Double Angle Formulas/Cosine/Proof 3

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

$\blacksquare$