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Use De Moivreβs theorem to express cos three π and sin three π in terms of cos π and sin π, respectively.
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First, let us recall De Moivreβs theorem.
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This states that cos π plus π sin π raised to the power of π is equal to cos of ππ plus π sin of ππ.
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Looking at the right-hand side of this equation, we see that the two terms are very similar to the two terms that we must express in terms of cos π and sin π, respectively, specifically in the case where π is equal to three.
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Let us rewrite De Moivreβs theorem in the case where π equals three to see this more clearly.
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Now that we have written out this equation, we can see that the right-hand side has a real part, cos three π, and an imaginary part, sin three π.
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We can express both of these parts in terms of cos π and sin π by isolating the real and imaginary parts of the left-hand side of the equation.
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In order to do this, we will need to multiply out our parentheses and separate the terms.
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Looking at the left-hand side of the equation, we can see that we have a binomial to the power of three.
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A general form for this would be π plus π cubed.
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In order to expand our parentheses, we could multiply out each of our terms by hand or use the binomial theorem.
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Instead, weβre going to use a slightly different shortcut.
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Looking at our parentheses, it is fairly easy to convince ourselves that each term will have some power of π and some power of π.
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By simple inspection, we can see that the highest possible power of π that can be reached by multiplying these parentheses is π cubed.
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By similar reasoning, the highest possible power of π that can be reached by multiplying out these parentheses is π cubed.
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Our parentheses will also give us the terms in between these two, which will have decreasing powers of π and increasing powers of π.
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There are multiple ways to reach these in between terms, giving the πs and πs within our parentheses.
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We will hence have a multiple of them.
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In other words, they will have a coefficient.
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In order to find these coefficients, we can take the numbers from the corresponding tier of Pascalβs triangle.
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Since we are dealing with a binomial to the power of three, we will use the third tier.
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The coefficients for our four terms are therefore one π cubed plus three π squared π plus three ππ squared plus one π cubed.
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Now that we have this expansion, letβs apply it to our equation.
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We can substitute in our terms to this expansion by taking π to be cos π and π to be π sin π.
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We first have π cubed, which becomes cos π cubed.
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We next have three π squared π, which becomes three times cos π squared times π sin π.
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We then have three ππ squared, which becomes three times cos π times π sin π squared.
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And finally, we have π cubed, which is π sin π cubed.
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As our next step, weβre going to multiply out the squares and the cubes of our parentheses and move all of our πs to the front of the term.
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Let us now simplify by recalling the definition of π.
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We know that π denotes an imaginary number.
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And itβs the square root of negative one.
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π squared is then the square root of negative one times the square root of negative one, in other words, negative one.
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To make the simplification easier, we can write π cubed as π squared times π.
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Since we now know that π squared is negative one, we can use this in our definition.
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And we can say that π cubed is equal to negative π.
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Let us use these identities to replace the π squared and the π cubed within our equation.
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Here we have simplified our equation.
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We can now see that, of our four terms, we have two that do not have a factor of π or real components.
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And we have two terms that do have a factor of π or imaginary components.
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Let us now group together the real and imaginary terms.
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Weβre now in a position to look back on our original equation derived from De Moivreβs theorem.
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Letβs review.
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We have expanded out the left-hand side of our equation and separated out the real and imaginary terms.
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To proceed, we can equate the real and imaginary parts of our expansion with the real and imaginary parts of the right-hand side of our equation.
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Let us show this now.
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We now have an equation for cos three π and an equation for sin three π.
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However, we have one final simplification to perform.
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Our question requires that we express cos three π in terms of cos π.
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But we still have a sin in this equation.
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Likewise, the question requires that we express sin three π in terms of sin π.
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But we still have a cos in this equation.
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We can substitute these terms by recalling one of the Pythagorean identities.
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This states that cos squared π plus sin squared π is equal to one.
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By subtracting cos squared π from both sides of this equation, we get that sin squared π is equal to one minus cos squared π.
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We can do exactly the same to find cos squared π by subtracting sin squared π from both sides of our original identity.
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Let us work on our equation for cos three π using the first rearrangement of the identity.
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We substitute sin squared π for one minus cos squared π.
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We then multiply out the parentheses.
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Finally, we collect our cos cubed π terms.
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We can now perform the exact same method on the equation for sin three π.
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We replace cos squared π with one minus sin squared π.
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We then multiply out the parentheses and collect our sin cubed π terms.
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Now that we have done this, we have completed our question.
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And we have expressed cos three π in terms of cos π and we have expressed sin three π in terms of sin π.