Proof of The Pythagorean Identity for Sine and Cosine

Each result cited below is universally quantified over the data in its own statement; it is applied to the data named at the point of use. Sums and products of functions are formed pointwise, as in the statements of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives and Continuity of Sums and Products of Real-Valued Functions on a Metric Space; those items state their hypotheses for a pair of functions, and where only one function is named below the second is taken to be that same function. Every point of $\mathbb{R}$ is an interior point of the interval $\mathbb{R}$ by Basic Facts about Intervals of the Real Line and Their Interior Points §whole-line.

Let $h:\mathbb{R}\to\mathbb{R}$ be the function $h=\cos\cos+\sin\sin$, that is, $h(t)=(\cos t)^{2}+(\sin t)^{2}$.

Claim 1 (the derivative of $h$ vanishes). Let $t\in\mathbb{R}$. By Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine §derivative, $\cos$ is differentiable at $t$ with derivative $-\sin t$, and $\sin$ is differentiable at $t$ with derivative $\cos t$. Claim 3 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives (the product rule), applied on the interval $\mathbb{R}$ at $t$ with both of the functions named there taken to be $\cos$, gives that $\cos\cos$ is differentiable at $t$ with derivative $(-\sin t)\cos t+\cos t(-\sin t)$; the same claim with both functions taken to be $\sin$ gives that $\sin\sin$ is differentiable at $t$ with derivative $(\cos t)\sin t+\sin t\cos t$. Claim 2 of that lemma then gives that $h$ is differentiable at $t$ with derivative

Multiplication in the field $\mathbb{R}$ is commutative, and claim 2 of Zero Products and Elementary Identities in a Field gives $(-s)r=-(sr)$ for all $s,r\in\mathbb{R}$; so each of the first two summands equals $-(\cos t\sin t)$ and each of the last two equals $\cos t\sin t$. By the additive inverse axiom of that field the four summands cancel in pairs, and the derivative of $h$ at $t$ is $0$.

Claim 2 ($h$ is continuous on $\mathbb{R}$). By Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine §continuous, $\cos$ and $\sin$ are continuous on $\mathbb{R}$. Claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, applied with $(X,d)=(\mathbb{R},d_{\mathbb{R}})$ and $A=\mathbb{R}$ — first to the pair $\cos,\cos$, then to the pair $\sin,\sin$, then to the pair $\cos\cos,\sin\sin$ — therefore gives in turn that $\cos\cos$, $\sin\sin$ and their sum $h$ are continuous on $\mathbb{R}$.

Claim 3 (clause 1). Fix $x\in\mathbb{R}$ and put $R=1+|x|$. By claim 1 of Properties of the Absolute Value in an Ordered Field one has $0\le|x|$, and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field. Claim 3 of the latter, applied with $0<1$ and $|x|\le|x|$, gives $|x|=0+|x|<1+|x|=R$; applied with $0<1$ and $0\le|x|$ it gives $0=0+0<R$. By claim 9 of Properties of the Absolute Value in an Ordered Field, $|x|<R$ means $-R<x<R$, and $0<R$ likewise means $-R<0<R$. So $x$ and $0$ both lie in the closed interval $[-R,R]$, which by Basic Facts about Intervals of the Real Line and Their Interior Points §closed-interval is an interval every point of which strictly between $-R$ and $R$ is an interior point of it.

By claim 1 of Restriction Stability of Continuity and of the Derivative and Claim 2, the restriction $h|_{[-R,R]}$ is continuous on $[-R,R]$; by claim 2 of that lemma and Claim 1 it is differentiable at every $t$ with $-R<t<R$, with derivative $0$ there. Hence A Continuous Function with Vanishing Derivative is Constant, applied with $a=-R$ and $b=R$, gives $h(t)=h(-R)$ for every $t\in[-R,R]$; in particular

By Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine §values, $\cos0=1$ and $\sin0=0$, so $h(0)=1\cdot1+0\cdot0=1+0=1$, using the multiplicative identity of $\mathbb{R}$, claim 1 of Zero Products and Elementary Identities in a Field and the additive identity. Therefore $(\cos x)^{2}+(\sin x)^{2}=1$, which is clause 1.

Claim 4 (clause 2). Let $x\in\mathbb{R}$. By claim 2 of Nonnegativity of Squares in an Ordered Field one has $0\le(\sin x)^{2}$, and by Claim 3 the difference $1-(\cos x)^{2}$ equals $(\sin x)^{2}$; so claim 3 of Elementary Arithmetic in an Ordered Field gives $(\cos x)^{2}\le1$.

By claim 1 of Nonnegativity of Squares in an Ordered Field, $|\cos x|^{2}=(\cos x)^{2}$, so $|\cos x|^{2}\le1=1\cdot1=1^{2}$. Since $0\le|\cos x|$ by claim 1 of Properties of the Absolute Value in an Ordered Field and $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $|\cos x|\le1$.

The same argument with $\cos$ and $\sin$ exchanged — using $0\le(\cos x)^{2}$, again by claim 2 of Nonnegativity of Squares in an Ordered Field, and, from Claim 3, $1-(\sin x)^{2}=(\cos x)^{2}$ — gives $|\sin x|\le1$. This is clause 2, and the proof is complete.