Trigonometric depth: domains, inverse branches and rational parametrization

Proof

Take as a geometric fact that a rotation about the origin preserves distances and orientation and is linear in coordinate components. Let $R_\alpha$ rotate counterclockwise through $\alpha$. On the unit circle, it sends the first and second coordinate basis vectors to

$$R_\alpha e_1=(\cos\alpha,\sin\alpha)^T,\qquad R_\alpha e_2=(-\sin\alpha,\cos\alpha)^T.$$

By linearity, its action on the point at angle $\beta$ is

$$R_\alpha(\cos\beta,\sin\beta)^T =(\cos\alpha\cos\beta-\sin\alpha\sin\beta,\ \sin\alpha\cos\beta+\cos\alpha\sin\beta)^T.$$

Geometrically, successive rotations add their angles, so this same point is $(\cos(\alpha+\beta),\sin(\alpha+\beta))^T$. Comparing coordinates proves both identities without assuming them. Setting $\beta=\alpha=\theta$ gives $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta=\cos^2\theta-\sin^2\theta$. An identity holds for every input in its domain; an equation asks which inputs make a particular equality true.

Trigonometric equations and branches

A trigonometric equation is not only an algebraic expression with sine and cosine inserted. Periodicity produces repeated solutions, inverse functions choose restricted branches, and divisions can remove valid points or create undefined ones. We organize those issues before applying a rational circle parametrization. Prerequisites are the unit-circle definitions, addition and double-angle identities, real inverse-trigonometric branch conventions and quadratic equations.

1. Maintain an equivalence ledger

A transformation is equivalent if the old and new equations have exactly the same solutions in a fixed allowed domain $D$. Adding the same real expression $h(x)$ to both sides is equivalent provided $h$ is defined at every point of $D$. Multiplying both sides by $h(x)$ is equivalent if it is also nonzero everywhere on $D$. If a defined multiplier may vanish, multiplication can introduce candidates, while division must restrict to points where it is nonzero and can discard solutions. Squaring a real equation preserves every original solution in $D$ but may add others because opposite signs acquire equal squares. An operation undefined at an allowed input is not justified by these rules: for example, adding $1/x$ to $x=0$ on $\mathbb R$ would remove its only solution.

Worked diagnosis. Solve $\sin x+\cos x=1$ on $[0,2\pi]$. Squaring gives $1+\sin2x=1$, so the candidates are $0,\pi/2,\pi,3\pi/2,2\pi$. Substitution into the original keeps $0,\pi/2,2\pi$ and rejects the two values where the left side is $-1$. The candidate equation was necessary but not sufficient. Alternatively, write $\sin x+\cos x=\sqrt2\sin(x+\pi/4)$ and solve with the correctly shifted interval. The two methods should agree after the checks.

The related expression $\sqrt{1-\sin^2x}$ equals $|\cos x|$, not always $\cos x$. The square root denotes the nonnegative branch; discarding its absolute value silently imposes a sign condition.

2. Inverse compositions are folded graphs

For $x\in[0,2\pi]$,

$$\arcsin(\sin x)= \begin{cases} x,&0\le x\le\pi/2,\\ \pi-x,&\pi/2\le x\le3\pi/2,\\ x-2\pi,&3\pi/2\le x\le2\pi. \end{cases}$$

Each branch has the same sine as $x$ and lies in the arcsine range; uniqueness within that range proves the formula. The overlapping endpoint formulas agree, so no ambiguity occurs. Similarly $\arccos(\cos x)=x$ on $[0,\pi]$ and $2\pi-x$ on $[\pi,2\pi]$.

Worked inverse equation. To solve $\arcsin(\sin x)=\arccos(\cos x)$ on $[0,2\pi]$, compare their branch formulas. On $[0,\pi/2]$ both equal $x$. On $[\pi/2,\pi]$, equality $\pi-x=x$ gives only the shared endpoint $\pi/2$. On $[\pi,3\pi/2]$, the arcsine value is nonpositive while the arccosine value is positive. On $[3\pi/2,2\pi]$, equality $x-2\pi=2\pi-x$ gives $2\pi$. Thus the full solution set is $[0,\pi/2]\cup\{2\pi\}$. Assuming both compositions cancel would incorrectly claim the entire interval.

Proof: inverse branches and complete periodic solutions

We use the unit-circle geometric definition of sine and cosine and the already proved addition formulas. The basic geometric convention is that one counterclockwise traversal of the unit circle has angle $2\pi$; each point has a unique angle modulo $2\pi$, and each point of a specified semicircle has the corresponding angle in its stated half-turn interval. The signs of coordinates are their ordinary quadrant signs. These are geometric prerequisites, not derivative formulas.

Strict monotonicity. Subtracting the addition formulas gives

$$\sin y-\sin x=2\cos\frac{x+y}{2}\sin\frac{y-x}{2},\qquad \cos x-\cos y=2\sin\frac{x+y}{2}\sin\frac{y-x}{2}.$$

For $-\pi/2\le x<y\le\pi/2$, the cosine factor in the first identity is positive, as is its sine factor, so sine is strictly increasing on that interval. For $0\le x<y\le\pi$, both sine factors in the second identity are positive, so cosine is strictly decreasing. If $-\pi/2<x<y<\pi/2$, then

$$\tan y-\tan x=\frac{\sin(y-x)}{\cos x\cos y}>0,$$

because $0<y-x<\pi$ and both cosines are positive. Thus tangent is strictly increasing on this open interval. These arguments include the finite endpoints for sine and cosine without dividing by a vanishing factor.

Surjectivity and exact ranges. For any $c\in[-1,1]$, the unit-circle point $(\sqrt{1-c^2},c)$ lies on the right semicircle and hence has an angle in $[-\pi/2,\pi/2]$ with sine $c$. The point $(c,\sqrt{1-c^2})$ lies on the upper semicircle and has an angle in $[0,\pi]$ with cosine $c$. No sine or cosine value can lie outside $[-1,1]$ because the point lies on the unit circle. For any real $t$, the point $(1,t)/\sqrt{1+t^2}$ lies on the open right semicircle, so its angle $\theta\in(-\pi/2,\pi/2)$ satisfies $\tan\theta=t$. Consequently the restricted functions are bijections onto $[-1,1]$, $[-1,1]$ and $\mathbb R$, respectively. Their inverse functions are precisely $\arcsin$, $\arccos$ and $\arctan$ with these domains and branch ranges.

All solutions, not just principal values. If $|c|\le1$ and $\alpha=\arcsin c$, a circle point with ordinate $c$ has abscissa either $\sqrt{1-c^2}$ or $-\sqrt{1-c^2}$. These are the points at angles $\alpha$ and $\pi-\alpha$. Uniqueness of angle modulo one full turn gives exactly $\theta=\alpha+2k\pi$ or $\theta=\pi-\alpha+2k\pi$, $k\in\mathbb Z$. At $c=\pm1$ the two families coincide, which creates no extra solutions. If $|c|>1$, there is no solution. Similarly, for $\beta=\arccos c$ all solutions of $\cos\theta=c$ are $\theta=\pm\beta+2k\pi$. For real $t$ and $\gamma=\arctan t$, a unit-circle point of slope $t$ is one of the antipodal pair $\pm(1,t)/\sqrt{1+t^2}$; hence all solutions of $\tan\theta=t$ are $\theta=\gamma+k\pi$. The tangent poles are never admitted. This proves the branch restrictions and completeness of each periodic solution family without calculus.

3. Tangent addition needs a branch decision

Let $\alpha=\arctan u$, $\beta=\arctan v$, so each lies in $(-\pi/2,\pi/2)$. Where $1-uv\ne0$, the tangent addition formula gives $\tan(\alpha+\beta)=(u+v)/(1-uv)$. Applying $\arctan$ returns the representative in its principal interval, not automatically the actual sum.

For $u,v>0$, $\alpha+\beta\in(0,\pi)$. Its cosine has the sign of $1-uv$, since $\cos\alpha\cos\beta>0$. Consequently

$$\arctan u+\arctan v= \begin{cases} \arctan\frac{u+v}{1-uv},&uv<1,\\ \pi/2,&uv=1,\\ \pi+\arctan\frac{u+v}{1-uv},&uv>1. \end{cases}$$

This is a derived case split, not a memorized unrestricted identity. At $uv=1$, the quotient formula is undefined, but the angle sum exists. For a single nonzero real $x$, the same branch analysis gives $\arctan x+\arctan(1/x)=\pi/2$ if $x>0$ and $-\pi/2$ if $x<0$.

4. Derive the tangent half-angle parametrization

Let a line through the fixed circle point $(-1,0)$ have slope $t$: $y=t(x+1)$. Substituting into $x^2+y^2=1$ yields

$$x^2+t^2(x+1)^2-1=(x+1)[(1+t^2)x+t^2-1]=0.$$

One intersection is the fixed point. The other is

$$x=\frac{1-t^2}{1+t^2},\qquad y=\frac{2t}{1+t^2},\qquad t\in\mathbb R.$$

The denominator is always positive. Conversely every circle point other than $(-1,0)$ has finite $t=y/(1+x)$, and substitution recovers its coordinates. Thus the formula describes exactly the circle minus that one point, not the entire circle with no exception.

For a circle point $(\cos\theta,\sin\theta)$ with $\theta\not\equiv\pi\pmod{2\pi}$, double-angle identities give

$$t=\frac{\sin\theta}{1+\cos\theta}=\tan(\theta/2), \quad \cos\theta=\frac{1-t^2}{1+t^2},\quad \sin\theta=\frac{2t}{1+t^2}.$$

The derivation explains the missing odd multiples of $\pi$ and provides an inverse, so it also explains when replacing a trigonometric equation by a rational one is exhaustive.

Worked rational transformation. Solve $\cos\theta+2\sin\theta=2$ on $[0,2\pi)$. Check the omitted point first: at $\theta=\pi$, the left side is $-1$, not two. For all other points, substitute the rational formulas:

$$1-t^2+4t=2(1+t^2),\qquad 3t^2-4t+1=0.$$

Thus $t=1$ or $1/3$. The corresponding circle points are $(0,1)$ and $(4/5,3/5)$, giving $\theta=\pi/2$ and $\theta=2\arctan(1/3)$. The first point lies on the positive $y$-axis and the second lies in the first quadrant; both satisfy the original equation. The omitted point was checked separately, so no branch is missing.

5. Choose methods by the question, not habit

For a single sine or cosine equation, use its full periodic solution family and then intersect with the requested interval. For an inverse-composition equation, derive the relevant piecewise folds. For a linear combination $A\sin x+B\cos x$, amplitude and phase expose both its range and number of solutions. For a rational expression in sine and cosine, half-angle substitution can convert the task to algebra, provided the omitted point and original denominators are checked. A squared equation gives candidates that must return to the original sign conditions.

For an inequality, branch intervals matter throughout, not only at isolated roots. For example $\sin x+\cos x>1$ on $[0,2\pi]$ becomes $\sin(x+\pi/4)>1/\sqrt2$ on the shifted interval $[\pi/4,9\pi/4]$. Its only solution interval is $0<x<\pi/2$; endpoints satisfying equality are excluded. A graph can suggest this interval, while the shifted sine inequality proves it.

Graded synthesis and diagnostics

1. Inverse domain. Determine all real $x$ for which $\arcsin(\sin x)=x$, and justify both inclusion directions without testing finitely many angles.
2. Original denominator. Solve $\sin x/(1+\cos x)=1$ on $[0,2\pi)$. Explain why an equation obtained by multiplying out can contain an invalid extra point.
3. Branch correction. Express $2\arctan2$ in terms of the principal value $\arctan(4/3)$ and an appropriate multiple of $\pi$.
4. Absolute value. Determine the full real solution set of $\sqrt{1-\sin^2x}=\cos x$.
5. Rational geometry. Prove that every rational circle point $(x,y)$ with $x^2+y^2=1$ and $(x,y)\ne(-1,0)$ is obtained from a rational parameter $t$ in the half-angle formulas, and conversely.
6. Parameter classification. For real $A,B,C$, classify the number of solutions modulo $2\pi$ of $A\sin x+B\cos x=C$, including $A=B=0$ and tangency cases.

Full solutions

1. If $x\in[-\pi/2,\pi/2]$, sine is one-to-one onto $[-1,1]$ there and its inverse returns $x$. Conversely an arcsine output always belongs to that interval, so equality forces $x$ to lie in it. Thus the interval is exactly the solution set, with no other periodic copies because the right side is the actual real number $x$.
2. The original denominator vanishes at $x=\pi$, which is excluded. The valid half-angle relation makes the equation $t=1$, giving $x=\pi/2$ in the specified interval. Multiplying out gives $\sin x=1+\cos x$, also satisfied at $\pi$ because both sides are zero there, but that point is not in the original domain. Returning to the original denominator removes it.
3. With $u=v=2$, the product $uv=4>1$. The addition formula gives $2\arctan2=\pi+\arctan(-4/3)=\pi-\arctan(4/3)$. The sum lies in $(\pi/2,\pi)$; the principal negative arctangent alone cannot represent it.
4. The left side equals $|\cos x|$, so equality holds exactly when $\cos x\ge0$. The full set is $\bigcup_{k\in\mathbb Z}[-\pi/2+2k\pi,\pi/2+2k\pi]$. Endpoints are included because both sides are zero there.
5. A rational parameter gives rational coordinates because both formulas use rational operations and $1+t^2\ne0$, and direct substitution gives the circle identity. Conversely for a rational point other than the omitted point, $1+x\ne0$ and $t=y/(1+x)$ is rational. The earlier inverse calculation recovers the point. The excluded point is itself rational but has no finite slope from itself in this parametrization; it must be listed separately.
6. If $A=B=0$, every angle is a solution when $C=0$, and none when $C\ne0$. Otherwise put $R=\sqrt{A^2+B^2}>0$ and choose $\phi$ with $\cos\phi=A/R$, $\sin\phi=B/R$. The equation is $R\sin(x+\phi)=C$. If $|C|>R$, there are no solutions. If $|C|=R$, there is one solution class modulo $2\pi$, at the top or bottom of sine. If $|C|<R$, there are two distinct classes. Counting modulo the period, or on a half-open interval of length $2\pi$, avoids double-counting an endpoint.

Summary and terms

An equivalent transformation preserves the full solution set on its declared domain. A candidate equation gives only a necessary condition. An inverse-composition fold reflects values into a principal range. A rational parametrization expresses coordinates as rational functions of one parameter and comes with an omitted-point check. The right method exposes the structure, while a domain-and-branch ledger keeps that structure mathematically valid.