So far we have solved the equations in $\mathbb{R}$, obtaining infinite families of solutions. Often, however, only solutions in a range are of interest — for example, the angles of a triangle, which are between $0$ and $\pi$. And certain equations appear so much that they have their own methods: they are classical equations.

The story

The equation $a\operatorname{sen} x + b\cos x = c$ is the equation of the sum of two oscillations of the same period: it appears in astronomy (the combination of two circular movements), in wave physics and in electricity (out-of-phase alternating currents). The trick of rewriting $a\operatorname{sen} x + b\cos x$ as a single displaced sinusoid — the auxiliary angle $\theta$ — was already used by 18th century astronomers. It gained industrial importance at the end of the 19th century, when the engineer Charles Steinmetz showed, in 1893, that alternating currents could be added as complex numbers: each sinusoid becomes an arrow with length (the amplitude) and slope (the phase), and adding oscillations becomes adding arrows. This is exactly what we do when writing $a\operatorname{sen} x + b\cos x = \sqrt{a^2 + b^2}\,\operatorname{sen}(x + \theta)$.

Steinmetz phasors: $a\operatorname{sen}x + b\cos x$ is the height of a rotating arrow, and this arrow has length $\sqrt{a^2 + b^2}$.

Sources: Charles Proteus Steinmetz (ETHW) · Trigonometric functions — History (MacTutor).

In everyday life

  • Springs and shock absorbers. A mass attached to a spring, released from an initial position and velocity, oscillates as $a\operatorname{sen}\omega t + b\cos\omega t = \sqrt{a^2 + b^2}\,\operatorname{sen}(\omega t + \theta)$: the auxiliary angle gives the amplitude of the movement.
  • Sound from two speakers. Two speakers with the same note, out of phase, add up to a single wave; the amplitude depends on the phase shift, and in certain parts of the room the sound almost disappears.
  • Only the answers that make sense. In a physical problem, the launch angle is between $0^\circ$ and $90^\circ$: of the infinite solutions of the equation, we are left with only those in the interval.
A mass on a spring: $a\operatorname{sen}t + b\cos t$ is a single sinusoid of amplitude $\sqrt{a^2 + b^2}$.

Theory

Solutions in a range

ResumoRoad map

  1. Solve the equation in $\mathbb{R}$, obtaining the families of solutions in function of $k \in \mathbb{Z}$.
  2. Assign $k$ integer values ​​and keep the solutions that fall in the requested range (or solve the inequality in $k$).

Exemplosin 2x = 1/2 in [0, 2π]

$2x = \frac{\pi}{6} + 2k\pi$ or $2x = \frac{5\pi}{6} + 2k\pi$, that is, $x = \frac{\pi}{12} + k\pi$ or $x = \frac{5\pi}{12} + k\pi$.

  • $k = 0$: $x = \frac{\pi}{12}$ or $x = \frac{5\pi}{12}$;
  • $k = 1$: $x = \frac{13\pi}{12}$ or $x = \frac{17\pi}{12}$;
  • $k = -1$ or $k = 2$: solutions are out of range.

$S = \left\{\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}\right\}$.

Ative o JavaScript para ver esta figura interativa.

The equation a sin x + b cos x = c

Let the equation be $a\operatorname{sen} x + b\cos x = c$, with $a$, $b$, $c$ non-zero reals. There are three classic methods; let’s apply them to the same equation, $\operatorname{sen} x + \cos x = 1$.

1st method — system. By doing $\operatorname{sen} x = u$ and $\cos x = v$, we solve the system

$$a u + b v = c, \qquad u^2 + v^2 = 1.$$

In the example: $u + v = 1$ and $u^2 + v^2 = 1$; squaring the first, $1 + 2uv = 1$, $uv = 0$. So $(u, v) = (0, 1)$, which gives $x = 2k\pi$, or $(u, v) = (1, 0)$, which gives $x = \frac{\pi}{2} + 2k\pi$.

2nd method — auxiliary angle. Dividing by $a$ and making $\frac{b}{a} = \operatorname{tg}\theta$:

$$\operatorname{sen} x + \frac{\operatorname{sen}\theta}{\cos\theta}\cos x = \frac{c}{a} \;\Longrightarrow\; \operatorname{sen} x\cos\theta + \operatorname{sen}\theta\cos x = \frac{c}{a}\cos\theta \;\Longrightarrow\; \operatorname{sen}(x + \theta) = \frac{c}{a}\cos\theta,$$

which is a fundamental equation. In the example, $\operatorname{tg}\theta = 1$, $\theta = \frac{\pi}{4}$ and $\operatorname{sen}\left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$: $x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi$ or $x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$ — the same solutions.

3rd method — half arc tangent. With $t = \operatorname{tg}\frac{x}{2}$, $\operatorname{sen} x = \frac{2t}{1 + t^2}$ and $\cos x = \frac{1 - t^2}{1 + t^2}$ (chapter VI):

$$a \cdot \frac{2t}{1 + t^2} + b \cdot \frac{1 - t^2}{1 + t^2} = c \;\Longrightarrow\; (c + b)t^2 - 2at + (c - b) = 0.$$

In the example, $2t^2 - 2t = 0$: $t = 0$, $\frac{x}{2} = k\pi$, $x = 2k\pi$; or $t = 1$, $\frac{x}{2} = \frac{\pi}{4} + k\pi$, $x = \frac{\pi}{2} + 2k\pi$.

AtençãoA solution may be lost

The substitution $t = \operatorname{tg}\frac{x}{2}$ is only valid when $\frac{x}{2} \ne \frac{\pi}{2} + k\pi$, that is, $x \ne \pi + 2k\pi$. Therefore, it is necessary to separately check whether $x = \pi + 2k\pi$ is a solution to the original equation: this happens when $-b = c$, and then the coefficient $c + b$ cancels out and the equation in $t$ drops in degree. In the example, $\operatorname{sen}\pi + \cos\pi = -1 \ne 1$, and nothing was lost.

PropriedadeWhen there is a solution

By the 2nd method, $a\operatorname{sen} x + b\cos x = \sqrt{a^2 + b^2}\operatorname{sen}(x + \theta)$, which varies between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$. Therefore, the equation has a solution if, and only if, $c^2 \le a^2 + b^2$.

Sums of sines and cosines equal to zero

Equations of the form $\operatorname{sen} p \pm \operatorname{sen} q = 0$, $\cos p \pm \cos q = 0$, or larger sums of these terms, are solved transforming the sum into a product and canceling each factor.

Exemplocos x + cos 2x + cos 3x = 0

Grouping the first and third terms: $\cos 3x + \cos x = 2\cos 2x\cos x$. Then

$$2\cos 2x\cos x + \cos 2x = 0 \;\Longrightarrow\; \cos 2x\,(2\cos x + 1) = 0.$$
  • $\cos 2x = 0$: $2x = \frac{\pi}{2} + k\pi$, $x = \frac{\pi}{4} + \frac{k\pi}{2}$;
  • $\cos x = -\frac{1}{2}$: $x = \pm\frac{2\pi}{3} + 2k\pi$.

Equations with sin⁴x + cos⁴x and sin⁶x + cos⁶x

TeoremaTwo useful identities

$$\operatorname{sen}^4 x + \cos^4 x = 1 - \frac{1}{2}\operatorname{sen}^2 2x$$$$\operatorname{sen}^6 x + \cos^6 x = 1 - \frac{3}{4}\operatorname{sen}^2 2x$$
View the demos

With $s = \operatorname{sen}^2 x$, $c = \cos^2 x$, $s + c = 1$ and $sc = \operatorname{sen}^2 x\cos^2 x = \frac{1}{4}\operatorname{sen}^2 2x$:

$$s^2 + c^2 = (s + c)^2 - 2sc = 1 - \frac{1}{2}\operatorname{sen}^2 2x,$$$$s^3 + c^3 = (s + c)^3 - 3sc(s + c) = 1 - \frac{3}{4}\operatorname{sen}^2 2x.$$

PropriedadeDiscussion

  • $\operatorname{sen}^4 x + \cos^4 x = a \iff \operatorname{sen}^2 2x = 2(1 - a)$. Like $0 \le \operatorname{sen}^2 2x \le 1$, there is a solution if and only if $\frac{1}{2} \le a \le 1$. Using $\operatorname{sen}^2 2x = \frac{1 - \cos 4x}{2}$, the equation becomes $\cos 4x = 4a - 3$.
  • $\operatorname{sen}^6 x + \cos^6 x = a \iff \operatorname{sen}^2 2x = \frac{4(1 - a)}{3}$. There is a solution if, and only if, $\frac{1}{4} \le a \le 1$, and the equation becomes $\cos 4x = \frac{8a - 5}{3}$.
$\operatorname{sen}^4 x + \cos^4 x$ (red) oscillates between $\frac{1}{2}$ and $1$; $\operatorname{sen}^6 x + \cos^6 x$ (blue), between $\frac{1}{4}$ and $1$.

The exercises in this section are numbers 215 to 245 from chapter list.