CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Parity and the Pythagorean identity for sine and cosine

Statement

For every real $x$ , $\sin (- x) = - \sin x, \cos (- x) = \cos x, \sin^{2} x + \cos^{2} x = 1.$ Consequently $∣ \sin x ∣ \leq 1$ and $∣ \cos x ∣ \leq 1$ .

Facts & Assumptions

Given: A real $x$ .

[L1]

For all real $u, v$ , $\sin (u + v) = \sin u \cos v + \cos u \sin v$ and $\cos (u + v) = \cos u \cos v - \sin u \sin v$ (The addition formulas for sine and cosine).

[L2]

The derivative identities and values at zero hold (The derivatives of sine and cosine are cosine and minus sine).

[L3]

The algebra of derivatives and the zero-derivative theorem hold (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

[L4]

Differentiability implies continuity (A function differentiable at $c$ is continuous at $c$ ).

Proof

technique · direct

1.1L2L3algebra

The derivative of $F (x) = \sin^{2} x + \cos^{2} x$ is $2 \sin x \cos x - 2 \cos x \sin x = 0$ .

2.1step 1.1L2L3L4algebra

Step 1.1 makes $F$ differentiable everywhere, hence continuous by [L4]. The zero-derivative theorem therefore makes $F$ constant, and $F (0) = 1$ , proving $\sin^{2} x + \cos^{2} x = 1$ ; each square is then at most $1$ .

3.1L1L2step 2.1algebra∎

Applying the addition formulas at $x + (- x) = 0$ gives $0 = \sin x \cos (- x) + \cos x \sin (- x), 1 = \cos x \cos (- x) - \sin x \sin (- x) .$ The coefficient matrix squares to the identity by step 2.1, so these equations give $\cos (- x) = \cos x$ and $\sin (- x) = - \sin x$ .

Parity and the Pythagorean identity for sine and cosine

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources