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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.
 2 years ago
 2 years ago
CHEMMIB Group Title
Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.
 2 years ago
 2 years 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 \(\sin\theta\).
 2 years 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?
 2 years 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: \[ \cos^5\theta10(\sin^2\theta)(\cos^3\theta)+5(\sin^4\theta)(\cos\theta)=\cos(5\theta) \]
 2 years ago
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