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In this lesson, we’ll learn how to use De Moivre’s theorem to prove trigonometric identities.
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There’s a very good chance you will have been using some of these identities for a significant period of time without really realising where they come from.
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And this lesson will give you some insight into this.
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We’ll begin by recapping how to apply De Moivre’s theorem and the binomial theorem for distributing parentheses before deriving a number of identities and looking at how to use them to solve equations involving trigonometric functions.
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Remember De Moivre’s theorem says that for integer values of 𝑛, a complex number written in polar form 𝑟 cos 𝜃 plus 𝑖 sin 𝜃 to the power of 𝑛 is equal to 𝑟 to the power of 𝑛 times cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃.
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Similarly, the binomial theorem shows us how to evaluate 𝑎 plus 𝑏 all to the power of 𝑛.
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You might also know the binomial series as the expansion of one plus 𝑥 to the power of 𝑛 though we won’t be applying this method during this video.
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Let’s see how we can use these two concepts to derive the multiple angle formula.
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1) Use De Moivre’s theorem to express sin five 𝜃 in terms of powers of sin 𝜃.
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2) By considering the solutions of sin five 𝜃 equals zero, find an exact representation for sin squared of 𝜋 by five.
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To answer part 1) of this question, we’ll actually use the inverse of De Moivre’s theorem to evaluate cos of five 𝜃 plus 𝑖 sin of five 𝜃.
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According to De Moivre’s theorem, this is equal to cos of 𝜃 plus 𝑖 sin 𝜃 all to the power of five.
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So let’s use the binomial theorem to distribute these parentheses.
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We’ll compare cos 𝜃 plus 𝑖 sin 𝜃 to the power of five to the binomial theorem.
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We see that 𝑎 is equal to cos 𝜃, 𝑏 is equal to 𝑖 sin 𝜃, and 𝑛 is equal to five.
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The first term in the expansion of cos 𝜃 plus 𝑖 sin 𝜃 to the power of five is therefore cos 𝜃 to the power of five.
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The second term is five choose one times cos 𝜃 to the power of four times 𝑖 sin 𝜃.
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The third term is five choose two cos cubed 𝜃 plus 𝑖 sin 𝜃 squared.
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And we can evaluate the remaining terms as shown.
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Then, we recall that five choose one is five, five choose two is 10, five choose three is 10 again, and five choose four is also five.
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We also know that 𝑖 squared is going to be negative one, 𝑖 cubed is going to be negative 𝑖, 𝑖 to the power of four is going to be one, and 𝑖 to the power of five is going to be 𝑖.
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And our expressions simplifies as shown.
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And since we initially said that cos five 𝜃 plus 𝑖 sin five 𝜃 was equal to cos 𝜃 plus 𝑖 sin 𝜃 to the power of five, we can equate this entire expression to cos five 𝜃 plus 𝑖 sin five 𝜃.
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Now, remember we’re trying to find an expression for sin five 𝜃.
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So we’re going to equate the imaginary parts on each side of our equation.
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On the left-hand side, that’s simply sin five 𝜃.
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And on the right, it’s five cos 𝜃 to the power of four times sin 𝜃 minus 10 cos squared 𝜃 sin cubed 𝜃 plus sin 𝜃 to the power of five.
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We are going to need to clear a little bit of space here.
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And at this stage, we’re going to recall the fact that sin squared 𝜃 plus cos squared 𝜃 is always equal to one.
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Rearranging this, we see that cos squared 𝜃 is equal to one minus sin squared 𝜃.
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And we’ve done this because it will allow us to replace cos squared 𝜃 and cos 𝜃 to the power of four in our expression because we’re trying to write it in terms of powers of sine.
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And when we do, we see that sin five 𝜃 is equal to five times one minus sin squared 𝜃 squared times sin 𝜃 minus 10 times one minus sin squared 𝜃 sin cubed 𝜃 plus sin 𝜃 to the power of five.
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And distributing these parentheses and simplifying, we see that sin five 𝜃 is equal to 16 sin 𝜃 to the power of five minus 20 sin cubed 𝜃 plus five sin 𝜃.
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Now that we have sin five 𝜃 expressed in terms of powers of sin 𝜃, we can answer part 2) of this question.
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We need to consider the solutions of sin five 𝜃 equals zero.
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We know that sin 𝜃 is equal to zero has solutions at integer multiples of 𝜋.
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So this must mean that sin of five 𝜃 equals zero has solutions when 𝜃 is equal to 𝑛𝜋 over five or integer multiples of 𝜋 over five.
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Now comparing this to our equation in part one, we see that 16 sin 𝜃 to the power of five minus 20 sin cubed 𝜃 plus five sin 𝜃 equals zero must also have solutions when 𝜃 is equal to 𝑛𝜋 over five.
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Let’s perform some manipulation to the expression on the left-hand side of our equation, remembering that our aim is to find the exact value of sin squared 𝜋 by five.
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We’ll begin by factoring sin 𝜃.
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And we see that sin 𝜃 times 16 sin 𝜃 to the power of four minus 20 sin squared 𝜃 plus five must be equal to zero.
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And let’s assume our value for 𝜃 is 𝜋 by five.
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In other words, 𝑛 is equal to one.
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Sin of 𝜋 by five is not equal to zero.
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So for the product of these two parentheses to be equal to zero, this must mean that 16 sin 𝜋 by five to the power of four minus 20 sin squared 𝜋 by five plus five is equal to zero.
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And notice this looks a little bit like a quadratic equation.
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We’ll set 𝑥 as sin squared 𝜋 by five.
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And our quadratic equation is now 16𝑥 squared minus 20𝑥 plus five equals zero.
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And we can solve this equation using any method we like for solving quadratic such as completing the square or the quadratic formula.
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And when we do, we see that 𝑥 is equal to five plus or minus root five all divided by eight.
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And of course, we said that 𝑥 is equal to sin squared of 𝜋 by five.
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So our solutions for sin squared of 𝜋 by five are five plus or minus root five over eight.
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Now in fact, if we check these on our calculator, we see that sin squared 𝜋 by five is five minus root five divided by eight.
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And we’ve answered question 2).
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The exact representation for sin squared of 𝜋 by five is five minus root five over eight.
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We’ve just seen how to use De Moivre’s theorem to evaluate sin of multiples of 𝜃.
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But we can also use the theorem to derive identities for sin 𝜃 to the power of 𝑛 and cos 𝜃 to the power of 𝑛.
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Let’s say we have a complex number 𝑧 which is simply cos 𝜃 plus 𝑖 sin 𝜃.
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We know that the reciprocal of 𝑧 is 𝑧 to the power of negative one.
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And we can use De Moivre’s theorem to evaluate this.
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It’s cos of negative 𝜃 plus 𝑖 sin of negative 𝜃.
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Now, remember cos is an even function.
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So cos of negative 𝜃 is simply the same as 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 the negative sin of 𝜃.
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And we can therefore see that the reciprocal of 𝑧 can be written as cos 𝜃 minus 𝑖 sin 𝜃.
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So why is this useful?
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Well, it allows us to evaluate 𝑧 plus the reciprocal of 𝑧.
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𝑧 plus the reciprocal of 𝑧 is simply two cos 𝜃.
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Similarly, it allows us to evaluate their difference.
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𝑧 minus the reciprocal of 𝑧 is two 𝑖 sin 𝜃.
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And we can actually generalize this for higher powers of 𝑧.
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Using De Moivre’s theorem and applying the odd and even identities for sine and cosine, we get these two equations.
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Rearranging these by dividing the first equation by two and the second by two 𝑖 for a complex number 𝑧 in exponential form, we see that cos of 𝑛𝜃 is equal to one-half 𝑒 to the 𝑖𝑛𝜃 plus 𝑒 to the negative 𝑖𝑛𝜃.
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And sin 𝑛𝜃 is equal to one over two 𝑖 multiplied by 𝑒 to the 𝑖𝑛𝜃 plus 𝑒 to the negative 𝑖𝑛𝜃.
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These formulae are incredibly powerful for deriving a number of trigonometric identities.
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And they should be committed to memory for easy recall.
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Let’s have a look at some examples for where these might be helpful.
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Using De Moivre’s theorem, find the exact value of the integral of sin 𝜃 to the power of seven with respect to 𝜃 between the limits 𝜋 by two and zero.
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We’ll begin by expressing sin 𝜃 to the power of seven in terms of multiple angles since those are quite straightforward to integrate.
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We’ll also use the fact that 𝑧 minus the reciprocal of 𝑧 is equal to two 𝑖 sin 𝜃.
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And since we’re interested in sin 𝜃 to the power of seven, we’re going to raise this entire equation to the power of seven.
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It’s 𝑧 minus one over 𝑧 all to the power of seven and that’s equal to two 𝑖 sin 𝜃 to the power of seven.
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Two to the power of seven is 128 and 𝑖 to the power of seven is negative 𝑖.
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We divide both sides of this equation by negative 128𝑖.
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We’re going to evaluate negative one over 128𝑖 by multiplying both the numerator and the denominator by 𝑖.
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When we do, we get negative 𝑖 over 128𝑖 squared.
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But since 𝑖 squared is equal to negative one, this simplifies to 𝑖 over 128.
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Our next step is to apply the binomial theorem to 𝑧 minus one over 𝑧 all to the power of seven.
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When we use the binomial theorem, we see that sin 𝜃 to the power of seven is as shown.
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We then recall the fact that 𝑧 to the power of 𝑛 minus one over 𝑧 to the power of 𝑛 is two 𝑖 sin 𝑛𝜃.
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And we can see that we can write this expression in terms of multiple angles by gathering powers of 𝑧.
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This means that sin 𝜃 to the power of seven is equal to 𝑖 over ~~28~~ [128] times two 𝑖 sin seven 𝜃 minus seven times two 𝑖 sin five 𝜃 plus 21 times two 𝑖 sin three 𝜃 minus 35 times two 𝑖 sin 𝜃.
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And then fully simplified, we see that this is equal to one over 64 times 35 sin 𝜃 minus 21 sin three 𝜃 plus seven sin five 𝜃 minus sin seven 𝜃.
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Let’s clear some space and replace sin 𝜃 to the power of seven in our integral with this expression.
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We then recall the fact that the integral of sin 𝑛𝜃 with respect to 𝜃 is negative one over 𝑛 times cos 𝑛𝜃 plus obviously that constant of integration.
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And this means our integral is negative 35 cos 𝜃 plus seven cos three 𝜃 minus seven-fifths times cos of five 𝜃 plus one-seventh of cos of seven 𝜃.
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And since we’ll be evaluating this between the limits of 𝜋 by two and zero, we don’t need the constant of integration.
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This becomes one sixty-fourth times 35 minus seven plus seven-fifths minus a seventh which is sixteen thirty-fifths.
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Notice how this process actually made what was quite a tricky integral really quite simple to evaluate.
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And this process isn’t limited to just powers of sine or cosine.
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We can also use it to find expressions for the products of powers of these functions.
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But they also work with the tangent function.
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By recalling the fact that tan 𝜃 is equal to sin 𝜃 divided by cos 𝜃, we can express tan of some integer multiple of 𝜃 in terms of powers of tan.
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Let’s see what this looks like.
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Express tan six 𝜃 in terms of powers of tan 𝜃.
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We begin by recalling the fact that tan 𝜃 is equal to sin 𝜃 divided by cos 𝜃.
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And this in turn means that tan six 𝜃 equals sin six 𝜃 divided by cos six 𝜃.
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So we’ll evaluate sin six 𝜃 and cos six 𝜃 in terms of powers of sine and cosine.
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De Moivre’s theorem says that cos six 𝜃 plus 𝑖 sin six 𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃 all to the power of six.
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We distribute this parenthesis by using the binomial theorem and simplify by evaluating powers of 𝑖.
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And we see that cos 𝜃 plus 𝑖 sin 𝜃 to the power of six is as shown.
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And of course, we said that cos six 𝜃 plus 𝑖 sin six 𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃 to the power of six.
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So we can equate this expansion to cos six 𝜃 plus 𝑖 sin six 𝜃.
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And now we see that we can equate the real and the imaginary parts of the equation.
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The real part of the left-hand side is cos six 𝜃.
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On the right-hand side, we got cos 𝜃 to the power of six, negative 15 cos 𝜃 to the power of four sin squared 𝜃, 15 cos squared 𝜃 times sin 𝜃 to the power of four, and negative sin 𝜃 to the power of six.
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And then equating the imaginary parts, on the left-hand side, we have sin six 𝜃.
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On the right-hand side, we have six cos 𝜃 to the power of five times sin 𝜃, negative 20 cos cubed 𝜃 sin cubed 𝜃, and six cos 𝜃 times sin 𝜃 to the power of five.
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And we’re now ready to evaluate tan of six 𝜃.
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It’s six cos 𝜃 to the power of five times sin 𝜃 minus 20 cos cubed 𝜃 sin cubed 𝜃 plus six cos 𝜃 times sin 𝜃 to the power of five all over cos 𝜃 to the power of six minus 15 cos 𝜃 to the power of four times sin squared 𝜃 plus 15 cos squared 𝜃 sin 𝜃 to the power of four minus sin 𝜃 to the power of six.
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To express this in terms of tan, we’re going to divide everything by cos 𝜃 to the power of six.
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On the numerator, six cos 𝜃 to the power of five times sin 𝜃 divided by cos 𝜃 to the power of six is simply six tan 𝜃.
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We have negative 20 tan cubed 𝜃 and six times tan 𝜃 to the power of five.
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On the denominator, we have one minus 15 tan squared 𝜃 plus 15 tan 𝜃 to the power of four minus tan 𝜃 to the power of six.
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And we’ve successfully expressed tan of six 𝜃 in terms of powers of tan 𝜃.
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In this video, we’ve seen that we can use De Moivre’s theorem and the binomial theorem to derive multiple angle formulae for different sine, cosine, and tangent functions.
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We’ve also seen that for a complex number 𝑧 equals cos 𝜃 plus 𝑖 sin 𝜃, 𝑧 plus the reciprocal of 𝑧 is two cos 𝜃 and 𝑧 minus the reciprocal of 𝑧 is two 𝑖 sin 𝜃.
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And we’ve also extended this idea into 𝑧 to the power of 𝑛.
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We’ve used these equations to find expressions for the powers of sine and cosine.
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And we said that we can even find it for their products.
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We’ve even seen that we can use these techniques for deriving trigonometric identities to simplify more complicated integrals.