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The cosine addition formula gives the horizontal coordinate after two successive rotations:The minus sign comes from the way a counterclockwise rotation mixes horizontal and vertical coordinates:The two rows recover the cosine addition formula and sine addition formula together.

With $x=\pi /6$ and $y=\pi /4$, the resulting angle is $5\pi /12$, andAlthough the horizontal coordinate becomes small, the vertical one approaches one; their squared sum still obeys the Pythagorean trigonometric identity. Taking $y=-x$ in the cosine addition formula, together with the even function property of cosine and odd function property of sine, also recovers that identity.

Mathematics studies patterns in quantity, structure, space, and change by giving objects precise mathematical definitions and examining what follows from them. Arithmetic starts with operations on numbers; algebra organizes relations and transformations; geometry and topology studies shapes and the structures preserved by maps; analysis studies limits and continuous change. These fields connect because the same structure can appear in different forms.

A small example shows the value of changing viewpoint. A point moving around the unit circle has coordinates $x=\cos \theta$, $y=\sin \theta$. Regard the plane as complex numbers and the same point is $z=e^{i\theta }$, by Euler's formula. Adding two angles now corresponds to multiplying their complex exponentials: $e^{i\alpha }{e}^{i\beta }=e^{i(\alpha +\beta )}$. A geometric rotation has become an algebraic operation.

The connection reaches calculus too. Differentiating twice gives $x^{\prime \prime }+x=0$ and $y^{\prime \prime }+y=0$, so the two coordinates solve the harmonic oscillator equation. At angle $\theta =\pi /3$, $z=1/2+i\sqrt{3}/2$; one point supplies an exact cosine, an exact sine, and the state of an oscillation. The Pythagorean trigonometric identity records that the modulus stays one. Seeing these as descriptions of one object makes each formula easier to remember and reuse.

This is why a useful mathematical explanation provides both a reliable mathematical proof and a way to see what the result is doing. In his 1994 essay, William Thurston discusses mathematical progress in terms of advancing understanding, and uses the many interpretations of a derivative to illustrate this aim. ^[ref] A diagram, a computed example, and a concise argument can make different parts of the same idea accessible.

Applications ask us to select a mathematical model of a physical or other system. Its assumptions determine which conclusions transfer to that system. Within mathematics, the same practice of making assumptions explicit lets a result about groups, vector spaces, or probability distributions apply to many concrete examples. The beauty is the combination of precision and reuse: an argument can reveal more than the particular case that first suggested it.

A real square matrix $M$ is an orthogonal matrix when $M^{T}M=I$. The planar rotationobeys $R^{T}R=I$ by the Pythagorean trigonometric identity.

For $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, the off-diagonal entry of $RA{R}^T$ is $\frac{1}{2}(a-c)\sin 2\theta +b\cos 2\theta$. ThusIf $(a-c,b)\ne (0,0)$, equivalently $\theta =\frac{1}{2}atan2(-2b,a-c)+k\pi /2$; if $A$ is scalar, every angle works.

Because $RA{R}^T$ is an orthogonal similarity, its diagonal entries are the eigenvaluesThe matrix $C=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ has eigenvalues $3,-1$ along $(1,1{)}^T,(1,-1{)}^T$. Raising its orthogonal diagonalization to the even power gives

Differentiation givesThe initial conditions set the constant to one, sothe Pythagorean trigonometric identity.

Since $|c|\le 1$, Taylor theorem with Lagrange remainder givesfor some $0<\xi <1$, whilefor some $0<\eta <2$. The intermediate value theorem supplies $k\in (1,2)$ with $c(k)=0$. The sine addition formula gives $s(2k)=2s(k)c(k)=0$. The analogous cosine addition formula gives $c(2k)=c(k{)}^2-s(k{)}^2=-1$, soThus $s$ is a periodic function.

The electric potential of a point charge $Q$ at the origin isFor the two charges forming the dipole,The first-order Taylor expansion at large $|\mathbf{x}|$ isand therefore, with the electric dipole moment $\mathbf{p}=Q\mathbf{d}$,to leading order. Taking the interaction of a second dipole with the corresponding electric field gives the stated electric dipole-dipole interaction

Let the lattice spacing be $d$ and write the central dipole asIts two horizontal neighbours have moment $p(\cos \theta ,-\sin \theta )$. For either one, the expression in parentheses, after extracting $d^{-3}$, isIts two vertical neighbours have moment $p(-\cos \theta ,\sin \theta )$, and each contributes insteadAdding all four nearest-neighbour interactions and using the Pythagorean trigonometric identity givesThe angle has cancelled, so the energy is independent of $\theta$.

Using $\gamma =\lambda$ and the Pythagorean trigonometric identity,so the speed is independent of $\theta$. Since $\gamma =\pi /\alpha >1$,and the speed tends to zero at the vertex. The steady irrotational Bernoulli equation givesThus the fluid pressure is also independent of $\theta$, withIn particular, the pressure difference between the vertex and $r_0$ is

By De Moivre's theorem, $\cos 5\theta$ is the real part of $(\cos \theta +i\sin \theta {)}^5$. Writing $c=\cos \theta$ and $s=\sin \theta$, the binomial theorem givesUse the Pythagorean trigonometric identity $s^2=1-c^2$ and expand:Therefore

Matching the two parametric equations at an intersection gives $\cos \varphi =-t$ and $\sin \varphi =1+t$. The Pythagorean trigonometric identity therefore requiresFor $t=0$, the sine is one and the cosine is zero; for $t=-1$, the cosine is one and the sine is zero. Keeping the stated half-open angular interval givesThe corresponding points are $(0,1)$ and $(2,0)$; both satisfy the original parametric equations.

For the first definite integral, use integration by substitution $t=\tan x$, with $dx=dt/(1+t^2)$, ${\cos }^{2}x=1/(1+t^2)$ and ${\sin }^{2}x=t^2/(1+t^2)$. This givesFor the second calculation, let $D=p{\cos }^{2}x+q{\sin }^{2}x$. Since $p,q>0$, this denominator is bounded away from zero; on compact positive parameter ranges the differentiated integrands have uniform bounds. Thus differentiation under the integral sign is justified and givesAdding the two equations and using the Pythagorean trigonometric identity in the numerators yields $J=-I_p-I_q$, henceFor $p=q$, this becomes $\pi /(2p^2)$, agreeing with direct integration of the constant integrand.

The Pythagorean trigonometric identity makes the second factor $1+{\cos }^2\theta$, which is strictly positive. It is therefore enough to solve $\sin \theta -\cos \theta \le 0$. The sine addition formula givesThe shifted angle lies between $\pi /4$ and $5\pi /4$. On this interval the sine is nonpositive precisely from $\pi$ onwards. Restoring the original angle and retaining its strict upper bound gives

Using the Pythagorean trigonometric identity and the double-angle formula,The required value $1/2$ therefore occurs when ${\sin }^2(2\theta )=1$. Within $0\le \theta \le \pi$, the only possibilities are $2\theta =\pi /2$ and $3\pi /2$. HenceEquivalently equality requires ${\sin }^2\theta ={\cos }^2\theta =1/2$.

The secant function is the reciprocal of the cosine, $\sec x=1/\cos x$, where the denominator is nonzero. Its derivative is $\sec x\tan x$, and ${\sec }^{2}x=1+{\tan }^{2}x$ follows from the Pythagorean trigonometric identity.

For a real angle $\theta$, $\sin \theta$ is the vertical coordinate on the unit circle reached by a counterclockwise rotation through $\theta$ radians. For an acute angle in a right triangle, it is the opposite side divided by the hypotenuse. In particular,The first-quarter rise and second-quarter fall are visible in the circle projection; the lower semicircle gives negative values.

Reflecting the unit circle across the horizontal axis gives $\sin (-\theta )=-\sin \theta$, so sine is an odd function. A full turn gives $\sin (\theta +2\pi )=\sin \theta$, making it a periodic function. For real arguments, its range is $[-1,1]$ and its zeros of a function occur at $\theta =k\pi$ for integers $k$. The cosine describes the perpendicular coordinate of the same point, and the Pythagorean trigonometric identity keeps their two values tied together.