14:39.650
Let’s now consider part two of this question.
00:14:39.940 --> 00:14:44.960
We begin by using our answer to part one and multiplying both sides by 16.
00:14:45.410 --> 00:14:50.850
We then subtract two sin 𝜃 from both sides and multiply through by negative one.
00:14:51.210 --> 00:14:56.660
And we can now see that we’ve got an equation in sin five 𝜃 minus sin three 𝜃.
00:14:56.880 --> 00:15:00.460
We’re told that sin five 𝜃 minus sin three 𝜃 is equal to zero.
00:15:00.690 --> 00:15:05.990
So we let two sin 𝜃 minus 16 sin cubed 𝜃 cos squared 𝜃 be equal to zero.
00:15:06.350 --> 00:15:09.000
And then, we factor by two sin 𝜃.
00:15:09.450 --> 00:15:14.870
Since the product of these two terms is equal to zero, this means that either of these terms must be equal to zero.
00:15:15.250 --> 00:15:24.180
So either two sin 𝜃 is equal to zero and dividing by two, we could see that sin 𝜃 is equal to zero or one minus eight sin squared 𝜃 cos squared 𝜃 is equal to zero.
00:15:24.690 --> 00:15:31.310
Given the interval 𝜃 is greater than or equal to zero and less than 𝜋, we can see that one of our solutions is when 𝜃 is equal to zero.
00:15:31.630 --> 00:15:34.100
We’re going to rewrite our other equations somewhat.
00:15:34.100 --> 00:15:37.320
We know that sin two 𝜃 is equal to two sin 𝜃 cos 𝜃.
00:15:37.720 --> 00:15:42.460
Squaring this, we get sin squared two 𝜃 equals four sin squared 𝜃 cos squared 𝜃.
00:15:42.890 --> 00:15:48.450
And that’s in turn means that our equation is one minus two sin squared two 𝜃 equals zero.
00:15:48.910 --> 00:15:55.550
Rearranging to make sin two 𝜃 the subject, we see that sin two 𝜃 is equal to plus or minus one over root two.
00:15:56.040 --> 00:16:05.080
Starting with the positive square root for 𝜃 in the interval given, we know that sin two 𝜃 is equal to one over root two when 𝜃 is equal to 𝜋 by eight or three 𝜋 by eight.
00:16:05.470 --> 00:16:07.860
Similarly, we can solve for the negative square root.
00:16:07.890 --> 00:16:10.450
And we get five 𝜋 by eight and seven 𝜋 by eight.
00:16:10.900 --> 00:16:20.280
And there are therefore five solutions to the equation sin five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal to zero and less than 𝜋.
00:16:20.670 --> 00:16:26.090
They are zero 𝜋 by eight, three 𝜋 by eight, five 𝜋 by eight, and seven 𝜋 by eight.
00:16:26.850 --> 00:16:32.690
In this video, we’ve seen that we can use Euler’s formula in conjunction with the properties of the exponential functions.
00:16:32.860 --> 00:16:39.160
And we can derive many trigonometric identities such as the Pythagorean identity and multiple angle formulae.
00:16:39.640 --> 00:16:46.730
We also saw the we can use the theorem to express sine and cosine in terms of the complex exponential function as shown.
00:16:47.160 --> 00:16:53.490
We’ve also seen that we can use the identities derived from Euler’s formula to help us simplify expressions.
00:16:53.660 --> 00:16:57.070
And these in turn can help us to solve trigonometric equations.