An equality between functions

A trigonometric identity is an equality valid for all entries of a declared domain. It is necessary to check both the values ​​and the points at which the expressions exist.

For example, $\sec^2x-\tan^2x=1$ is valid when $\cos x\ne0$. The right side also exists in the excluded points, but this does not extend the domain of the expression to the left.

First way: transform one of the members

To demonstrate $\sin x\tan x+\cos x=\sec x$, with $\cos x\ne0$, we start from the left:

$$\begin{aligned} \sin x\tan x+\cos x &=\frac{\sin^2x}{\cos x}+\frac{\cos^2x}{\cos x}\\ &=\frac{\sin^2x+\cos^2x}{\cos x}=\frac1{\cos x}. \end{aligned}$$

We recognize the right-hand member by the definition of secant.

Second way: transform the two into a common expression

Consider:

$$\frac{1-\cos^2x}{\sin x}=\sin x.$$

In the domain $\sin x\ne0$, the left member reduces to $\sin^2x/\sin x=\sin x$. We could also transform both sides to a third expression, as long as each transformation respects the domain.

Third way: reversible equivalences

We can work on the proposed equality if each step is reversible. For $\sin x\ne0$:

$$\begin{aligned} \frac{1-\cos^2x}{\sin x}=\sin x &\iff1-\cos^2x=\sin^2x\\ &\iff1=\sin^2x+\cos^2x. \end{aligned}$$

The last equality is known and all steps can be undone. Squaring, on the contrary, is not always reversible: $u=v$ implies $u^2=v^2$, but the return can give $u=-v$.

Numerical example is not demonstration

Testing $x=0$ and $x=\pi/2$ may help discover an error, but it does not prove a statement about all reals. A counterexample is enough to refute; a proof requires a general argument.

In everyday life

  • Maps. The haversine formula, used to calculate distances on the Earth’s surface, depends on the identity $\operatorname{sen}^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$.
  • Rms voltage. An identification shows that 127 Rms equals a peak of about 180 V at the outlet.
  • Check formulas. Engineers check a long formula by testing whether it reduces to a known identity in simple cases.
The identity $\operatorname{sen}^2x = \frac{1 - \cos 2x}{2}$ at the socket: the peak of 180 V and the rms voltage of 127 V.

Exercises with resolution

1. Demonstrate $(1-\sin x)(1+\sin x)=\cos^2x$.

Check resolution 1

By difference of squares, the left member is $1-\sin^2x=\cos^2x$. There is no denominator; applies to every real.

2. Simplify $(1-\sin x)/\cos x$ by multiplying by the conjugate expression. State the conditions of the new form.

Check resolution 2

The form $\cos x/(1+\sin x)$ results from multiplication by $(1+\sin x)/(1+\sin x)$. In the original domain $\cos x\ne0$, also $\sin x\ne-1$, so the factor used exists. The two forms are equal in this domain.

In the interactive book

The same subjects, with all the demonstrations and the complete list of exercises, are in the interactive Trigonometry book:

This class is a shorter introduction to the topic; In the book, the theory is accompanied by solved and proposed exercises from the chapter.

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