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Express sin cubed π in terms of the sines of multiples of π.
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In this question, we need to find an expression for the sin cubed of π in terms of sines of multiples of our angle π.
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We might be tempted to try using things like our double-angle formula for sine to try and rewrite sine cubed in terms of this.
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And this might work; however, itβs difficult and will require lots of experimentation.
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And in fact, thereβs an easier method.
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And to do this, weβre going to first need to recall de Moivreβs theorem.
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This tells us for any integer value of π and real value π, the cos of π plus π sin of π all raised to the πth power is equal to the cos of ππ plus π sin of ππ.
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And this is just one version of the theorem.
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There are several different equivalent statements.
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And there are some very useful direct results of this theorem that will help us answer this question.
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And these are worth committing to memory.
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If we call our complex number cos π plus π sin π equal to π, then by de Moivreβs theorem π to the πth of power is the cos of ππ plus π times the sin of ππ.
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And we can prove the following two useful results for any integer value of π.
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π to the πth power plus one over π to the πth power will be equal to two times the cos of ππ and π to the πth power minus one over π to the πth power will be two π sin of ππ.
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To prove these two results, we just remember that one over π to the πth power is equal to π to the power of negative π.
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And we can apply de Moivreβs theorem for any integer value of π.
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The result weβre interested in here is the bottom result.
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And the reason for this is itβs going to help us find an expression for the sin cubed of π.
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Letβs start by setting our value of π equal to one.
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By setting our value of π equal to one in our bottom result, we get two π sin of π will be equal to π minus one over π.
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Of course, in the question, weβre looking for an expression for the sin cubed of π.
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So, weβre going to have to cube both sides of this expression.
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We can then rearrange to find an expression for the sin cubed of π.
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Letβs start by simplifying the left-hand side of this expression.
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Weβre going to want to cube each of our factors separately.
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This gives us eight times π cubed multiplied by sin cubed of π.
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And on the right-hand side of our expression, we can see inside our parentheses we have two numbers, and weβre raising this to an exponent.
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This is a binomial expression.
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So, we could write this out in full and multiply this by using the FOIL method or we could use the binomial theorem.
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Remember, this tells us π plus π all raised to the πth power will be equal to the sum from π equals zero to π of π choose π times π to the πth power multiplied by π to the power of π minus π, where π is a positive integer.
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Weβll do this by using the binomial theorem.
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Expanded out, this gives us three choose zero times π cubed plus three choose one times π squared multiplied by negative one over π plus three choose two times π multiplied by negative one over π all squared plus three choose three multiplied by negative one over π all cubed.
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And now, we can start simplifying.
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First, remember that π is the square root of negative one.
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So, π squared is equal to negative one.
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This means that π cubed is just equal the negative π.
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So, we can simplify the left-hand side of our equation to give us negative eight π times the sin cubed of π.
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And we can simplify each term on the right-hand side of our equation separately.
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First, three choose zero is just equal to one, so our first term is π cubed.
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To simplify our next term, recall that negative one over π can be written as negative π to the power of negative one.
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And remember, to multiply two terms of the same base together, we just add their exponents.
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And in our exponent, two minus one is equal to one.
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So, we get π to the first power, which is just π.
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And the coefficient will be three choose one multiplied by negative one, which is negative three.
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We can do exactly the same to evaluate our next term.
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However, this time, we have negative one over π all squared.
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This is just going to simplify to give us π to the power of negative two.
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So, π times π to the power of negative two is equal to π to the power of negative one.
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And our coefficient is just three choose two, which we know is equal to three.
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So, our third term is three π to the power of negative one.
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Finally, in our last term, we can distribute the cube over our parentheses.
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We get negative one over π cubed.
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And three choose three is just equal to one.
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So, our final term is negative one over π cubed.
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So, our equation now reads negative eight π sin cubed π is equal to π cubed minus three π plus three π to the power of negative one minus one over π cubed.
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And we could notice something interesting.
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We have π cubed minus one over π cubed.
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And this is exactly one of our statements for de Moivreβs theorem with our value of π set to be three.
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And in fact, this isnβt the only time this appears.
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Remember, we can rewrite π to the power of negative one as one over π.
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So, we can rewrite our third term as three divided by π.
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And we want to write this in the form π to the πth power minus one over π to the πth power.
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We want our value of π to be one.
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And we can either take out a factor of three or a factor of negative three.
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Weβll take out the factor of negative three.
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Taking out our factor of negative three and rearranging our equation, we get negative eight π sin cubed π is equal to π cubed minus one over π cubed minus three times π minus one over π.
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Weβre now ready to start writing this expression in terms of sines of multiples of π.
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And to do this, weβre going to once again need to use De Moivreβs theorem.
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By setting our value of π equal to three, we know that π cubed minus one over π cubed is two π times the sin of three π.
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And by setting our value of π equal to one, we know π minus one over π is two π sin π.
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Substituting these expressions in and remembering we need to multiply two π sin π by negative three, we get the following equation.
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We get that negative eight π sin cubed of π is equal to two π sin of three π minus six π sin π.
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Now, all we need to do is rearrange this equation for sin cubed π.
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To do this, weβre going to need to divide through both sides of our equation by negative eight π.
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To make this easier, to simplify, weβll divide each term on the right-hand side of our equation separately by negative eight π.
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We get sin cubed π is equal to two π sin three π all over negative eight π plus negative six π sin π all divided by negative eight π.
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And now, we can start simplifying.
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In both terms, we have a shared factor of π in our numerator and denominator.
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Similarly, both the first and second term share a factor of two in both the numerator and our denominator.
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So, we can cancel this out.
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Finally, in our second term, we can cancel the shared factor of negative one in the numerator and denominator.
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This leaves us with sin three π all over negative four plus three sin π all over four.
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We could leave our answer like this.
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However, weβll rearrange these two terms and weβll take out the shared factor of one-quarter, giving us our final answer of one-quarter multiplied by three sin π minus the sin of three π.
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Therefore, in this question, we were able to use de Moivreβs theorem to express sin cubed π in terms of the sines of multiples of π.
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We showed sin cubed π was equal to one-quarter times three sin π minus the sin of three π.