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:
\(\ds \cos^2 \theta - \sin^2\theta\) | \(=\) | \(\ds \paren {\frac 1 2 \paren {e^{i \theta} + e^{-i \theta} } }^2 - \paren {\frac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} } }^2\) | Euler's Cosine Identity, Euler's Sine Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {e^{i \theta} + e^{-i \theta} }^2 + \frac 1 4 \paren {e^{i \theta} - e^{-i \theta} }^2\) | $i$ is the imaginary unit | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {e^{2 i \theta} + 2 + e^{-2 i \theta} + e^{2 i \theta} - 2 + e^{-2 i \theta} }\) | Square of Sum, Square of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {e^{2 i \theta} + e^{-2 i \theta} }\) | Simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos 2 \theta\) | Euler's Cosine Identity |
$\blacksquare$