Double Angle Formulas/Cosine/Proof 3

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Theorem

$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$


Proof

Starting from the right, we have:

\(\displaystyle \cos^2 \theta - \sin^2\theta\) \(=\) \(\displaystyle \left({\frac 1 2 \left({e^{i \theta} + e^{-i \theta} }\right)}\right)^2 - \left({\frac 1 {2 i} \left({e^{i \theta} - e^{-i \theta} }\right)}\right)^2\) Cosine Exponential Formulation, Sine Exponential Formulation
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 4 \left({e^{i \theta} + e^{-i \theta} }\right)^2 + \frac 1 4 \left({e^{i \theta} - e^{-i \theta} }\right)^2\) $i$ is the imaginary unit
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 4 \left({e^{2 i \theta} + 2 + e^{-2 i \theta} + e^{2 i \theta} - 2 + e^{-2 i \theta} }\right)\) Square of Sum, Square of Difference
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({e^{2 i \theta} + e^{-2 i \theta} }\right)\) Simplifying
\(\displaystyle \) \(=\) \(\displaystyle \cos 2 \theta\) Cosine Exponential Formulation

$\blacksquare$