High School Mathematics · Grades 10–12

24Trigonometric Functions

The functions $\cos$ and $\sin$ turn the geometry of the circle into analysis. Defined by winding the real line around the unit circle, they are the archetype of periodic functions, and their derivatives make them solutions of the equation of oscillations $y'' = -y$ .

24.1 The unit circle

Definition 24.1 (Cosine and sine)

Let $\mathcal{C}$ be the circle of radius 1 centered at the origin of an orthonormal coordinate system. To each real $t$ , associate the point $M(t)$ obtained by winding a length $\abs{t}$ along $\mathcal{C}$ from the point $I(1,0)$ , counterclockwise if $t \geq 0$ and clockwise otherwise. Then $\cos t$ and $\sin t$ are the coordinates of $M(t)$ :

M(t) = (\cos t, \sin t).

Since the circle has circumference $2\pi$ , the point $M(t)$ is unchanged when $t$ increases by $2\pi$ ; and since $M(t)$ lies on the unit circle:

Proposition 24.2 (Fundamental identities)

For all $t \in \R$ and $k \in \Z$ :

\cos^2 t + \sin^2 t = 1, \qquad \cos(t + 2k\pi) = \cos t, \qquad \sin(t + 2k\pi) = \sin t,

\cos(-t) = \cos t, \qquad \sin(-t) = -\sin t,

\cos(\pi - t) = -\cos t, \quad \sin(\pi - t) = \sin t, \quad \cos\left(\tfrac{\pi}{2} - t\right) = \sin t, \quad \sin\left(\tfrac{\pi}{2} - t\right) = \cos t .

Proof. The first identity is the equation of the unit circle; the others express the symmetries of the circle: $M(-t)$ is the reflection of $M(t)$ in the $x$ -axis, $M(\pi - t)$ in the $y$ -axis, and $M(\frac\pi2 - t)$ in the line $y = x$ . ∎

The values to know by heart:

$t$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$
[6pt] $\cos t$	$1$	$\dfrac{\sqrt3}{2}$	$\dfrac{\sqrt2}{2}$	$\dfrac12$	$0$
[6pt] $\sin t$	$0$	$\dfrac12$	$\dfrac{\sqrt2}{2}$	$\dfrac{\sqrt3}{2}$	$1$

Proposition 24.3 (Addition formulas)

For all $a, b \in \R$ :

\begin{align*} \cos(a+b) &= \cos a \cos b - \sin a \sin b, & \sin(a+b) &= \sin a \cos b + \cos a \sin b,\\ \cos(a-b) &= \cos a \cos b + \sin a \sin b, & \sin(a-b) &= \sin a \cos b - \cos a \sin b. \end{align*}

In particular $\cos 2a = 2\cos^2 a - 1 = 1 - 2\sin^2 a$ and $\sin 2a = 2 \sin a \cos a$ .

Proof. The vectors $\vect{u} = (\cos a, \sin a)$ and $\vect{v} = (\cos b, \sin b)$ are unit vectors making an angle $a - b$ ; their scalar product equals both $\cos a\cos b + \sin a \sin b$ (coordinates) and $\norm{\vect u}\,\norm{\vect v}\cos(a-b) = \cos(a - b)$ (geometry), which is the third formula. Replacing $b$ by $-b$ gives the first; the sine formulas follow using $\sin x = \cos(\frac\pi2 - x)$ . The duplication formulas are the case $b = a$ combined with $\cos^2 + \sin^2 = 1$ . ∎

24.2 Analytic study

Lemma 24.4 (Fundamental limit)

\lim_{t \to 0} \frac{\sin t}{t} = 1 \qquad\text{and}\qquad \lim_{t \to 0} \frac{\cos t - 1}{t} = 0 .

Proof. For $0 < t < \frac{\pi}{2}$ , compare three areas in the unit circle: the triangle $OIM(t)$ , the circular sector $OIM(t)$ , and the right triangle with base $OI$ and height $\tan t$ :

\frac{\sin t}{2} \leq \frac{t}{2} \leq \frac{\tan t}{2}.

The three nested areas: the triangle OIM (blue, area t/2), the circular sector (bounded by the red arc, area t/2), and the right triangle of height t (orange, area t/2). — The three nested areas: the triangle $OIM$ (blue, area $\frac{\sin t}{2}$ ), the circular sector (bounded by the red arc, area $\frac{t}{2}$ ), and the right triangle of height $\tan t$ (orange, area $\frac{\tan t}{2}$ ).

The first inequality gives $\frac{\sin t}{t} \leq 1$ ; the second gives $\cos t \leq \frac{\sin t}{t}$ . As $t \to 0^+$ , $\cos t \to 1$ (continuity), and the squeeze theorem yields $\frac{\sin t}{t} \to 1$ ; parity extends this to $t \to 0$ . For the second limit,

\frac{\cos t - 1}{t} = \frac{\cos^2 t - 1}{t(\cos t + 1)} = -\frac{\sin t}{t}\cdot\frac{\sin t}{\cos t + 1} \xrightarrow[t\to0]{} -1 \cdot \frac{0}{2} = 0. \qedhere

∎

Theorem 24.5 (Derivatives of sine and cosine)

The functions $\sin$ and $\cos$ are differentiable on $\R$ , with

(\sin)' = \cos, \qquad (\cos)' = -\sin .

More generally, $(\sin(ax+b))' = a\cos(ax+b)$ and $(\cos(ax+b))' = -a\sin(ax+b)$ .

Proof. By the addition formulas,

\frac{\sin(x+h) - \sin x}{h} = \sin x\,\frac{\cos h - 1}{h} + \cos x\,\frac{\sin h}{h} \xrightarrow[h\to0]{} \sin x \cdot 0 + \cos x \cdot 1 = \cos x,

using Lemma 24.4. The computation for $\cos$ is identical, and the general form follows from the chain rule. ∎

Proposition 24.6 (Variations)

On $\intcc{0}{\pi}$ , $\cos$ decreases from $1$ to $-1$ . On $\intcc{-\frac\pi2}{\frac\pi2}$ , $\sin$ increases from $-1$ to $1$ . Both functions are $2\pi$ -periodic and have ranges $\intcc{-1}{1}$ ; $\cos$ is even, $\sin$ is odd.

Proof. On $\intoo{0}{\pi}$ , $(\cos)' = -\sin < 0$ since the point $M(t)$ has positive ordinate there; similarly $(\sin)' = \cos > 0$ on $\intoo{-\frac\pi2}{\frac\pi2}$ . Periodicity and parity come from Proposition 24.2. ∎

The curves of (blue) and (red). — The curves of $\cos$ (blue) and $\sin$ (red).

Method 24.7 (Solving $\cos x = a$ and $\sin x = a$ )

For $a \in \intcc{-1}{1}$ , find one particular solution $\alpha$ (from the table of values or a calculator). Then all solutions are:

\cos x = \cos\alpha \iff x = \alpha + 2k\pi \text{ or } x = -\alpha + 2k\pi, \quad k \in \Z;

\sin x = \sin\alpha \iff x = \alpha + 2k\pi \text{ or } x = \pi - \alpha + 2k\pi, \quad k \in \Z.

For inequalities, locate the solution arcs on the unit circle and read off the intervals within one period.

Example 24.8

Solve $\cos x = \frac12$ in $\intoc{-\pi}{\pi}$ : a particular solution is $\frac\pi3$ , so the solutions are $x = \frac\pi3$ and $x = -\frac\pi3$ . In all of $\R$ : $x = \pm\frac\pi3 + 2k\pi$ , $k \in \Z$ .

24.3 Exercises

Exercise 24.1 ★

Compute $\cos\frac{2\pi}{3}$ , $\sin\frac{5\pi}{6}$ , $\cos\frac{7\pi}{4}$ and $\sin\left(-\frac{\pi}{3}\right)$ using the identities of Proposition 24.2.

Solution

Solution of Exercise 24.1.

$\cos\frac{2\pi}{3} = \cos\left(\pi - \frac\pi3\right) = -\cos\frac\pi3 = -\frac12$ .

$\sin\frac{5\pi}{6} = \sin\left(\pi - \frac\pi6\right) = \sin\frac\pi6 = \frac12$ .

$\cos\frac{7\pi}{4} = \cos\left(2\pi - \frac{\pi}{4}\right) = \cos\left(-\frac\pi4\right) = \frac{\sqrt2}{2}$ .

$\sin\left(-\frac\pi3\right) = -\sin\frac\pi3 = -\frac{\sqrt3}{2}$ .

Exercise 24.2 ★

Differentiate $f(x) = \sin^2 x$ , $g(x) = \cos(3x) + x\sin x$ and $h(x) = \dfrac{\sin x}{2 + \cos x}$ .

Solution

Solution of Exercise 24.2.

$f'(x) = 2\sin x\cos x = \sin 2x$ (chain rule).

$g'(x) = -3\sin(3x) + \sin x + x\cos x$ (chain and product rules).

Quotient rule:

h'(x) = \frac{\cos x\,(2 + \cos x) - \sin x\,(-\sin x)}{(2+\cos x)^2} = \frac{2\cos x + \cos^2 x + \sin^2 x}{(2+\cos x)^2} = \frac{2\cos x + 1}{(2+\cos x)^2}.

Exercise 24.3 ★

Solve in $\intoc{-\pi}{\pi}$ , then in $\R$ :

\text{(a) } \sin x = \frac{\sqrt3}{2}; \qquad \text{(b) } \cos\left(2x\right) = \frac{\sqrt2}{2} .

Solution

Solution of Exercise 24.3.

(a) A particular solution of $\sin x = \frac{\sqrt3}{2}$ is $\frac\pi3$ . General solutions: $x = \frac\pi3 + 2k\pi$ or $x = \pi - \frac\pi3 + 2k\pi = \frac{2\pi}{3} + 2k\pi$ . In $\intoc{-\pi}{\pi}$ : $x \in \left\{\frac\pi3, \frac{2\pi}{3}\right\}$ .

(b) $\cos 2x = \frac{\sqrt2}{2} = \cos\frac\pi4$ gives $2x = \pm\frac\pi4 + 2k\pi$ , so $x = \pm\frac\pi8 + k\pi$ . In $\intoc{-\pi}{\pi}$ : $x \in \left\{-\frac{7\pi}{8}, -\frac\pi8, \frac\pi8, \frac{7\pi}{8}\right\}$ .

Exercise 24.4 ★★

Using the addition formulas, compute the exact value of $\cos\frac{\pi}{12}$ and $\sin\frac{\pi}{12}$ . (Hint: $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ .)

Solution

Solution of Exercise 24.4.

With $a = \frac\pi3$ , $b = \frac\pi4$ :

\cos\frac{\pi}{12} = \cos a\cos b + \sin a \sin b = \frac12\cdot\frac{\sqrt2}{2} + \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} = \frac{\sqrt2 + \sqrt6}{4},

\sin\frac{\pi}{12} = \sin a\cos b - \cos a \sin b = \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} - \frac12\cdot\frac{\sqrt2}{2} = \frac{\sqrt6 - \sqrt2}{4}.

Exercise 24.5 ★★

Solve in $\intcc{0}{2\pi}$ the inequality $2\sin^2 x - \sin x - 1 \geq 0$ . (Hint: factor the quadratic in $\sin x$ .)

Solution

Solution of Exercise 24.5.

Set $s = \sin x$ : $2s^2 - s - 1 = (2s + 1)(s - 1) \geq 0$ iff $s \leq -\frac12$ or $s = 1$ .

$\sin x = 1$ gives $x = \frac\pi2$ . $\sin x \leq -\frac12$ : on the unit circle, the arc below the horizontal line $y = -\frac12$ , which in $\intcc{0}{2\pi}$ is $x \in \intcc{\frac{7\pi}{6}}{\frac{11\pi}{6}}$ . Hence

S = \left\{\frac{\pi}{2}\right\} \cup \intcc{\frac{7\pi}{6}}{\frac{11\pi}{6}} .

Exercise 24.6 ★★

Compute the limits

\lim_{x\to0} \frac{\sin 3x}{x}, \qquad \lim_{x\to0} \frac{1 - \cos x}{x^2}, \qquad \lim_{x\to+\infty} x \sin\frac{1}{x}.

Solution

Solution of Exercise 24.6.

$\dfrac{\sin 3x}{x} = 3\,\dfrac{\sin 3x}{3x} \to 3 \cdot 1 = 3$ (composition with $u = 3x \to 0$ ).

Using $1 - \cos x = \dfrac{\sin^2 x}{1 + \cos x}$ :

\frac{1 - \cos x}{x^2} = \left(\frac{\sin x}{x}\right)^{\!2} \frac{1}{1 + \cos x} \xrightarrow[x\to0]{} 1 \cdot \frac12 = \frac12 .

Substituting $u = \frac1x \to 0^+$ : $x \sin\frac1x = \frac{\sin u}{u} \to 1$ .

Exercise 24.7 ★★

Study the function $f(x) = x + \cos x$ on $\intcc{0}{2\pi}$ : variations and extrema. Show that $f$ is increasing on $\R$ although $f'$ vanishes at infinitely many points.

Solution

Solution of Exercise 24.7.

$f'(x) = 1 - \sin x \geq 0$ for all $x$ , with equality exactly when $\sin x = 1$ , i.e. $x = \frac\pi2 + 2k\pi$ . On $\intcc{0}{2\pi}$ , $f$ is increasing from $f(0) = 1$ to $f(2\pi) = 2\pi + 1$ (the zero of $f'$ at $\frac{\pi}{2}$ is isolated, so the increase is even strict); minimum $1$ at $0$ , maximum $2\pi + 1$ at $2\pi$ . On $\R$ , $f' \geq 0$ everywhere with only isolated zeros, so $f$ is (strictly) increasing by Theorem 22.7, although $f'$ vanishes at every $\frac\pi2 + 2k\pi$ .

Exercise 24.8 ★★★

Let $f(x) = A\cos x + B \sin x$ with $(A,B) \neq (0,0)$ .

Show that $f$ can be written $f(x) = R\cos(x - \varphi)$ with $R = \sqrt{A^2 + B^2}$ and a suitable $\varphi$ .
Deduce the maximum and minimum of $f$ , and solve $\cos x + \sin x = 1$ in $\intoc{-\pi}{\pi}$ .

Solution

Solution of Exercise 24.8.

1. Let $R = \sqrt{A^2 + B^2} > 0$ . The point $\left(\frac AR, \frac BR\right)$ lies on the unit circle, so there exists $\varphi$ with $\cos\varphi = \frac AR$ and $\sin\varphi = \frac BR$ . Then

R\cos(x - \varphi) = R\bigl(\cos x\cos\varphi + \sin x \sin\varphi\bigr) = A\cos x + B\sin x = f(x).

2. Since $\cos$ has range $\intcc{-1}{1}$ , $f$ has maximum $R$ and minimum $-R$ . For $\cos x + \sin x$ : $A = B = 1$ , $R = \sqrt2$ , $\varphi = \frac\pi4$ , so the equation becomes $\sqrt2\cos\left(x - \frac\pi4\right) = 1$ , i.e. $\cos\left(x - \frac\pi4\right) = \frac{\sqrt2}{2}$ , giving $x - \frac\pi4 = \pm\frac\pi4 + 2k\pi$ : in $\intoc{-\pi}{\pi}$ , $x \in \left\{0, \frac\pi2\right\}$ .

Exercise 24.9 ★★★

Show that $\sin$ and $\cos$ both satisfy the differential equation $y'' = -y$ . Conversely, let $y$ be a solution of $y'' = -y$ with $y(0) = a$ and $y'(0) = b$ ; show that $y = a\cos + b\sin$ . (Hint: consider $g = (y - a\cos - b\sin)$ and the “energy” $E = g^2 + (g')^2$ .)

Solution

Solution of Exercise 24.9.

$(\sin)'' = (\cos)' = -\sin$ and $(\cos)'' = (-\sin)' = -\cos$ : both solve $y'' = -y$ .

Let $g = y - a\cos - b\sin$ . Then $g'' = -g$ (linearity of differentiation), $g(0) = y(0) - a = 0$ and $g'(0) = y'(0) - b = 0$ . Set $E = g^2 + (g')^2$ . Then

E' = 2g g' + 2g' g'' = 2g g' - 2g' g = 0,

so $E$ is constant equal to $E(0) = 0$ . A sum of two squares vanishing identically forces $g = 0$ , hence $y = a\cos + b\sin$ .

24.4 Problem: The pendulum’s secret

Problem 24.1

Weekend problem — one geometric limit powers the derivatives of sine and cosine, and from them all of oscillation: springs, pendulums, beats and the meter that almost was

Every clock that ever ticked, every string that ever sounded, obeys the same mathematics: a restoring force proportional to displacement, hence the equation $y'' = -\omega^2 y$ , hence cosines. At the base of it all sits one innocent-looking limit, $\frac{\sin x}{x} \to 1$ — promised in grade 9, used by Theorem 24.5, and proved here with a picture. This problem climbs from that limit to the swing of a one-meter pendulum, and explains why the meter is, by a whisker, not defined by it.

Part I — The fundamental limit.

Warm up numerically: compute $\frac{\sin x}{x}$ for $x = 0.1$ and $x = 0.01$ (radians!).
The sandwich, on the unit circle with $0 < x < \frac\pi2$ : express by elementary geometry the areas of (i) the triangle with vertices $O$ , $A(1,0)$ and the circle point $M(\cos x, \sin x)$ ; (ii) the circular sector $OAM$ ; (iii) the right triangle $OAT$ where $T(1, \tan x)$ .
From the area sandwich deduce $\sin x < x < \tan x$ , then $\cos x < \frac{\sin x}{x} < 1$ , and conclude by the squeeze theorem that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ (parity handles $x < 0$ ).
Deduce $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac12$ (multiply by the conjugate).
Explain why this limit is the engine of Theorem 24.5 (what is $\sin'(0)$ ?), and why it also settles an old debt: in what sense are radians the only angle unit for which $\sin' = \cos$ holds without a constant?

Part II — Harmonic motion.

Verify by the chain rule that $y(t) = R\cos(\omega t - \varphi)$ satisfies $y'' = -\omega^2 y$ .
A mass on a spring obeys $y'' = -4y$ , with $y(0) = 3$ and $y'(0) = 8$ . Admitting (from Exercise 24.9) that all solutions are $A\cos 2t + B\sin 2t$ , determine $A$ and $B$ .
Rewrite the solution as $R\cos(2t - \varphi)$ (Exercise 24.8): give the amplitude $R$ , the period, and the phase $\varphi$ (three decimals).
At what time does the mass first pass through the equilibrium position $y = 0$ ?
The pendulum: a bob on a string of length $L$ obeys exactly $\theta'' = -\frac gL \sin\theta$ — unsolvable by elementary functions. For small swings, replace $\sin\theta$ by $\theta$ (Part I justifies it!) and deduce the period $T = 2\pi\sqrt{\frac Lg}$ . Compute $T$ for $L = 1$ m ( $g = 9.81$ ): Galileo’s famous discovery — what does the formula not depend on?
The seconds pendulum: what length gives $T = 2$ s exactly? The French Academy of 1791 hesitated between this length and the ten-millionth of the quarter-meridian for defining the meter — by how many millimeters do the two candidates differ?

Part III — Addition formulas at work.

Show with the addition formulas (Proposition 24.3) that $\cos\left(x + \frac\pi2\right) = -\sin x$ and $\sin\left(x + \frac\pi2\right) = \cos x$ : circular differentiation is a quarter-turn — check consistency with Theorem 24.5.
Derive the product-to-sum identity $\cos a \cos b = \frac12\left(\cos(a - b) + \cos(a + b)\right)$ and the linearization $\cos^2 x = \frac{1 + \cos 2x}{2}$ (the integration chapter will thank you).
Beats: derive $\sin p + \sin q = 2 \sin\frac{p + q}{2} \cos\frac{p - q}{2}$ , and apply it to two tuning forks at $440$ and $444$ Hz: what tone does the ear hear, and how many throbs per second? (This is how orchestras tune.)
Derive the triple-angle identity $\cos 3x = 4\cos^3 x - 3\cos x$ (write $3x = 2x + x$ ). Setting $x = 20^\circ$ , what cubic equation must $\cos 20^\circ$ satisfy? (That this cubic cannot be solved with square roots alone is precisely why the trisection of the angle defeated ruler and compass for two millennia — the full story is told in the university volumes.)
An alternating current reads $I(t) = 3\cos(100\pi t) + 3\sqrt3 \sin(100\pi t)$ . Rewrite it as $R\cos(100\pi t - \varphi)$ : peak current and phase?

Part IV — The oscillating world.

Periods: what is the (smallest) period of $\sin^2 t$ (use question 13)? Of $\sin(2t) + \sin(3t)$ ?
A real pendulum damps: $y(t) = \eu^{-t/10}\cos(2\pi t)$ . Describe the motion, compute the envelope’s value at $t = 10$ (an old friend appears, Problem 23.1), and name the chapter where equations with such solutions are solved.
Why do sines and cosines rule every vibration? Give the physics-to-mathematics dictionary in two sentences (restoring force proportional to displacement $\rightarrow$ which equation $\rightarrow$ which solutions), and say what Exercise 24.9 contributes to the claim.
Finale: the chapter’s arc in four steps — one geometric limit, two derivatives, one differential equation, all of oscillation. And the honest asterisk: for large swings the pendulum’s true period involves an elliptic integral, computed at lightning speed by …Gauss’s arithmetic–geometric mean, the dessert of Problem 20.1. The series’ threads tie themselves.

Solution

Solution of Problem 24.1.

1. $0.99833\ldots$ and $0.99998\ldots$ : creeping to $1$ .

2. Triangle $OAM$ : base $1$ , height $\sin x$ : area $\frac{\sin x}{2}$ . Sector $OAM$ : fraction $\frac{x}{2\pi}$ of the unit disk: area $\frac x2$ . Triangle $OAT$ : base $1$ , height $\tan x$ : area $\frac{\tan x}{2}$ .

3. The triangle sits inside the sector inside the big triangle: $\frac{\sin x}{2} < \frac x2 < \frac{\tan x}{2}$ , i.e. $\sin x < x < \tan x$ . From $\sin x < x$ : $\frac{\sin x}{x} < 1$ ; and from $x < \tan x = \frac{\sin x}{\cos x}$ , multiplying by $\frac{\cos x}{x} > 0$ : $\cos x < \frac{\sin x}{x}$ . As $x \to 0^+$ , $\cos x \to 1$ : squeezed, $\frac{\sin x}{x} \to 1$ ; and $\frac{\sin x}{x}$ is even, so the two-sided limit is $1$ .

4. $\frac{1 - \cos x}{x^2} = \frac{1 - \cos^2 x}{x^2 (1 + \cos x)} = \left(\frac{\sin x}{x}\right)^2 \frac{1}{1 + \cos x} \to 1 \times \frac12$ .

5. $\sin'(0) = \lim_{h \to 0} \frac{\sin h - 0}{h} = 1$ : the whole differentiation of sine and cosine flows from this one limit. Measured in degrees, the limit would be $\frac{\pi}{180}$ , and every derivative would drag that constant: radians are exactly the unit that makes the calculus of oscillation clean.

6. $y' = -R\omega\sin(\omega t - \varphi)$ , $y'' = -R\omega^2\cos(\omega t - \varphi) = -\omega^2 y$ .

7. $\omega = 2$ . $y(0) = A = 3$ ; $y'(0) = 2B = 8$ : $B = 4$ : $y = 3\cos 2t + 4\sin 2t$ .

8. $R = \sqrt{9 + 16} = 5$ ; period $\frac{2\pi}{2} = \pi$ ; phase $\varphi = \arctan\frac43 \approx 0.927$ : $y = 5\cos(2t - 0.927)$ .

9. $5\cos(2t - \varphi) = 0$ first when $2t - \varphi = \frac\pi2$ : $t = \frac{\pi/2 + 0.927}{2} \approx 1.25$ s.

10. With $\sin\theta \approx \theta$ : $\theta'' = -\frac gL \theta$ : harmonic with $\omega = \sqrt{\frac gL}$ , period $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac Lg}$ . For $L = 1$ : $T \approx 2.006$ s. The period does not depend on the amplitude (for small swings) — Galileo’s isochronism, the fact that makes pendulum clocks possible.

11. $L = g\left(\frac{T}{2\pi}\right)^2 = \frac{9.81}{\pi^2} \approx 0.994$ m: the seconds pendulum is about $6$ mm short of the meridian meter. The Academy chose the meridian (gravity varies from place to place, betraying the pendulum); had $g$ been a whisker larger, our meter would tick.

12. $\cos\left(x + \frac\pi2\right) = \cos x \cos\frac\pi2 - \sin x \sin\frac\pi2 = -\sin x$ ; $\sin\left(x + \frac\pi2\right) = \cos x$ . So differentiating sine gives sine shifted a quarter turn ( $\cos$ ), and again ( $-\sin$ ), and again, and home in four steps: differentiation rotates the circle.

13. Adding the two expansions of $\cos(a \mp b)$ : $\cos(a-b) + \cos(a+b) = 2\cos a\cos b$ . With $a = b = x$ : $1 + \cos 2x = 2\cos^2 x$ , i.e. $\cos^2 x = \frac{1 + \cos 2x}{2}$ .

14. With $p = \frac{p+q}{2} + \frac{p-q}{2}$ and $q$ its mirror, expanding and adding: $\sin p + \sin q = 2\sin\frac{p+q}{2}\cos\frac{p-q}{2}$ . For $440$ and $444$ Hz: a tone at $442$ Hz whose loudness is modulated by $\cos(2\pi \cdot 2t)$ — the loudness peaks $4$ times per second (the envelope’s absolute value): four beats a second, vanishing as the forks agree.

15. $\cos 3x = \cos 2x\cos x - \sin 2x \sin x = (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x = 4\cos^3 x - 3\cos x$ . At $x = 20^\circ$ : $\cos 60^\circ = \frac12$ , so $c = \cos 20^\circ$ satisfies $8c^3 - 6c - 1 = 0$ — a cubic with no square-root solution: constructing $20^\circ$ , i.e. trisecting $60^\circ$ , is beyond ruler and compass.

16. $R = \sqrt{9 + 27} = 6$ ; $\tan\varphi = \frac{3\sqrt3}{3} = \sqrt3$ : $\varphi = \frac\pi3$ : $I(t) = 6\cos\left(100\pi t - \frac\pi3\right)$ : peak $6$ amperes, phase lag $60^\circ$ .

17. $\sin^2 t = \frac{1 - \cos 2t}{2}$ : period $\pi$ . $\sin 2t$ has period $\pi$ , $\sin 3t$ has $\frac{2\pi}{3}$ : the sum repeats after the smallest common multiple, $2\pi$ .

18. An oscillation of period $1$ inside the shrinking envelope $\pm\eu^{-t/10}$ : each swing about $10\,\%$ lower. At $t = 10$ the envelope is $\eu^{-1} \approx 0.368$ — the constant of Problem 23.1 yet again. Equations with such damped solutions ( $y'' + a y' + b y = 0$ ) are solved in the differential equations chapter.

19. A force pulling back proportionally to the displacement gives $y'' = -\omega^2 y$ ; and Exercise 24.9 shows the solutions of that equation are exactly the combinations of $\cos$ and $\sin$ — existence by exhibition, uniqueness by the exercise: vibration has no choice but to be sinusoidal.

20. One limit ( $\frac{\sin x}{x} \to 1$ , by sandwiched areas), two derivatives ( $\sin' = \cos$ , $\cos' = -\sin$ ), one equation ( $y'' = -\omega^2 y$ ), a world of clocks, strings, currents and tides. Asterisk: for wide swings, $T = 4\sqrt{\frac Lg} \cdot \frac{\pi/2}{M(1, \cos(\theta_0/2))}$ — the arithmetic–geometric mean of Problem 20.1 computes the elliptic integral in a handful of iterations: Gauss’s diary entry and Galileo’s pendulum, one formula apart.