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Question: How to use de moivre’s theorem to express \( \dfrac{{\sin 6\theta }}{{\sin \theta }} \) as a polynom...

How to use de moivre’s theorem to express sin⁡6θsin⁡θ\dfrac{{\sin 6\theta }}{{\sin \theta }} as a polynomial in cos⁡θ\cos \theta ?

Explanation

Solution

Hint : Here, we will use the De Moivre’s theorem and the correlation between the cosine and sine. And the value of the imaginary number and its equivalent value and substitute its value using the theorem and simplify the equation accordingly.

Complete step by step solution:
By using the Pythagoras identity, cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
As per our requirement we need polynomial in cosine form. sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta
[When we move any term from one side to another, the sign of the term also changes. Positive term becomes negative term and vice-versa.]
By using De Moivre’s theorem in the above equation -
r(cos⁡θ+isin⁡θ)n=rn(cos⁡nθ+isin⁡nθ)r{(\cos \theta + i\sin \theta )^n} = {r^n}(\cos n\theta + i\sin n\theta )
Let us assume that cos⁡θ\cos \theta is denoted by “c” and sin⁡θ\sin \theta is denoted by “s” ….(A)
So, by using de Moivre’s theorem
We get,
cos⁡6θ+isin⁡6θ=(c+is)6\cos 6\theta + i\sin 6\theta = {(c + is)^6}
By using the formula for the expansion for algebraic with sixth power-
cos⁡6θ+isin⁡6θ=c6+6ic5s−15c4s2−20ic3s3+15c2s4+6ics5−s6\cos 6\theta + i\sin 6\theta = {c^6} + 6i{c^5}s - 15{c^4}{s^2} - 20i{c^3}{s^3} + 15{c^2}{s^4} + 6ic{s^5} - {s^6}
Arrange all the like terms together, means all real terms together and the imaginary terms together.
cos⁡6θ+isin⁡6θ=(c6−s6−15c4s2+15c2s4)+is(6cs4+6c5−20c3s2)\cos 6\theta + i\sin 6\theta = ({c^6} - {s^6} - 15{c^4}{s^2} + 15{c^2}{s^4}) + is(6c{s^4} + 6{c^5} - 20{c^3}{s^2})
Comparing the real and imaginary parts in the above equation:
We get –
sin⁡6θ=s(6c5−20c3s2+6cs4)\sin 6\theta = s(6{c^5} - 20{c^3}{s^2} + 6c{s^4})
Substitute, s2=1−c2{s^2} = 1 - {c^2}
sin⁡6θ=s(6c5−20c3(1−c2)+6c(1−c2)2)\sin 6\theta = s(6{c^5} - 20{c^3}(1 - {c^2}) + 6c{(1 - {c^2})^2})
Open the brackets, when there is a positive sign outside the bracket then the sign of the terms inside the bracket remains the same, while if there is a negative sign outside the bracket then the sign of the terms inside the bracket changes.
sin⁡6θ=s(6c5−20c3+20c5+6c−12c3+6c5)\sin 6\theta = s(6{c^5} - 20{c^3} + 20{c^5} + 6c - 12{c^3} + 6{c^5})
Take like terms together,
sin⁡6θ=s(6c5+20c5+6c‾5−20c3−12c3‾+6c)\sin 6\theta = s({\underline {6{c^5} + 20{c^5} + 6c} ^5}\underline { - 20{c^3} - 12{c^3}} + 6c)
Simplify the like terms-
sin⁡6θ=s(32c5−32c3+6c)\sin 6\theta = s(32{c^5} - 32{c^3} + 6c) ….(B)
Take the given expression-
sin⁡6θsin⁡θ\dfrac{{\sin 6\theta }}{{\sin \theta }}
Placing the values from equation (A) and (B)
sin⁡6θsin⁡θ=s(32c5−32c3+6c)s\dfrac{{\sin 6\theta }}{{\sin \theta }} = \dfrac{{s(32{c^5} - 32{c^3} + 6c)}}{s}
Common factors from the numerator and the denominator cancel each other.
⇒sin⁡6θsin⁡θ=(32c5−32c3+6c)\Rightarrow \dfrac{{\sin 6\theta }}{{\sin \theta }} = (32{c^5} - 32{c^3} + 6c)
Replace the values from equation (A)
⇒sin⁡6θsin⁡θ=(32cos⁡5θ−32cos⁡3θ+6cos⁡θ)\Rightarrow \dfrac{{\sin 6\theta }}{{\sin \theta }} = (32{\cos ^5}\theta - 32{\cos ^3}\theta + 6\cos \theta )
This is the required solution.
So, the correct answer is “sin⁡6θsin⁡θ=(32cos⁡5θ−32cos⁡3θ+6cos⁡θ)\dfrac{{\sin 6\theta }}{{\sin \theta }} = (32{\cos ^5}\theta - 32{\cos ^3}\theta + 6\cos \theta )”.

Note : Be careful while expansion of the sixth power. Remember the de moivre’s theorem and different identities to get the sixth term for the sum of two terms. Be careful while using the law of exponents and power while simplification.