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CHEMMIB
Group Title
Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.
 one year ago
 one year ago
CHEMMIB Group Title
Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.
 one year ago
 one year ago

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LolWolf Group TitleBest ResponseYou've already chosen the best response.2
So, by de Moivre's we know: \[ (\cos \theta+i \sin\theta)^n=\cos(n\theta)+i\sin(n\theta) \]Therefore, we know: \[ \Re\left(\left(\cos\theta+i\sin\theta\right)^5\right)=\cos(5\theta)\\ \Im\left(\left(\cos\theta+i\sin\theta\right)^5\right)=\sin(5\theta) \]Expand the previous identities, and find the real part to give a closedform expression for \(\cos\theta\), it follows similarly (for the imaginary part) for \(\sin\theta\).
 one year ago

CHEMMIB Group TitleBest ResponseYou've already chosen the best response.0
Would my answer then be (cos θ + i sin θ)^5 = cos 5θ + i sin 5θ?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.2
Although the original answer I gave *is* a correct answer. It should not be the final statement, it was more of a hint. Let's solve for \(\cos(\theta)\) and leave the other for you. Using binomial theorem: \[ \left(\cos\theta+i\sin(\theta)\right)^5=\\ \cos^5\theta+C_1i(\sin\theta)(\cos^4\theta)+C_2(i^2\sin^2\theta)(\cos^3\theta)+\cdots+C_5(i^5\sin^5\theta) \]We can ignore all of the odd exponents (excluding the first), since we are looking for the real part, and, thus, we get: \[ \cos^5\theta10(\sin^2\theta)(\cos^3\theta)+5(\sin^4\theta)(\cos\theta)=\cos(5\theta) \]
 one year ago
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