Identities and their demonstration
An identity is an equality that holds for all values where both sides make sense. Demonstrating identities is the great training in manipulating trigonometry: each step uses an already proven relationship.
The story
To demonstrate identities it is necessary to write the functions in short form. Until the 17th century this did not exist: each relationship was said in words — “the sine of the complement of the arc”, “the tangent of the bent arc” — and a one-line identity occupied a paragraph. The abbreviations appeared little by little: Gunter used sin in a figure in 1624, the Frenchman Pierre Hérigone printed it in a book in 1634, and the Englishman William Oughtred wrote something like s co arc for cosine. It was Euler, in Introductio (1748), who established the use of sin., cos., tang. as quantities that are added, multiplied and squared — and it is this use that allows you to manipulate an identity as an algebraic expression.
- 1620Gunter proposes co-sinus and co-tangens.
- 1624Gunter uses the abbreviation sin in a figure.
- 1626Albert Girard writes tan and sec on his tablets.
- 1634Hérigone prints sin in a book.
- 1657Oughtred, in Trigonometry, writes s co arc for the cosine.
- 1748Euler fixes sin., cos., tang. as quantity symbols.
Sources: Oughtred (MacTutor) · Trigonometric functions (MacTutor) · F. Cajori, A History of Mathematical Notations, vol. II (1929).
In everyday life
- Distances on the map. Navigation applications calculate distances on the Earth’s surface with the haversine formula, which uses the identity $\operatorname{sen}^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$ to avoid rounding errors at short distances.
- 127 V socket. The socket voltage oscillates like a sinusoid; the identity $\operatorname{sen}^2 x = \frac{1 - \cos 2x}{2}$ shows that the mean square is half the peak squared. Therefore “effective” 127 V corresponds to a peak of about $127\sqrt{2} \approx 180$ V.
- Compressed photos and music. JPEG and MP3 formats decompose images and sounds into sums of cosines; trigonometric identities guarantee that this decomposition can be undone.
Theory
Definition
DefiniçãoIdentical functions
Let $f$ and $g$ be two functions of domains $D_1$ and $D_2$, respectively. We say that $f$ is identical to $g$, and we indicate $f \equiv g$ if and only if $f(x) = g(x)$ for all $x$ where both functions are defined:
$$f \equiv g \iff f(x) = g(x),\ \forall x \in D_1 \cap D_2.$$Examples
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$f\colon \mathbb{R} \to \mathbb{R}$ such that $f(x) = (x + 1)^2 - (x - 1)^2$ and $g\colon \mathbb{R} \to \mathbb{R}$ such that $g(x) = 4x$ are identical, since
$$f(x) = x^2 + 2x + 1 - x^2 + 2x - 1 = 4x = g(x),\ \forall x \in \mathbb{R}.$$ -
$f\colon \mathbb{R} \to \mathbb{R}$ such that $f(x) = x + 1$ and $g\colon \mathbb{R} - \{1\} \to \mathbb{R}$ such that $g(x) = \dfrac{x^2 - 1}{x - 1}$ are identical, since
$$g(x) = \frac{(x + 1)(x - 1)}{x - 1} = x + 1 = f(x),\ \forall x \in \mathbb{R} - \{1\}.$$Note: the domains are different, but equality applies at their intersection.
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$f(x) = \operatorname{sen}^2 x$ and $g(x) = 1 - \cos^2 x$, both from $\mathbb{R}$ to $\mathbb{R}$, are identical: $f(x) = \operatorname{sen}^2 x = 1 - \cos^2 x = g(x)$ to all $x \in \mathbb{R}$.
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$f\colon \left\{x \in \mathbb{R} \;\middle|\; x \ne \frac{\pi}{2} + k\pi\right\} \to \mathbb{R}$ such that $f(x) = \sec^2 x - \operatorname{tg}^2 x$ and $g\colon \mathbb{R} \to \mathbb{R}$ such that $g(x) = 1$ are identical, since
$$f(x) = \sec^2 x - \operatorname{tg}^2 x = (1 + \operatorname{tg}^2 x) - \operatorname{tg}^2 x = 1 = g(x),\ \text{para todo } x \ne \frac{\pi}{2} + k\pi.$$
How to demonstrate an identity
To demonstrate a trigonometric identity we can apply any already established formula — which is also an identity: the fundamental relationships and, in the next chapters, the formulas for reduction, addition, multiplication, division and transformation into a product. In the exercises in this chapter, we can only use fundamental relations.
PropriedadeThree processes
Depending on the difficulty, we choose the most appropriate method:
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We start with one of the members (generally the most complicated) and transform it into the other.
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We transform separately the 1st member ($f$) and the 2nd member ($g$), both arriving at the same expression $h$. The method is valid because
$$f \equiv h \ \text{ e } \ g \equiv h \;\Longrightarrow\; f \equiv g.$$ -
We construct the difference $h = f - g$ and prove that $h \equiv 0$. The method is valid because
$$f - g \equiv 0 \iff f \equiv g.$$
ExemploThe three methods in action
- 1st method — exercise 93: we start from $(1 + \operatorname{cotg}^2 x)(1 - \cos^2 x)$ and arrive at 1.
- 2nd method — exercise 95: the two members of $(1 - \operatorname{tg} x)^2 + (1 - \operatorname{cotg} x)^2 = (\sec x - \operatorname{cossec} x)^2$ reduce to $\dfrac{1 - 2\operatorname{sen} x \cos x}{\operatorname{sen}^2 x \cos^2 x}$.
- 3rd method — exercise 96: the difference between the two members cancels out.
See the resolutions in chapter list.
AtençãoBe careful with domains
An identity only needs to hold where both members are defined. When dividing by $\operatorname{sen} x$ or $\cos x$, for example, it is implied that $x \ne k\pi$ or $x \ne \frac{\pi}{2} + k\pi$. And never “pass to the other side” the equality you want to prove: starting from the thesis and arriving at something true does not demonstrate anything, unless all passages are reversible.