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CHEMMIB

  • 3 years ago

Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

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  1. LolWolf
    • 3 years ago
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    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 closed-form expression for \(\cos\theta\), it follows similarly (for the imaginary part) for \(\sin\theta\).

  2. CHEMMIB
    • 3 years ago
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    Would my answer then be (cos θ + i sin θ)^5 = cos 5θ + i sin 5θ?

  3. LolWolf
    • 3 years ago
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    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: \[ \cos^5\theta-10(\sin^2\theta)(\cos^3\theta)+5(\sin^4\theta)(\cos\theta)=\cos(5\theta) \]

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