Sum of Squares of Sine and Cosine/Proof 5

Theorem

$\cos^2 x + \sin^2 x = 1$

Proof

\(\ds \cos^2 x + \sin^2 x\)

\(=\)

\(\ds \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \sin^2 x\)

Euler's Cosine Identity

\(\ds \)

\(=\)

\(\ds \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2\)

Euler's Sine Identity

\(\ds \)

\(=\)

\(\ds \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2\)

Square of Sum

\(\ds \)

\(=\)

\(\ds \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \frac {\paren {e^{i x} }^2 - e^{-i x} e^{i x} + \paren {e^{-i x} }^2} {-4}\)

Square of Difference and $i^2 = -1$

\(\ds \)

\(=\)

\(\ds \frac {e^{2 i x} + 2 + e^{-2 i x} } 4 + \frac {e^{2 i x} - 2 + e^{-2 i x} } {-4}\)

Exponential of Sum

\(\ds \)

\(=\)

\(\ds \frac {e^{2 i x} + 2 + e^{-2 i x} - e^{2 i x} + 2 - e^{-2 i x} } 4\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac 4 4\)

simplifying

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $2$: Functions, Limits and Continuity: The Elementary Functions: $4$