Sum of Squares of Sine and Cosine/Proof 1

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Theorem

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


Proof

\(\ds 1\) \(=\) \(\ds \cos 0\) Cosine of Zero is One
\(\ds \) \(=\) \(\ds \map \cos {x - x}\)
\(\ds \) \(=\) \(\ds \cos x \map \cos {-x} - \sin x \map \sin {-x}\) Cosine of Sum‎
\(\ds \) \(=\) \(\ds \cos x \cos x - \paren {-\sin x \sin x}\) Cosine Function is Even and Sine Function is Odd
\(\ds \) \(=\) \(\ds \cos^2 x + \sin^2 x\)

$\blacksquare$


Warning

Note that we need to start from the algebraic definitions of sine and cosine:

$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

and then use the proofs of the Cosine of Sum that derive directly from these.

Otherwise these proofs are circular.


Sources

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