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CommeUnJeu · L1 MPSI

Trigonometry

⌚ ~123 min ▢ 15 blocks ✓ 50 exercises ➣ Prerequisites : Radians and the Unit Circle, Trigonometric Functions
Trigonometry, in lycée, is a recipe: \(\cos\), \(\sin\), \(\tan\) are box outputs whose values one memorizes at \(0, \pi/6, \pi/4, \pi/3, \pi/2\). This course rebuilds them from a single picture --- the trigonometric circle --- and rewrites the « modulo \(2\pi\) » hand-wave as a precise relation, the congruence modulo \(2\pi\). From this picture, every formula of the chapter (symmetries, standard values, addition, duplication, product-to-sum, derivatives, \(\tan(t/2)\) parameterization) becomes either a direct reading of the circle or a two-line consequence of one master identity, \(\cos(a - b) = \cos a \cos b + \sin a \sin b\).
The plan has six sections. The first section introduces the circle, the radian, and the congruence relation \(a \equiv b \,[2\pi]\). The second section collects the symmetries and standard values --- the picture does most of the work. The third section derives the addition, duplication, and product-to-sum formulas from the dot-product proof of \(\cos(a - b)\). The fourth section proves the geometric inequality \(|\sin x| \le |x|\) and uses it to establish the limit \(\sin x / x \to 1\) and the derivatives \(\sin' = \cos\), \(\cos' = -\sin\). The fifth section gives the full treatment of the tangent function: \(\pi\)-period, parity, addition, derivative, variations, graph, and the program-mandated \(\tan(t/2)\) parameterization. The last section solves the trigonometric equations \(\cos x = a\), \(\sin x = a\), \(\tan x = a\) and the matching inequations.
Three notions deliberately wait for later chapters. The systematic linearization of \(\cos^p \theta \sin^q \theta\), the sum-to-product formulas, the amplitude-phase form \(a \cos x + b \sin x = R \cos(x - \varphi)\), and the trigonometric sums \(\sum \cos(k\theta)\) are all postponed to Complex numbers, where Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\) makes them one-line. The inverse functions \(\arccos, \arcsin, \arctan\) as differentiable objects are postponed to Standard functions; here they appear only as names for an angle on the standard interval.
I Trigonometric circle\(\virgule\) radians\(\virgule\) congruence modulo \(2\pi\)
The trigonometric circle is the unit circle of the plane, oriented counter-clockwise. To each real number \(\theta\) we associate the point of the circle obtained by walking an arc of length \(\theta\) from \((1, 0)\), counter-clockwise if \(\theta > 0\), clockwise if \(\theta < 0\). The \(x\)-coordinate of that point is \(\cos\theta\), the \(y\)-coordinate is \(\sin\theta\). The full picture --- circle, point, projections --- is the only diagram one really needs for the rest of the chapter. The relation « two real numbers \(a\) and \(b\) correspond to the same point of the circle » is the congruence modulo \(2\pi\), which we now make precise.
Definition — Trigonometric circle\(\virgule\) radian\(\virgule\) cosine\(\virgule\) sine
The trigonometric circle is the unit circle of the plane \(\mathbb{R}^2\) centred at the origin, oriented counter-clockwise. The radian measure of an angle is the length of the corresponding arc on the trigonometric circle. For \(\theta \in \mathbb{R}\), the point \(M(\theta)\) of the circle reached by walking an arc of (signed) length \(\theta\) from \((1, 0)\) has coordinates $$ M(\theta) = (\cos\theta, \, \sin\theta). $$ This defines two functions \(\cos, \sin : \mathbb{R} \to [-1, 1]\).
Definition — Congruence modulo \(2\pi\)
For \(a, b \in \mathbb{R}\), we say that \(a\) is congruent to \(b\) modulo \(2\pi\), and we write $$ a \equiv b \,[2\pi], $$ when \(a - b \in 2\pi\mathbb{Z}\), i.e.\ when there exists \(k \in \mathbb{Z}\) such that \(a = b + 2k\pi\). The same notation, with \(\pi\) in place of \(2\pi\), defines congruence modulo \(\pi\) (used later for \(\tan\)).
Proposition — Algebra of congruences
Let \(\alpha > 0\). The relation \(\equiv \,[\alpha]\) is an equivalence relation on \(\mathbb{R}\) (reflexive, symmetric, transitive). For all \(a, a', b, b' \in \mathbb{R}\):
  • if \(a \equiv b \,[\alpha]\) and \(a' \equiv b' \,[\alpha]\), then \(a + a' \equiv b + b' \,[\alpha]\) and \(a - a' \equiv b - b' \,[\alpha]\);
  • if \(a \equiv b \,[\alpha]\) and \(r \in \mathbb{R}^*\), then \(r a \equiv r b \,[\,|r|\,\alpha\,]\) (the modulus is multiplied by \(|r|\); one keeps \(|r|\), not \(r\), because a modulus must be \(> 0\)). For \(r = 0\) the statement is trivial: both sides equal \(0\).
In doubt, fall back to the parametric form \(a = b + k\alpha\) (\(k \in \mathbb{Z}\)).

  • Equivalence relation. Reflexive: \(a - a = 0 \in \alpha\mathbb{Z}\). Symmetric: \(a - b \in \alpha\mathbb{Z} \Rightarrow b - a = -(a - b) \in \alpha\mathbb{Z}\). Transitive: \((a - b) + (b - c) = a - c \in \alpha\mathbb{Z}\).
  • Sum / difference. If \(a = b + k\alpha\) and \(a' = b' + k'\alpha\) with \(k, k' \in \mathbb{Z}\), then \(a \pm a' = (b \pm b') + (k \pm k')\alpha\).
  • Multiplication by \(r\). Let \(r \in \mathbb{R}^*\). If \(a = b + k\alpha\) with \(k \in \mathbb{Z}\), then \(ra = rb + k r \alpha = rb + (\operatorname{sgn}(r)\, k)\,|r|\alpha\). Since \(\operatorname{sgn}(r)\, k\) ranges over \(\mathbb{Z}\) as \(k\) does, this says exactly \(ra \equiv rb \,[\,|r|\alpha\,]\); the new modulus is \(|r|\alpha > 0\), not \(\alpha\). (For \(r = 0\) both sides are \(0\), so the congruence holds trivially.)
Warning. Do not multiply a congruence by an arbitrary real keeping the same modulus: \(a \equiv b \,[\alpha]\) does not imply \(ra \equiv rb \,[\alpha]\) in general.

Congruence \(\ne\) equality
Note carefully: \(a \equiv b \,[2\pi]\) does not mean \(a = b\). It means « \(a\) and \(b\) correspond to the same point of the trigonometric circle ». Concretely, \(\frac{13\pi}{6} \ne \frac{\pi}{6}\) (the two are different real numbers), but \(\frac{13\pi}{6} \equiv \frac{\pi}{6} \,[2\pi]\) since \(\frac{13\pi}{6} - \frac{\pi}{6} = 2\pi\). Throughout the chapter, congruence is the right tool for « location on the circle », equality is the right tool for « value of a real number ».
Proposition — Fundamental identity
For all \(\theta \in \mathbb{R}\), $$ \cos^2\theta + \sin^2\theta = 1. $$

By definition, \(M(\theta) = (\cos\theta, \sin\theta)\) lies on the unit circle, whose Cartesian equation is \(x^2 + y^2 = 1\). Substituting \(x = \cos\theta\), \(y = \sin\theta\) gives \(\cos^2\theta + \sin^2\theta = 1\). *(In other words: the identity is the equation of the unit circle, applied to the point \((\cos\theta, \sin\theta)\).)*

Proposition — Periodicity and parity
For all \(\theta \in \mathbb{R}\):
  • \(\cos(\theta + 2\pi) = \cos\theta\) and \(\sin(\theta + 2\pi) = \sin\theta\) (both are \(2\pi\)-periodic);
  • \(\cos(-\theta) = \cos\theta\) (\(\cos\) is even) and \(\sin(-\theta) = -\sin\theta\) (\(\sin\) is odd).

Read directly off the circle. Walking an extra full turn (\(+2\pi\)) returns to the same point, so coordinates are unchanged: \(\cos\) and \(\sin\) are \(2\pi\)-periodic. Walking an arc \(-\theta\) instead of \(\theta\) reflects the point across the \(x\)-axis: the abscissa is preserved (\(\cos\) is even), the ordinate is negated (\(\sin\) is odd).

Method — Reading off the circle\(\virgule\) reducing a large angle
Given a real number \(\theta\), to compute \(\cos\theta\) and \(\sin\theta\):
  1. Reduce \(\theta\) modulo \(2\pi\) to a representative \(\theta' \in [0, 2\pi[\) (or \(]-\pi, \pi]\) if more convenient): subtract or add multiples of \(2\pi\).
  2. Place \(M(\theta')\) on the trigonometric circle.
  3. Read the abscissa (\(\cos\theta'\)) and the ordinate (\(\sin\theta'\)).
  4. By the \(2\pi\)-periodicity, \(\cos\theta = \cos\theta'\) and \(\sin\theta = \sin\theta'\).
Example
Compute \(\cos\!\big(\frac{11\pi}{6}\big)\) and \(\sin\!\big(\frac{11\pi}{6}\big)\).

We have \(\frac{11\pi}{6} = 2\pi - \frac{\pi}{6}\), so \(\frac{11\pi}{6} \equiv -\frac{\pi}{6} \,[2\pi]\). By \(2\pi\)-periodicity then parity: $$ \cos\!\left(\frac{11\pi}{6}\right) = \cos\!\left(-\frac{\pi}{6}\right) = \cos\!\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \qquad \sin\!\left(\frac{11\pi}{6}\right) = \sin\!\left(-\frac{\pi}{6}\right) = -\sin\!\left(\frac{\pi}{6}\right) = -\frac{1}{2}. $$

Example
Show that \(\frac{13\pi}{6} \equiv \frac{\pi}{6} \,[2\pi]\).

The difference \(\frac{13\pi}{6} - \frac{\pi}{6} = \frac{12\pi}{6} = 2\pi \in 2\pi\mathbb{Z}\), hence the congruence holds (with \(k = 1\) in the parametric form \(a = b + 2k\pi\)).

Example
The standard reference circle, with the cardinal angles at multiples of \(\pi/6\) and \(\pi/4\):
This figure should be re-drawn from memory at each use, not consulted as a table; reading the abscissa and ordinate directly off the circle is more robust than memorizing 16 pairs of numbers.
Skills to practice
  • Reading values at remarkable angles
  • Manipulating congruence modulo \(2\pi\)
II Symmetries and standard values
Reflecting a point of the trigonometric circle across an axis of symmetry, the origin, or the diagonal \(y = x\) produces a new point whose \((\cos, \sin)\) coordinates are predictable. This single observation gives eight identities at once. The standard values at \(0, \pi/6, \pi/4, \pi/3, \pi/2\) then come from two reference triangles: the isoceles right triangle (for \(\pi/4\)) and half of an equilateral triangle (for \(\pi/6\) and \(\pi/3\)). The combination « standard table + symmetry » gives every value at any multiple of \(\pi/6\) or \(\pi/4\) on the circle.
Proposition — Axial and origin symmetries
For all \(\theta \in \mathbb{R}\): $$ \begin{aligned} \cos(-\theta) &= \cos\theta, & \sin(-\theta) &= -\sin\theta & &\text{(symmetry across the \(x\)-axis)}, \\ \cos(\pi - \theta) &= -\cos\theta, & \sin(\pi - \theta) &= \sin\theta & &\text{(symmetry across the \(y\)-axis)}, \\ \cos(\pi + \theta) &= -\cos\theta, & \sin(\pi + \theta) &= -\sin\theta & &\text{(symmetry through the origin)}. \end{aligned} $$

Read off the circle. The three operations (negate the angle, take \(\pi - \theta\), take \(\pi + \theta\)) are exactly the three reflections of \(M(\theta)\) across the \(x\)-axis, \(y\)-axis, and origin respectively. The coordinates transform predictably:
  • Reflection across the \(x\)-axis preserves the abscissa and negates the ordinate: \(\cos(-\theta) = \cos\theta\), \(\sin(-\theta) = -\sin\theta\).
  • Reflection across the \(y\)-axis negates the abscissa and preserves the ordinate: \(\cos(\pi - \theta) = -\cos\theta\), \(\sin(\pi - \theta) = \sin\theta\).
  • Reflection through the origin negates both: \(\cos(\pi + \theta) = -\cos\theta\), \(\sin(\pi + \theta) = -\sin\theta\).

Example
The three axial / origin symmetries on the unit circle:
Proposition — Complementary angles\(\virgule\) \(\pi/2\) shifts
For all \(\theta \in \mathbb{R}\): $$ \begin{aligned} \cos\!\left(\frac{\pi}{2} - \theta\right) &= \sin\theta, & \sin\!\left(\frac{\pi}{2} - \theta\right) &= \cos\theta & &\text{(reflection across \(y = x\))}, \\ \cos\!\left(\frac{\pi}{2} + \theta\right) &= -\sin\theta, & \sin\!\left(\frac{\pi}{2} + \theta\right) &= \cos\theta & &\text{(quarter-turn rotation)}. \end{aligned} $$

The angle \(\frac{\pi}{2} - \theta\) is the reflection of \(\theta\) across the diagonal \(y = x\) on the trigonometric circle (this swaps abscissa and ordinate); hence the first pair. The angle \(\frac{\pi}{2} + \theta\) is the rotation of \(M(\theta)\) by a quarter-turn counter-clockwise: \((x, y) \mapsto (-y, x)\); hence the second pair.

Two operational mnemonics for symmetries
The eight symmetry identities boil down to two rules of thumb that students retain quickly:
  • Adding \(\pi\) multiplies \(\cos\) and \(\sin\) by \(-1\). \(\cos(\pi + \theta) = -\cos\theta\), \(\sin(\pi + \theta) = -\sin\theta\) --- both negated by the same shift. Use this when a problem asks « flip the sign of a cos / sin ».
  • The map \(\theta \mapsto \pi/2 - \theta\) swaps \(\cos\) and \(\sin\). This is THE transformation to use whenever a problem asks « replace a \(\cos\) by a \(\sin\) » or vice-versa.
Example
The complementary-angle and quarter-turn symmetries on the unit circle:
Proposition — Standard values
$$ \begin{array}{c|ccccc} \theta & 0 & \pi/6 & \pi/4 & \pi/3 & \pi/2 \\ \hline \cos\theta & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{\sqrt{2}}{2} & \dfrac{1}{2} & 0 \\ \sin\theta & 0 & \dfrac{1}{2} & \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{3}}{2} & 1 \end{array} $$

The values at \(0\) and \(\pi/2\) are read off the circle: \(M(0) = (1, 0)\) and \(M(\pi/2) = (0, 1)\).
Value at \(\pi/4\). The point \(M(\pi/4)\) lies on the diagonal \(y = x\) in the first quadrant. Setting \(M(\pi/4) = (a, a)\) with \(a > 0\) and using \(a^2 + a^2 = 1\) gives \(a = \sqrt{2}/2\).
Values at \(\pi/3\) and \(\pi/6\). Consider the equilateral triangle \(OAB\) with \(O = (0, 0)\), \(A = (1, 0)\), and \(B\) on the upper unit circle. By construction, the angle at \(O\) is \(\pi/3\), so \(B = M(\pi/3)\). Drop the perpendicular from \(B\) to \(OA\): the foot is the midpoint of \(OA\) (the equilateral triangle is isoceles, so its altitude bisects the base), so \(B\) has abscissa \(1/2\), hence \(\cos(\pi/3) = 1/2\). The Pythagorean identity then gives \(\sin(\pi/3) = \sqrt{1 - 1/4} = \sqrt{3}/2\). The values at \(\pi/6\) follow by complementary angles: \(\cos(\pi/6) = \sin(\pi/2 - \pi/6) = \sin(\pi/3) = \sqrt{3}/2\) and \(\sin(\pi/6) = \cos(\pi/3) = 1/2\).

Method — Reach any value at a multiple of \(\pi/6\) or \(\pi/4\)
For any angle \(\theta\) that is a multiple of \(\pi/6\) or \(\pi/4\):
  1. Reduce \(\theta\) modulo \(2\pi\) to a representative in \([0, 2\pi[\).
  2. Identify the « base angle » \(\theta_0 \in [0, \pi/2]\) from the standard table by applying one of the symmetry formulas (Propositions above).
  3. Read \(\cos\theta\) and \(\sin\theta\) from \(\cos\theta_0\) and \(\sin\theta_0\) with the correct signs.
Example
Compute \(\cos(7\pi/6)\).

We have \(7\pi/6 = \pi + \pi/6\). The « \(+ \pi\) » symmetry gives \(\cos(\pi + \theta) = -\cos\theta\), so $$ \cos\!\left(\frac{7\pi}{6}\right) = -\cos\!\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}. $$

Skills to practice
  • Reducing an angle to the standard range
III Addition\(\virgule\) duplication\(\virgule\) product-to-sum
The single most productive identity of trigonometry is the addition formula for \(\cos\): $$ \cos(a - b) = \cos a \cos b + \sin a \sin b. $$ Once it is in hand, every other addition formula (for \(\cos(a + b)\), \(\sin(a \pm b)\)), every duplication formula (for \(\cos 2a\), \(\sin 2a\)), and every product-to-sum formula (for \(\cos a \cos b\), \(\sin a \sin b\), \(\sin a \cos b\)) is a two-line consequence. The proof of the master identity uses the lycée formula for the dot product of two unit vectors --- a single geometric idea, no chapters of vector-space machinery required.
The sum-to-product formulas (\(\cos p + \cos q = \dots\)), the systematic linearization of \(\cos^p\theta\sin^q\theta\) for general \(p, q\), and the amplitude-phase form \(a\cos x + b\sin x = R\cos(x - \varphi)\) are deferred to Complex numbers, where the imaginary exponential makes them one-line.
Proposition — Master addition formula for \(\cos\)
For all \(a, b \in \mathbb{R}\), $$ \cos(a - b) = \cos a \cos b + \sin a \sin b. $$

Place \(A = M(a) = (\cos a, \sin a)\) and \(B = M(b) = (\cos b, \sin b)\) on the trigonometric circle. Compute the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB}\) in two ways:
  • Coordinate-wise. \(\overrightarrow{OA} \cdot \overrightarrow{OB} = \cos a \cos b + \sin a \sin b\).
  • Geometric (lycée formula \(u \cdot v = \|u\| \|v\| \cos(\widehat{u, v})\)). Both vectors have norm \(1\) (they end on the unit circle), and the angle between them, measured from \(\overrightarrow{OB}\) to \(\overrightarrow{OA}\), is \(a - b\). Hence \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 1 \cdot 1 \cdot \cos(a - b) = \cos(a - b)\).
Equating the two expressions gives \(\cos(a - b) = \cos a \cos b + \sin a \sin b\).
Remark (rigour). The lycée formula \(u \cdot v = \|u\| \|v\| \cos(\widehat{u, v})\) is itself a consequence of the law of cosines, which is \(\cos(a - b)\) under another name --- so this proof has a circular flavour at the foundational level. A non-circular algebraic proof will be given in Complex numbers via \(|e^{ia} - e^{ib}|^2 = 2 - 2\cos(a - b)\). At the present level we accept the lycée scalar-product formula as a working axiom; no forward reference to general inner-product spaces is made.

Proposition — Other addition formulas
For all \(a, b \in \mathbb{R}\): $$ \begin{aligned} \cos(a + b) &= \cos a \cos b - \sin a \sin b, \\ \sin(a + b) &= \sin a \cos b + \cos a \sin b, \\ \sin(a - b) &= \sin a \cos b - \cos a \sin b. \end{aligned} $$

  • \(\cos(a + b)\). Replace \(b\) by \(-b\) in the master formula: \(\cos(a - (-b)) = \cos a \cos(-b) + \sin a \sin(-b) = \cos a \cos b - \sin a \sin b\), using parity of \(\cos\) and \(\sin\).
  • \(\sin(a + b)\). Use \(\sin x = \cos(\pi/2 - x)\). Then \(\sin(a + b) = \cos(\pi/2 - a - b) = \cos((\pi/2 - a) - b)\). Apply the master formula to the right-hand side: \(\cos(\pi/2 - a) \cos b + \sin(\pi/2 - a) \sin b = \sin a \cos b + \cos a \sin b\).
  • \(\sin(a - b)\). Replace \(b\) by \(-b\) in the previous identity and use parity.

Proposition — Duplication formulas
For all \(a \in \mathbb{R}\): $$ \cos 2a = \cos^2 a - \sin^2 a = 2\cos^2 a - 1 = 1 - 2\sin^2 a, \qquad \sin 2a = 2 \sin a \cos a. $$

Set \(b = a\) in the addition formulas. For \(\cos\): \(\cos(a + a) = \cos a \cos a - \sin a \sin a = \cos^2 a - \sin^2 a\). The two alternative forms come from \(\cos^2 a + \sin^2 a = 1\): replace \(\sin^2 a\) by \(1 - \cos^2 a\) to get \(2\cos^2 a - 1\), or replace \(\cos^2 a\) by \(1 - \sin^2 a\) to get \(1 - 2\sin^2 a\). For \(\sin\): \(\sin(a + a) = 2 \sin a \cos a\).

Proposition — Basic linearization
For all \(x \in \mathbb{R}\): $$ \cos^2 x = \frac{1 + \cos 2x}{2}, \qquad \sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos x \sin x = \frac{\sin 2x}{2}. $$

Direct rearrangement of the duplication formulas: from \(\cos 2x = 2\cos^2 x - 1\) one isolates \(\cos^2 x = (1 + \cos 2x)/2\); from \(\cos 2x = 1 - 2\sin^2 x\) one isolates \(\sin^2 x = (1 - \cos 2x)/2\); from \(\sin 2x = 2 \sin x \cos x\) one isolates \(\cos x \sin x = (\sin 2x)/2\).

Forward reference. The systematic linearization of \(\cos^p \theta \sin^q \theta\) for general \(p, q \in \mathbb{N}\) (recipe: write \(\cos\theta = (e^{i\theta} + e^{-i\theta})/2\), \(\sin\theta = (e^{i\theta} - e^{-i\theta})/(2i)\), expand by the binomial formula, group conjugate exponentials) is the subject of Complex numbers.
Proposition — Product-to-sum
For all \(a, b \in \mathbb{R}\): $$ \begin{aligned} \cos a \cos b &= \tfrac{1}{2}\big[\cos(a - b) + \cos(a + b)\big], \\ \sin a \sin b &= \tfrac{1}{2}\big[\cos(a - b) - \cos(a + b)\big], \\ \sin a \cos b &= \tfrac{1}{2}\big[\sin(a - b) + \sin(a + b)\big]. \end{aligned} $$

Add and subtract pairs of addition formulas.
For \(\cos a \cos b\): add \(\cos(a - b) = \cos a \cos b + \sin a \sin b\) and \(\cos(a + b) = \cos a \cos b - \sin a \sin b\) to get \(\cos(a - b) + \cos(a + b) = 2 \cos a \cos b\), then divide by \(2\).
For \(\sin a \sin b\): subtract the same two formulas: \(\cos(a - b) - \cos(a + b) = 2 \sin a \sin b\).
For \(\sin a \cos b\): add \(\sin(a + b) = \sin a \cos b + \cos a \sin b\) and \(\sin(a - b) = \sin a \cos b - \cos a \sin b\): \(\sin(a + b) + \sin(a - b) = 2 \sin a \cos b\).

Method — Choosing the right form of \(\cos 2a\)
The three equivalent forms \(\cos 2a = \cos^2 a - \sin^2 a = 2\cos^2 a - 1 = 1 - 2\sin^2 a\) are not interchangeable in practice. To eliminate \(\sin a\) (e.g.\ to integrate \(\cos^2 a\)), use \(\cos 2a = 2\cos^2 a - 1\), then \(\cos^2 a = (1 + \cos 2a)/2\). To eliminate \(\cos a\), use \(\cos 2a = 1 - 2 \sin^2 a\), then \(\sin^2 a = (1 - \cos 2a)/2\). The third form \(\cos^2 a - \sin^2 a\) is symmetric and useful when both \(\cos a\) and \(\sin a\) are kept.
Example
Compute \(\cos(\pi/12)\) using \(\pi/12 = \pi/3 - \pi/4\).

By the master addition formula, $$ \begin{aligned} \cos\!\left(\frac{\pi}{12}\right) &= \cos\!\left(\frac{\pi}{3} - \frac{\pi}{4}\right) \\ &= \cos\!\left(\frac{\pi}{3}\right) \cos\!\left(\frac{\pi}{4}\right) + \sin\!\left(\frac{\pi}{3}\right) \sin\!\left(\frac{\pi}{4}\right) \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} + \sqrt{6}}{4}. \end{aligned} $$

Two further forward references to Complex numbers.
  • The sum-to-product formulas \(\cos p \pm \cos q = \dots\), \(\sin p \pm \sin q = \dots\) follow in one line from \(e^{ip} \pm e^{iq} = (e^{i(p-q)/2} \pm e^{-i(p-q)/2}) e^{i(p+q)/2}\).
  • The amplitude-phase form \(a \cos x + b \sin x = R \cos(x - \varphi)\) follows from the polar form of \(a + ib\), with \(R = \sqrt{a^2 + b^2}\) and \(\varphi\) an argument of \(a + ib\).
Skills to practice
  • Computing remarkable sums of angles
  • Linearizing the basic squares
  • Using product-to-sum
IV The inequality \(|\sin x| \le |x|\) and derivatives of \(\sin\)\(\virgule\) \(\cos\)
Analysis tools used in this section
We use here the lycée notion of derivative and the standard derivative rules (sum, product, quotient), without rebuilding the full theory of differentiability. The rigorous general theory will be developed in the chapter Differentiability.
Two facts the analysis chapters will cite often: \(\sin x\) behaves like \(x\) near \(0\) (\(|\sin x| \le |x|\)), and \((\sin)' = \cos\), \((\cos)' = -\sin\). Both come from one geometric picture: chord \(\le\) arc on the unit circle. The deduction is: geometric inequality \(\Rightarrow\) corollary \(|\sin x| \le |x|\) \(\Rightarrow\) continuity of \(\cos\) at \(0\) \(\Rightarrow\) limit \(\sin x / x \to 1\) \(\Rightarrow\) derivatives of \(\sin\) and \(\cos\). The section on tangent below then uses these derivatives to compute \(\tan'\) via the quotient rule, with no acyclic dependency.
Pedagogical note (deliberate departure from the standard textbooks). The standard textbooks admit the derivatives \(\sin' = \cos\) and \(\cos' = -\sin\) at the start, and deduce \(\sin x / x \to 1\) as the value of the derivative of \(\sin\) at \(0\). We take here the historical, more rigorous direction: prove the geometric inequality on the circle, deduce \(\sin x / x \to 1\) by squeeze, then deduce \(\sin' = \cos\). This avoids any postulate on the derivative of \(\sin\) and gives the inequality \(|\sin x| \le |x|\) as a free side-product. The price is a slightly longer development here; the gain is a fully self-contained derivation matching the program's mandate « Démonstration : \(\sin x / x \to 1\) et \(|\sin x| \le |x|\) ».
Definition — Tangent (provisional)
For \(x \in \mathbb{R}\) with \(x \not\equiv \pi/2 \,[\pi]\), set $$ \tan x = \frac{\sin x}{\cos x}. $$ The full study of \(\tan\) (period, parity, addition formula, derivative, graph, \(\tan(t/2)\) parameterization) is the subject of the section on tangent below; here we use only this notation.
Proposition — Geometric inequality
For all \(x \in [0, \pi/2[\), $$ \sin x \le x \le \tan x. $$

Place on the unit circle the points \(A = (1, 0)\), \(P = M(x) = (\cos x, \sin x)\), and \(T = (1, \tan x)\) --- the intersection of the half-line \(OP\) with the vertical line \(x = 1\). Compare three nested regions:
  • the triangle \(OAP\) with base \(OA\) of length \(1\) and height \(\sin x\), area \(\tfrac{1}{2} \sin x\);
  • the circular sector \(OAP\) subtending an arc of length \(x\) on the unit circle, area \(\tfrac{1}{2} x\) (the radian definition above --- arc length on the unit circle equals the angle in radians --- gives sector area \(= \tfrac{1}{2} \times \text{radius} \times \text{arc} = \tfrac{1}{2} \times 1 \times x\) for \(r = 1\); this will be re-derived rigorously in Integration);
  • the right triangle \(OAT\) with leg \(OA = 1\) and leg \(AT = \tan x\), area \(\tfrac{1}{2} \tan x\).
The three regions are nested: \(\text{(triangle \)OAP\()} \subset \text{(sector \)OAP\()} \subset \text{(triangle \)OAT\()}\). Their areas are therefore ordered: $$ \frac{1}{2} \sin x \le \frac{1}{2} x \le \frac{1}{2} \tan x, $$ which is the announced inequality after multiplying by \(2\).

Proposition — Universal bound
For all \(x \in \mathbb{R}\), $$ |\sin x| \le |x|. $$

  • For \(x \in [0, \pi/2[\): from the geometric inequality, \(\sin x \le x\), hence \(|\sin x| = \sin x \le x = |x|\).
  • For \(x \ge \pi/2\): \(|\sin x| \le 1 < \pi/2 \le x = |x|\).
  • For \(x \le 0\): by parity, \(|\sin x| = |\sin(-x)| \le |-x| = |x|\) (use the previous two cases on \(-x \ge 0\)).

Proposition — Continuity of \(\cos\) at \(0\)
\(\displaystyle \lim_{x \to 0} \cos x = 1\).

By the duplication formula \(1 - \cos x = 2 \sin^2(x/2)\). Combined with \(|\sin(x/2)| \le |x/2|\) from the previous Proposition, $$ 0 \le 1 - \cos x = 2 \sin^2(x/2) \le 2 \cdot \left(\frac{x}{2}\right)^2 = \frac{x^2}{2}. $$ The right-hand bound tends to \(0\) as \(x \to 0\), hence by the squeeze theorem \(1 - \cos x \to 0\), i.e.\ \(\cos x \to 1\).

Proposition — Limit of \(\sin x / x\) at \(0\)
\(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\).

For \(x \in \,]0, \pi/2[\), divide the geometric inequality \(\sin x \le x \le \tan x = \sin x / \cos x\) by \(\sin x > 0\) to get \(1 \le x / \sin x \le 1/\cos x\), then take reciprocals (all terms positive): $$ \cos x \le \frac{\sin x}{x} \le 1. $$ The left bound \(\cos x\) tends to \(1\) as \(x \to 0^+\) by the previous Proposition, and the right bound is constantly \(1\). By the squeeze theorem, \(\sin x / x \to 1\) as \(x \to 0^+\).
The function \(x \mapsto \sin x / x\) is even (ratio of an odd by an odd), so the same limit holds as \(x \to 0^-\). Hence \(\sin x / x \to 1\) as \(x \to 0\).

Proposition — Derivatives of \(\sin\) and \(\cos\)
The functions \(\sin\) and \(\cos\) are differentiable on \(\mathbb{R}\), and for all \(x \in \mathbb{R}\): $$ \sin'(x) = \cos x, \qquad \cos'(x) = -\sin x. $$

Fix \(x \in \mathbb{R}\). By the addition formula \(\sin(x + h) = \sin x \cos h + \cos x \sin h\), the difference quotient is $$ \frac{\sin(x + h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}. $$ As \(h \to 0\): the second term \(\frac{\sin h}{h}\) tends to \(1\) (previous Proposition), and the first factor \(\frac{\cos h - 1}{h} = -\frac{1 - \cos h}{h}\) tends to \(0\) since \(|1 - \cos h| \le h^2/2\) gives \(|(1 - \cos h)/h| \le |h|/2 \to 0\). Hence $$ \sin'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x. $$ Symmetrically, by the addition formula \(\cos(x + h) = \cos x \cos h - \sin x \sin h\), $$ \frac{\cos(x + h) - \cos x}{h} = \cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h} \xrightarrow[h \to 0]{} 0 - \sin x = -\sin x. $$

Method — « \(\sin x \approx x\) for small \(x\) »
The intuition « \(\sin x \approx x\) near \(0\) », constantly invoked in physics and analysis, is rigorously justified by \(|\sin x - x| \le |x|^3 / 6\) (a sharper bound, treated later in Limited expansions) or, at the present level, by the limit \(\sin x / x \to 1\). In practice: whenever a problem asks for the leading-order behaviour of an expression involving \(\sin x\) near \(0\), replace \(\sin x\) by \(x\) and check the consistency by computing the limit of the resulting ratio.
Example
A numerical use of \(|\sin x| \le |x|\): bound the error in the approximation \(\sin(0.1) \approx 0.1\).

Write \(x = 0.1 > 0\). Lower estimate. The chapter inequality \(\sin x \le |x| = x\) (valid for \(x \ge 0\)) gives at once $$ x - \sin x \ge 0, \qquad \text{i.e.} \quad \sin(0.1) \le 0.1. $$ So the approximation \(\sin(0.1) \approx 0.1\) over-estimates, never under-estimates. Upper estimate of the error. The two tools proved in this section, \(|\sin x| \le |x|\) alone, only yield the crude bound \(0 \le x - \sin x \le x\), far too weak to read four decimals. The sharp bound that makes « \(\sin x \approx x\) » quantitative is $$ 0 \le x - \sin x \le \frac{x^3}{6}, $$ announced in the Method above and proved later in Limited expansions (it is not available with the elementary tools of this chapter alone). Granting it, at \(x = 0.1\): $$ 0 \le 0.1 - \sin(0.1) \le \frac{(0.1)^3}{6} = \frac{10^{-3}}{6} \approx 1.67 \cdot 10^{-4}. $$ Hence \(\sin(0.1) \in [\,0.1 - 1.67 \cdot 10^{-4},\ 0.1\,]\), so \(\sin(0.1) = 0.0998\) to four decimals, and the approximation \(\sin(0.1) \approx 0.1\) is correct to within \(2 \cdot 10^{-4}\). (With the chapter's own tools, only the qualitative conclusion \(\sin(0.1) \le 0.1\) and \(\sin(0.1) \to 0.1\) in relative terms is rigorous; the four-digit accuracy rests on the deferred cubic bound.) This is the working version of « \(\sin x \approx x\) for small \(x\) ».

Skills to practice
  • Proving the geometric inequality
  • Computing limits and derivatives of \(\sin\)\(\virgule\) \(\cos\)
V Tangent: full treatment
Geometrically, the tangent of an angle \(\theta\) is the \(y\)-coordinate of the intersection of the line through \(O\) at angle \(\theta\) with the vertical line \(x = 1\) tangent to the unit circle at \((1, 0)\). From this single picture, the period, parity, addition formula, derivative, variations, and the \(\tan(t/2)\) parameterization all follow.
Definition — Tangent
For \(\theta \in \mathbb{R}\) with \(\theta \not\equiv \pi/2 \,[\pi]\), the tangent of \(\theta\) is $$ \tan\theta = \frac{\sin\theta}{\cos\theta}. $$ The domain of \(\tan\) is \(\mathbb{R} \setminus \{ \pi/2 + k\pi \mid k \in \mathbb{Z}\}\).
Proposition — Period\(\virgule\) parity\(\virgule\) shifts\(\virgule\) standard values
On its domain:
  • \(\tan\) is \(\pi\)-periodic: \(\tan(\theta + \pi) = \tan\theta\).
  • \(\tan\) is odd: \(\tan(-\theta) = -\tan\theta\).
  • Identities at \(\pi\)-shifts: \(\tan(\pi + \theta) = \tan\theta\) (\(\pi\)-periodicity restated) and \(\tan(\pi - \theta) = -\tan\theta\) (different sign, do not write the two as « \(\tan(\pi \pm \theta) = \pm \tan\theta\) »).
  • Standard values: $$ \begin{array}{c|ccccc} \theta & 0 & \pi/6 & \pi/4 & \pi/3 & \pi/2 \\ \hline \tan\theta & 0 & \dfrac{1}{\sqrt{3}} & 1 & \sqrt{3} & \text{undefined} \end{array} $$

  • \(\pi\)-periodicity: \(\tan(\theta + \pi) = \sin(\theta + \pi)/\cos(\theta + \pi) = (-\sin\theta)/(-\cos\theta) = \sin\theta/\cos\theta = \tan\theta\).
  • Parity: \(\tan(-\theta) = \sin(-\theta)/\cos(-\theta) = -\sin\theta/\cos\theta = -\tan\theta\).
  • \(\tan(\pi - \theta) = \sin(\pi - \theta)/\cos(\pi - \theta) = \sin\theta/(-\cos\theta) = -\tan\theta\).
  • Standard values: read from the table of \(\sin\) and \(\cos\) above.

Proposition — Addition formula for \(\tan\)
For \(a, b \in \mathbb{R}\) with \(a, b \notin \pi/2 + \pi\mathbb{Z}\) and \(1 \mp \tan a \tan b \ne 0\), one has \(a \pm b \notin \pi/2 + \pi\mathbb{Z}\) and $$ \tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}. $$

We treat the « \(+\) » sign; the « \(-\) » sign is analogous (replace \(b\) by \(-b\) and use parity of \(\tan\)). By the addition formulas above: $$ \tan(a + b) = \frac{\sin(a + b)}{\cos(a + b)} = \frac{\sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b}. $$ The hypothesis \(a, b \notin \pi/2 + \pi\mathbb{Z}\) ensures \(\cos a \ne 0\) and \(\cos b \ne 0\), so we can divide numerator and denominator by \(\cos a \cos b\) (non-zero): $$ \tan(a + b) = \frac{\sin a / \cos a + \sin b / \cos b}{1 - (\sin a / \cos a)(\sin b / \cos b)} = \frac{\tan a + \tan b}{1 - \tan a \tan b}, $$ which is well-defined precisely when the denominator \(1 - \tan a \tan b\) is non-zero --- the second hypothesis. The non-vanishing of \(\cos(a + b)\) then forces \(a + b \notin \pi/2 + \pi\mathbb{Z}\).

Proposition — Duplication of \(\tan\)
For \(a \in \mathbb{R}\) with \(a \notin \pi/2 + \pi\mathbb{Z}\) and \(1 - \tan^2 a \ne 0\): $$ \tan 2a = \frac{2 \tan a}{1 - \tan^2 a}. $$

Set \(b = a\) in the addition formula for \(\tan\).

Proposition — Derivative of \(\tan\)
\(\tan\) is differentiable on its domain \(\mathbb{R} \setminus (\pi/2 + \pi\mathbb{Z})\), and $$ \tan'(x) = 1 + \tan^2 x = \frac{1}{\cos^2 x}. $$

On the domain of \(\tan\), \(\cos x \ne 0\). Apply the quotient rule to \(\tan = \sin / \cos\), using \(\sin' = \cos\) and \(\cos' = -\sin\) from the previous section: $$ \tan'(x) = \frac{\sin'(x) \cos x - \sin x \cos'(x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}. $$ The alternative form \(1 + \tan^2 x\) comes from dividing the numerator \(\cos^2 x + \sin^2 x\) by \(\cos^2 x\): $$ \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = 1 + \frac{\sin^2 x}{\cos^2 x} = 1 + \tan^2 x. $$ Practical note. The identity \(\tan'(x) = 1 + \tan^2 x = 1/\cos^2 x\) is less useful to compute \(\tan'\) than to convert \(\cos\) into \(\tan\) or vice-versa: \(1/\cos^2 x = 1 + \tan^2 x\). It is in this « \(\cos \leftrightarrow \tan\) bridge » form that one should remember it.

Proposition — Variations and graph of \(\tan\) on the principal interval
On \(]-\pi/2, \pi/2[\), \(\tan\) is strictly increasing, \(\tan(-\pi/2^+) = -\infty\), \(\tan(\pi/2^-) = +\infty\), and \(\tan(0) = 0\). $$ \begin{array}{c|ccccc} x & -\pi/2 & & 0 & & \pi/2 \\ \hline \tan'(x) & & + & + & + & \\ \hline \tan(x) & -\infty & \nearrow & 0 & \nearrow & +\infty \end{array} $$

Strict monotonicity follows from \(\tan'(x) = 1/\cos^2 x > 0\) on the open interval \(]-\pi/2, \pi/2[\). The limits at the endpoints come from \(\sin(\pm\pi/2) = \pm 1\) and \(\cos(\pm\pi/2^{\mp}) = 0^+\), so \(\tan(\pm\pi/2^{\mp}) = \pm 1/0^+ = \pm \infty\).

Proposition — The \(\tan(t/2)\) parameterization
For \(t \in \mathbb{R}\) with \(t \not\equiv \pi \,[2\pi]\), set \(u = \tan(t/2)\). Then $$ \cos t = \frac{1 - u^2}{1 + u^2}, \qquad \sin t = \frac{2u}{1 + u^2}. $$ If furthermore \(t \notin \pi/2 + \pi\mathbb{Z}\) (so \(1 - u^2 \ne 0\)), $$ \tan t = \frac{2u}{1 - u^2}. $$

By the duplication formulas applied to \(a = t/2\): $$ \cos t = 2 \cos^2(t/2) - 1 = 1 - 2 \sin^2(t/2), \qquad \sin t = 2 \sin(t/2) \cos(t/2). $$ The condition \(t \not\equiv \pi \,[2\pi]\) gives \(t/2 \not\equiv \pi/2 \,[\pi]\), so \(\cos(t/2) \ne 0\) and \(u = \tan(t/2)\) is well-defined. Use the identity \(1 + \tan^2(t/2) = 1/\cos^2(t/2)\) from the previous Proposition to extract \(\cos^2(t/2) = 1/(1 + u^2)\).
  • For \(\cos t\): \(\cos t = 2 \cos^2(t/2) - 1 = 2/(1 + u^2) - 1 = (2 - 1 - u^2)/(1 + u^2) = (1 - u^2)/(1 + u^2)\).
  • For \(\sin t\): \(\sin t = 2 \sin(t/2) \cos(t/2) = 2 \tan(t/2) \cos^2(t/2) = 2u \cdot 1/(1 + u^2) = 2u/(1 + u^2)\).
  • For \(\tan t\): \(\tan t = \sin t / \cos t = (2u/(1 + u^2))/((1 - u^2)/(1 + u^2)) = 2u/(1 - u^2)\), valid when the denominator \(1 - u^2\) is non-zero.

Method — When to use the \(\tan(t/2)\) substitution
The substitution \(u = \tan(t/2)\) converts any rational function of \(\sin t\) and \(\cos t\) into a rational function of \(u\). This is the canonical move when the trigonometric content of an expression is « rational in \(\sin t, \cos t\) »: the substitution removes the trigonometry and reduces the problem to algebra. The application to integration (Weierstrass substitution) belongs to the chapter Antiderivatives.
Example
Verify the \(\tan(t/2)\) formulas at \(t = \pi/3\).

We have \(u = \tan(\pi/6) = 1/\sqrt{3}\), hence \(u^2 = 1/3\) and \(1 + u^2 = 4/3\). Then $$ \frac{1 - u^2}{1 + u^2} = \frac{2/3}{4/3} = \frac{1}{2} = \cos(\pi/3) \quad \checkmark, \qquad \frac{2u}{1 + u^2} = \frac{2/\sqrt{3}}{4/3} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2} = \sin(\pi/3) \quad \checkmark. $$

Skills to practice
  • Computing with \(\tan\)\(\virgule\) addition formula
  • Sketching variations\(\virgule\) derivative\(\virgule\) graph of \(\tan\)
  • Using the \(\tan(t/2)\) parameterization
VI Trigonometric equations and inequations
Solving \(\cos x = a\) asks: which angles \(x\) project to \(a\) on the \(x\)-axis of the trigonometric circle? The picture says: at most two angles in any interval of length \(2\pi\), then add multiples of \(2\pi\). The notation that captures « add multiples of \(2\pi\) » is the congruence \(\equiv \,[2\pi]\) introduced earlier in the chapter. Inequations follow the same picture: « \(\cos x \ge a\) » becomes « which angles project to \(\ge a\)? », whose answer is an arc of the circle.
Theorem — Solutions of \(\cos x \equal a\)
Let \(a \in \mathbb{R}\).
  • If \(a \notin [-1, 1]\), the equation \(\cos x = a\) has no solution in \(\mathbb{R}\).
  • If \(a \in [-1, 1]\), denote by \(\alpha\) the unique angle of \([0, \pi]\) whose cosine is \(a\). (At this point \(\alpha\) is a named angle, not the value of an « \(\arccos\) » function --- whose differentiability and graph live in Standard functions.) Then $$ \cos x = a \iff x \equiv \alpha \,[2\pi] \quad \text{or} \quad x \equiv -\alpha \,[2\pi]. $$

The first case follows from \(\cos x \in [-1, 1]\) for all \(x\).
For the second case, the function \(\cos\) is continuous and strictly decreasing on \([0, \pi]\) from \(\cos 0 = 1\) to \(\cos \pi = -1\), hence (by the intermediate value theorem and strict monotonicity, both lycée results --- formalised in Differentiability) it is a bijection \([0, \pi] \to [-1, 1]\). So there is a unique \(\alpha \in [0, \pi]\) with \(\cos \alpha = a\). By the symmetry \(\cos(-\theta) = \cos\theta\) and \(2\pi\)-periodicity, the full set of solutions of \(\cos x = a\) in \(\mathbb{R}\) is the union of \(\alpha + 2\pi\mathbb{Z}\) and \(-\alpha + 2\pi\mathbb{Z}\), which is the announced congruence statement.

Proposition — Equality of two cosines\(\virgule\) sines\(\virgule\) tangents
For all \(a, b \in \mathbb{R}\): $$ \begin{aligned} \cos a = \cos b &\iff a \equiv b \,[2\pi] \quad \text{or} \quad a \equiv -b \,[2\pi], \\ \sin a = \sin b &\iff a \equiv b \,[2\pi] \quad \text{or} \quad a \equiv \pi - b \,[2\pi], \\ \tan a = \tan b &\iff a \equiv b \,[\pi] \quad (\text{provided \(a, b \notin \pi/2 + \pi\mathbb{Z}\)}). \end{aligned} $$ This « two-named-angle » form is the one most often used in practice: every time a problem reduces to comparing two trigonometric values, this is the rule that produces the parametric family of solutions. It is the symmetric counterpart of the equation Theorems for \(\cos\), \(\sin\), and \(\tan\) stated nearby.

  • Cosine. \(\cos a = \cos b\) means \(a\) is a solution of « \(\cos x = \cos b\) ». By the previous Theorem (with the value \(\cos b \in [-1, 1]\) and the named angle \(\alpha\) from \([0, \pi]\) such that \(\cos\alpha = \cos b\)), the solution set is \(\{a : a \equiv \alpha \,[2\pi]\) or \(a \equiv -\alpha \,[2\pi]\}\). Re-applying \(\cos\alpha = \cos b\) with the same Theorem to \(b\) gives \(b \equiv \alpha \,[2\pi]\) or \(b \equiv -\alpha \,[2\pi]\), hence the announced two cases up to congruence.
  • Sine. Same argument with the symmetry \(\sin(\pi - \theta) = \sin\theta\) in place of \(\cos(-\theta) = \cos\theta\). (This implicitly uses the sin Theorem stated below; the dependency is acyclic since the sin Theorem itself does not use the present Proposition.)
  • Tangent. The \(\pi\)-periodicity of \(\tan\) collapses the two congruence families into a single \(a \equiv b \,[\pi]\). (Same caveat: this uses the tan Theorem stated below.)

Theorem — Solutions of \(\sin x \equal a\)
Let \(a \in \mathbb{R}\).
  • If \(a \notin [-1, 1]\), the equation \(\sin x = a\) has no solution in \(\mathbb{R}\).
  • If \(a \in [-1, 1]\), denote by \(\beta\) the unique angle of \([-\pi/2, \pi/2]\) whose sine is \(a\) (named angle, not the « \(\arcsin\) » function). Then $$ \sin x = a \iff x \equiv \beta \,[2\pi] \quad \text{or} \quad x \equiv \pi - \beta \,[2\pi]. $$

Same structure as for \(\cos\): \(\sin\) is a bijection \([-\pi/2, \pi/2] \to [-1, 1]\) (continuous and strictly increasing --- both lycée results, formalised in Differentiability; the strict monotonicity comes from \(\sin'(x) = \cos x > 0\) on \(]-\pi/2, \pi/2[\) which is established earlier in the chapter). The symmetry \(\sin(\pi - \theta) = \sin\theta\) then gives the second family of solutions.

Theorem — Solutions of \(\tan x \equal a\)
Let \(a \in \mathbb{R}\). Denote by \(\gamma\) the unique angle of \(]-\pi/2, \pi/2[\) whose tangent is \(a\) (named angle, not the « \(\arctan\) » function). Then $$ \tan x = a \iff x \equiv \gamma \,[\pi]. $$ The congruence is modulo \(\pi\) (\(\pi\)-periodicity of \(\tan\)), not modulo \(2\pi\).

By the previous section, \(\tan\) is a bijection \(]-\pi/2, \pi/2[\, \to \mathbb{R}\), so \(\gamma\) exists and is unique. The \(\pi\)-periodicity of \(\tan\) then gives the announced family.

Method — Solving an equation that reduces to \(\cos x \equal a\)\(\virgule\) etc.
  1. Reduce algebraically to \(\cos x = a\), \(\sin x = a\), or \(\tan x = a\) --- typical tools: addition / duplication / linearization formulas, factor a \(\cos\) or \(\sin\), substitute \(\sin^2 = 1 - \cos^2\).
  2. Apply the matching theorem to write the solution set as a union of two (or one for \(\tan\)) congruence classes modulo \(2\pi\) (modulo \(\pi\) for \(\tan\)).
  3. If the problem restricts to a bounded interval \([a, b]\), intersect each congruence class with \([a, b]\) and list the finitely many concrete solutions.
Method — Solving a simple inequation \(\cos x \ge a\)\(\virgule\) etc.
  1. Draw the trigonometric circle and the level line \(x = a\) (vertical) for \(\cos x \ge a\), \(y = a\) (horizontal) for \(\sin x \ge a\), or the tangent graph for \(\tan x \ge a\).
  2. Identify the arc of the circle (or the union of intervals on the \(\tan\) graph) on which the inequality holds. Read off two endpoints in \([0, 2\pi[\) (or \([-\pi/2, \pi/2[\) for \(\tan\)).
  3. Lift to \(\mathbb{R}\) by union over \(2k\pi\) (or \(k\pi\) for \(\tan\)): the solution set is a union of intervals. Do not write « \(x \equiv \alpha \,[2\pi]\) with \(\alpha \in [-\beta, \beta]\) » --- a congruence is on a number, not a range.
Example
Solve \(\cos 2x = \sin x\) on \([0, 2\pi]\).

By duplication, \(\cos 2x = 1 - 2 \sin^2 x\). The equation becomes $$ 1 - 2 \sin^2 x = \sin x \iff 2 \sin^2 x + \sin x - 1 = 0. $$ Set \(u = \sin x\) and solve the quadratic \(2u^2 + u - 1 = 0\): discriminant \(\Delta = 1 + 8 = 9\), roots \(u = (-1 \pm 3)/4\), hence \(u = 1/2\) or \(u = -1\).
  • \(\sin x = 1/2\): \(\beta = \pi/6\), so \(x \equiv \pi/6 \,[2\pi]\) or \(x \equiv 5\pi/6 \,[2\pi]\). On \([0, 2\pi]\): \(x = \pi/6\) or \(x = 5\pi/6\).
  • \(\sin x = -1\): \(\beta = -\pi/2\), so \(x \equiv -\pi/2 \,[2\pi]\). On \([0, 2\pi]\): \(x = 3\pi/2\).
The solutions are \(x \in \{\pi/6, \, 5\pi/6, \, 3\pi/2\}\).

Example
Solve \(\cos x \ge 1/2\) on \(\mathbb{R}\).

On the trigonometric circle, the condition \(\cos x \ge 1/2\) describes the arc of points whose abscissa is at least \(1/2\). The two boundary angles in \([-\pi, \pi]\) are \(-\pi/3\) and \(\pi/3\) (since \(\cos(\pm \pi/3) = 1/2\)), and the favourable arc is \([-\pi/3, \pi/3]\). Lifting to \(\mathbb{R}\) by \(2\pi\)-periodicity: $$ \{ x \in \mathbb{R} \mid \cos x \ge 1/2 \} = \bigcup_{k \in \mathbb{Z}} \left[2k\pi - \frac{\pi}{3}, \, 2k\pi + \frac{\pi}{3}\right]. $$

Skills to practice
  • Solving \(\cos x \equal a\)\(\virgule\) \(\sin x \equal a\)\(\virgule\) \(\tan x \equal a\)
  • Solving by reduction to \(\cos\)\(\virgule\) \(\sin\)\(\virgule\) \(\tan\)
  • Proving the parametric form of solutions
  • Solving simple inequations