---
title: "Trigonometric Functions"
book: "High School Mathematics"
subject: math
language: en
chapter: 24
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions
---

# Chapter 24 — Trigonometric Functions

The [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $\cos$ 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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function), and their [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) make them solutions of the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-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 $M(t)$*:

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

![](https://one-course.com/images/onecourse/chapters/math-2/g12-trigo/fig-600a6e9a7eb2.svg)

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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) $\vect{u} = (\cos a, \sin a)$ and $\vect{v} = (\cos b, \sin b)$ are unit [vectors](https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane#def-g11-vect-vector) making an angle $a - b$; their [scalar product](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#def-g11-scal-dot) equals both $\cos a\cos b + \sin a \sin b$ ([coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system)) 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](#def-g12-trigo-cossin) 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).](https://one-course.com/images/onecourse/chapters/math-2/g12-trigo/fig-58dd2424af55.svg)

*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](https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity#def-g12-limcont-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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $\sin$ and $\cos$ are [differentiable](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) 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](#lem-g12-trigo-sinxx). 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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) there; similarly $(\sin)' = \cos > 0$ on $\intoo{-\frac\pi2}{\frac\pi2}$. Periodicity and parity come from [Proposition 24.2](#prop-g12-trigo-identities). ∎

![The curves of (blue) and (red).](https://one-course.com/images/onecourse/chapters/math-2/g12-trigo/fig-0ab6284eb634.svg)

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

**Method 24.7 (Solving cos⁡x=a\cos x = acosx=a and sin⁡x=a\sin x = asinx=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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) 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](#prop-g12-trigo-identities).

**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 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 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 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 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 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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $f(x) = x + \cos x$ on $\intcc{0}{2\pi}$: variations and extrema. Show that $f$ is [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) on $\R$ although $f'$ vanishes at infinitely many points.

**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](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) 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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $1$ at $0$, [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $2\pi + 1$ at $2\pi$. On $\R$, $f' \geq 0$ everywhere with only isolated zeros, so $f$ is (strictly) [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) by [Theorem 22.7](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#thm-g12-deriv-variations), although $f'$ vanishes at every $\frac\pi2 + 2k\pi$.

**Exercise 24.8 ★★★.**

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

1. Show that $f$ can be written $f(x) = R\cos(x - \varphi)$ with $R = \sqrt{A^2 + B^2}$ and a suitable $\varphi$ .
2. Deduce the [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) and [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) of $f$ , and solve $\cos x + \sin x = 1$ in $\intoc{-\pi}{\pi}$ .

**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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $R$ and [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $-R$. For $\cos x + \sin x$: $A = B = 1$, $R = \sqrt2$, $\varphi = \frac\pi4$, so the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $y'' = -\omega^2 y$, hence [cosines](#def-g12-trigo-cossin). 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](#thm-g12-trigo-derivatives), 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.**

1. Warm up numerically: compute $\frac{\sin x}{x}$ for $x = 0.1$ and $x = 0.01$ ( [radians](https://one-course.com/books/math/2/en/chapter/14-trigonometry-the-unit-circle#def-g11-trigo-radian) !).
2. 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 $T(1, \tan x)$ .
3. 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 $x < 0$ ).
4. Deduce $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac12$ (multiply by the conjugate).
5. Explain why this limit is the engine of [Theorem 24.5](#thm-g12-trigo-derivatives) (what is $\sin'(0)$ ?), and why it also settles an old debt: in what sense are *[radians](https://one-course.com/books/math/2/en/chapter/14-trigonometry-the-unit-circle#def-g11-trigo-radian)* the only angle unit for which $\sin' = \cos$ holds without a constant?

**Part II — Harmonic motion.**

6. Verify by the chain rule that $y(t) = R\cos(\omega t - \varphi)$ satisfies $y'' = -\omega^2 y$ .
7. A mass on a spring obeys $y'' = -4y$ , with $y(0) = 3$ and $y'(0) = 8$ . Admitting (from [Exercise 24.9](#exo-g12-trigo-9) ) that all solutions are $A\cos 2t + B\sin 2t$ , determine $A$ and $B$ .
8. Rewrite the solution as $R\cos(2t - \varphi)$ ( [Exercise 24.8](#exo-g12-trigo-8) ): give the amplitude $R$ , the period, and the phase $\varphi$ (three decimals).
9. At what time does the mass first pass through the equilibrium position $y = 0$ ?
10. The pendulum: a bob on a string of length $L$ obeys *exactly* $\theta'' = -\frac gL \sin\theta$ — unsolvable by elementary [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) . 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?
11. 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.**

12. Show with the addition formulas ( [Proposition 24.3](#prop-g12-trigo-addition) ) 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](#thm-g12-trigo-derivatives) .
13. 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).
14. 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.)
15. 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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.)
16. 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.**

17. Periods: what is the (smallest) period of $\sin^2 t$ (use question 13)? Of $\sin(2t) + \sin(3t)$ ?
18. 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](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#pb-g12-exp-1) ), and name the chapter where [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) with such solutions are solved.
19. Why do [sines](#def-g12-trigo-cossin) and [cosines](#def-g12-trigo-cossin) rule every vibration? Give the physics-to-mathematics dictionary in two sentences (restoring force proportional to displacement $\rightarrow$ which [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $\rightarrow$ which solutions), and say what [Exercise 24.9](#exo-g12-trigo-9) contributes to the claim.
20. Finale: the chapter’s arc in four steps — one geometric limit, two [derivatives](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) , one differential [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) , the dessert of [Problem 20.1](https://one-course.com/books/math/2/en/chapter/20-sequences#pb-g12-seq-1) . The series’ threads tie themselves.

**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](#def-g12-trigo-cossin) and [cosine](#def-g12-trigo-cossin) flows from this one limit. Measured in degrees, the limit would be $\frac{\pi}{180}$, and every [derivative](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) would drag that constant: [radians](https://one-course.com/books/math/2/en/chapter/14-trigonometry-the-unit-circle#def-g11-trigo-radian) 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](#def-g12-trigo-cossin) gives [sine](#def-g12-trigo-cossin) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-abs)): 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](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) 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](https://one-course.com/books/math/2/en/chapter/23-exponential-and-logarithm#pb-g12-exp-1) yet again. [Equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) with such damped solutions ($y'' + a y' + b y = 0$) are solved in the differential [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) chapter.

**19.** A force pulling back proportionally to the displacement gives $y'' = -\omega^2 y$; and [Exercise 24.9](#exo-g12-trigo-9) shows the solutions of that [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#def-g12-deriv-derivative) ($\sin' = \cos$, $\cos' = -\sin$), one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) of [Problem 20.1](https://one-course.com/books/math/2/en/chapter/20-sequences#pb-g12-seq-1) computes the elliptic integral in a handful of iterations: Gauss’s diary entry and Galileo’s pendulum, one formula apart.