Complex Reciprocal Proof

https://chatgpt.com/share/6825e065-658c-8004-84f4-9d4269d07dad

Yes — the equation

\[ (\cos x + i\sin x)^5 = \frac{1}{\cos(-5x) + i\sin(-5x)} \]

is true, and can be proven using Euler’s formula and basic properties of complex numbers.

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Step-by-step reasoning:

Step 1: Use Euler’s formula

Euler’s identity:

\[ e^{ix} = \cos x + i\sin x \]

So,

\[ (\cos x + i\sin x)^5 = (e^{ix})^5 = e^{i5x} \]

Step 2: Use properties of exponentials

Recall:

\[ \frac{1}{e^{-i5x}} = e^{i5x} \]

Also note:

\[ \cos(-5x) + i\sin(-5x) = e^{-i5x} \]

So:

\[ \frac{1}{\cos(-5x) + i\sin(-5x)} = \frac{1}{e^{-i5x}} = e^{i5x} \]

Step 3: Conclude equality

Therefore, both sides are equal:

\[ (\cos x + i\sin x)^5 = e^{i5x} = \frac{1}{e^{-i5x}} = \frac{1}{\cos(-5x) + i\sin(-5x)} \]

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Bonus: Alternate form using de Moivre’s Theorem

De Moivre’s Theorem says:

\[ (\cos x + i \sin x)^n = \cos(nx) + i\sin(nx) \]

So:

\[ (\cos x + i \sin x)^5 = \cos(5x) + i\sin(5x) \]

And:

\[ \cos(-5x) + i\sin(-5x) = \cos(5x) - i\sin(5x) \]

Then:

\[ \frac{1}{\cos(-5x) + i\sin(-5x)} = \frac{1}{\cos(5x) - i\sin(5x)} = \cos(5x) + i\sin(5x) \]

because the reciprocal of a unit complex number \(z\) is \(\frac{1}{z} = \bar{z}\) (complex conjugate), and for unit modulus: \(z^{-1} = \bar{z}\), and \(\bar{z} = \cos(5x) + i\sin(5x)\).